How to solve logarithmic equations using examples. Definition in mathematics
In this lesson we will review the basic theoretical facts about logarithms and consider solving the simplest logarithmic equations.
Let us recall the central definition - the definition of a logarithm. It is related to the decision exponential equation. This equation has a single root, it is called the logarithm of b to base a:
Definition:
The logarithm of b to base a is the exponent to which base a must be raised to get b.
Let us remind you basic logarithmic identity.
The expression (expression 1) is the root of the equation (expression 2). Substitute the value x from expression 1 instead of x into expression 2 and get the main logarithmic identity:
So we see that each value is associated with a value. We denote b by x(), c by y, and thus obtain a logarithmic function:
For example: 
Let us recall the basic properties of the logarithmic function.
Let us pay attention once again, here, since under the logarithm there can be a strictly positive expression, as the base of the logarithm.
Rice. 1. Graph of a logarithmic function with different bases
The graph of the function at is shown in black. Rice. 1. If the argument increases from zero to infinity, the function increases from minus to plus infinity.
The graph of the function at is shown in red. Rice. 1.
Properties of this function:
Domain: ;
Range of values: ;
The function is monotonic throughout its entire domain of definition. When monotonically (strictly) increases, a larger value of the argument corresponds to a larger value of the function. When monotonically (strictly) decreases, a larger value of the argument corresponds to a smaller value of the function.
The properties of the logarithmic function are the key to solving a variety of logarithmic equations.
Let's consider the simplest logarithmic equation; all other logarithmic equations, as a rule, are reduced to this form.
Since the bases of logarithms and the logarithms themselves are equal, the functions under the logarithm are also equal, but we must not miss the domain of definition. The logarithm can only stand positive number, we have:
We found out that the functions f and g are equal, so it is enough to choose any one inequality to comply with the ODZ.
Thus, we have a mixed system in which there is an equation and an inequality:
As a rule, it is not necessary to solve an inequality; it is enough to solve the equation and substitute the found roots into the inequality, thus performing a check.
Let us formulate a method for solving the simplest logarithmic equations:
Equalize the bases of logarithms;
Equate sublogarithmic functions;
Perform check.
Let's look at specific examples.
Example 1 - solve the equation:
The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we choose the first logarithm to compose the inequality:
Example 2 - solve the equation:
This equation differs from the previous one in that the bases of the logarithms are less than one, but this does not affect the solution in any way:
Let's find the root and substitute it into the inequality:
We received an incorrect inequality, which means that the found root does not satisfy the ODZ.
Example 3 - solve the equation:
The bases of logarithms are initially equal, we have the right to equate sublogarithmic expressions, do not forget about the ODZ, we choose the second logarithm to compose the inequality:
Let's find the root and substitute it into the inequality:
Obviously, only the first root satisfies the ODZ.
Logarithmic equations. From simple to complex.
Attention!
There are additional
materials in Special Section 555.
For those who are very "not very..."
And for those who “very much…”)
What is a logarithmic equation?
This is an equation with logarithms. I’m surprised, right?) Then I’ll clarify. This is an equation in which the unknowns (x's) and expressions with them are found inside logarithms. And only there! It is important.
Here are some examples logarithmic equations:
log 3 x = log 3 9
log 3 (x 2 -3) = log 3 (2x)
log x+1 (x 2 +3x-7) = 2
lg 2 (x+1)+10 = 11lg(x+1)
Well, you understand... )
Note! The most diverse expressions with X's are located exclusively within logarithms. If, suddenly, an X appears somewhere in the equation outside, For example:
log 2 x = 3+x,
this will be an equation mixed type. Such equations do not have clear rules for solving them. We will not consider them for now. By the way, there are equations where inside the logarithms only numbers. For example:
What can I say? You're lucky if you come across this! Logarithm with numbers is some number. That's all. It is enough to know the properties of logarithms to solve such an equation. Knowledge of special rules, techniques adapted specifically for solving logarithmic equations, not required here.
So, what is a logarithmic equation- we figured it out.
How to solve logarithmic equations?
Solution logarithmic equations- the thing is actually not very simple. So our section is a four... A decent amount of knowledge on all sorts of related topics is required. In addition, there is a special feature in these equations. And this feature is so important that it can safely be called the main problem in solving logarithmic equations. We will deal with this problem in detail in the next lesson.
For now, don't worry. We'll go the right way from simple to complex. On specific examples. The main thing is to delve into simple things and don’t be lazy to follow the links, I put them there for a reason... And everything will work out for you. Necessarily.
Let's start with the most elementary, simplest equations. To solve them, it is advisable to have an idea of the logarithm, but nothing more. Just no idea logarithm, take on a decision logarithmic equations - somehow even awkward... Very bold, I would say).
The simplest logarithmic equations.
These are equations of the form:
1. log 3 x = log 3 9
2. log 7 (2x-3) = log 7 x
3. log 7 (50x-1) = 2
Solution process any logarithmic equation consists in the transition from an equation with logarithms to an equation without them. In the simplest equations this transition is carried out in one step. That's why they are the simplest.)
And such logarithmic equations are surprisingly easy to solve. See for yourself.
Let's solve the first example:
log 3 x = log 3 9
To solve this example, you don’t need to know almost anything, yes... Purely intuition!) What do we need especially don't like this example? What-what... I don't like logarithms! Right. So let's get rid of them. We look closely at the example, and a natural desire arises in us... Downright irresistible! Take and throw out logarithms altogether. And what’s good is that Can do! Mathematics allows. Logarithms disappear the answer is:
Great, right? This can (and should) always be done. Eliminating logarithms in this manner is one of the main ways to solve logarithmic equations and inequalities. In mathematics this operation is called potentiation. Of course, there are rules for such liquidation, but they are few. Remember:
You can eliminate logarithms without any fear if they have:
a) the same numerical bases
c) logarithms from left to right are pure (without any coefficients) and are in splendid isolation.
Let me clarify the last point. In the equation, let's say
log 3 x = 2log 3 (3x-1)
Logarithms cannot be removed. The two on the right doesn't allow it. The coefficient, you know... In the example
log 3 x+log 3 (x+1) = log 3 (3+x)
It is also impossible to potentiate the equation. There is no lone logarithm on the left side. There are two of them.
In short, you can remove logarithms if the equation looks like this and only like this:
log a (.....) = log a (.....)
In parentheses, where there is an ellipsis, there may be any expressions. Simple, super complex, all kinds. Whatever. The important thing is that after eliminating logarithms we are left with simpler equation. It is assumed, of course, that you already know how to solve linear, quadratic, fractional, exponential and other equations without logarithms.)
Now you can easily solve the second example:
log 7 (2x-3) = log 7 x
Actually, it’s decided in the mind. We potentiate, we get:
Well, is it very difficult?) As you can see, logarithmic part of the solution to the equation is only in eliminating logarithms... And then comes the solution to the remaining equation without them. A trivial matter.
Let's solve the third example:
log 7 (50x-1) = 2
We see that there is a logarithm on the left:
Let us remember that this logarithm is a number to which the base must be raised (i.e. seven) to obtain a sublogarithmic expression, i.e. (50x-1).
But this number is two! According to Eq. That is:
That's basically all. Logarithm disappeared, What remains is a harmless equation:
We solved this logarithmic equation based only on the meaning of the logarithm. Is it still easier to eliminate logarithms?) I agree. By the way, if you make a logarithm from two, you can solve this example through elimination. Any number can be made into a logarithm. Moreover, the way we need it. A very useful technique in solving logarithmic equations and (especially!) inequalities.
Don't know how to make a logarithm from a number!? It's OK. Section 555 describes this technique in detail. You can master it and use it to the fullest! It greatly reduces the number of errors.
The fourth equation is solved in a completely similar way (by definition):
That's it.
Let's summarize this lesson. We looked at the solution of the simplest logarithmic equations using examples. It is very important. And not only because such equations appear in tests and exams. The fact is that even the most evil and complicated equations are necessarily reduced to the simplest!
Actually, the simplest equations are the final part of the solution any equations. And this final part must be understood strictly! And further. Be sure to read this page to the end. There's a surprise there...)
Now we decide for ourselves. Let's get better, so to speak...)
Find the root (or sum of roots, if there are several) of the equations:
ln(7x+2) = ln(5x+20)
log 2 (x 2 +32) = log 2 (12x)
log 16 (0.5x-1.5) = 0.25
log 0.2 (3x-1) = -3
ln(e 2 +2x-3) = 2
log 2 (14x) = log 2 7 + 2
Answers (in disarray, of course): 42; 12; 9; 25; 7; 1.5; 2; 16.
What, not everything works out? Happens. Don't worry! Section 555 explains the solution to all of these examples in a clear and detailed manner. You'll definitely figure it out there. You will also learn useful practical techniques.
Everything worked out!? All examples of “one left”?) Congratulations!
It's time to reveal the bitter truth to you. Successful solving of these examples does not guarantee success in solving all other logarithmic equations. Even the simplest ones like these. Alas.
The fact is that the solution to any logarithmic equation (even the most elementary!) consists of two equal parts. Solving the equation and working with ODZ. We have mastered one part - solving the equation itself. It's not that hard right?
For this lesson, I specially selected examples in which DL does not affect the answer in any way. But not everyone is as kind as me, right?...)
Therefore, it is imperative to master the other part. ODZ. This is the main problem in solving logarithmic equations. And not because it’s difficult - this part is even easier than the first. But because people simply forget about ODZ. Or they don't know. Or both). And they fall out of the blue...
In the next lesson we will deal with this problem. Then you can confidently decide any simple logarithmic equations and approach quite solid tasks.
If you like this site...
By the way, I have a couple more interesting sites for you.)
You can practice solving examples and find out your level. Testing with instant verification. Let's learn - with interest!)
You can get acquainted with functions and derivatives.
We are all familiar with equations primary classes. There we also learned to solve the simplest examples, and we must admit that they find their application even in higher mathematics. Everything is simple with equations, including quadratic equations. If you are having trouble with this topic, we highly recommend that you review it.
You've probably already gone through logarithms too. However, we consider it important to tell what it is for those who do not yet know. A logarithm is equated to the power to which the base must be raised to obtain the number to the right of the logarithm sign. Let's give an example based on which everything will become clear to you.
If you raise 3 to the fourth power, you get 81. Now substitute the numbers by analogy, and you will finally understand how logarithms are solved. Now all that remains is to combine the two concepts discussed. Initially, the situation seems extremely complicated, but upon closer examination the weight falls into place. We are sure that after this short article you will not have problems in this part of the Unified State Exam.
Today there are many ways to solve such structures. We will tell you about the simplest, most effective and most applicable in the case of Unified State Examination tasks. Solving logarithmic equations must start from the very beginning. simple example. The simplest logarithmic equations consist of a function and one variable in it.
It's important to note that x is inside the argument. A and b must be numbers. In this case, you can simply express the function in terms of a number to a power. It looks like this.
Of course, solving a logarithmic equation using this method will lead you to the correct answer. The problem for the vast majority of students in this case is that they do not understand what comes from where. As a result, you have to put up with mistakes and not get the desired points. The most offensive mistake will be if you mix up the letters. To solve the equation this way, you need to memorize this standard school formula because it is difficult to understand.
To make it easier, you can resort to another method - the canonical form. The idea is extremely simple. Turn your attention back to the problem. Remember that the letter a is a number, not a function or variable. A is not equal to one and greater than zero. There are no restrictions on b. Now, of all the formulas, let us remember one. B can be expressed as follows.
It follows from this that all original equations with logarithms can be represented in the form:
Now we can drop the logarithms. The result is a simple design, which we have already seen earlier.
The convenience of this formula is that it can be used in the most different cases, and not just for the simplest designs.
Don't worry about OOF!
Many experienced mathematicians will notice that we have not paid attention to the domain of definition. The rule boils down to the fact that F(x) is necessarily greater than 0. No, we did not miss this point. Now we are talking about another serious advantage of the canonical form.
There will be no extra roots here. If a variable will only appear in one place, then a scope is not necessary. It is done automatically. To verify this judgment, try solving several simple examples.
How to solve logarithmic equations with different bases
These are already complex logarithmic equations, and the approach to solving them must be special. Here it is rarely possible to limit ourselves to the notorious canonical form. Let's begin our detailed story. We have the following construction.
Pay attention to the fraction. It contains the logarithm. If you see this in a task, it’s worth remembering one interesting trick.
What does it mean? Each logarithm can be represented as the quotient of two logarithms with a convenient base. And this formula has a special case that is applicable with this example (we mean if c=b).
This is exactly the fraction we see in our example. Thus.
Essentially, we turned the fraction around and got a more convenient expression. Remember this algorithm!
Now it is necessary that the logarithmic equation does not contain different bases. Let's represent the base as a fraction.
In mathematics there is a rule based on which you can derive a degree from a base. The following construction results.
It would seem that what is preventing us from now turning our expression into the canonical form and simply solving it? Not so simple. There should be no fractions before the logarithm. Let's fix this situation! Fractions are allowed to be used as degrees.
Respectively.
If the bases are the same, we can remove the logarithms and equate the expressions themselves. This way the situation will become much simpler than it was. What will remain is an elementary equation that each of us knew how to solve back in 8th or even 7th grade. You can do the calculations yourself.
We have obtained the only true root of this logarithmic equation. Examples of solving a logarithmic equation are quite simple, aren't they? Now you will be able to independently deal with even the most complex tasks for preparing for and passing the Unified State Exam.
What's the result?
In the case of any logarithmic equations, we start from one very important rule. It is necessary to act in such a way as to bring the expression to the maximum simple view. In this case, you will have a better chance of not only solving the task correctly, but also doing it in the simplest and most logical way possible. This is exactly how mathematicians always work.
We strongly advise you not to search difficult paths, especially in this case. Remember a few simple rules, which will allow you to transform any expression. For example, reduce two or three logarithms to the same base or derive a power from the base and win on this.
It is also worth remembering that solving logarithmic equations requires constant practice. Gradually you will move on to more and more complex structures, and this will lead you to confidently solving all variants of problems on the Unified State Exam. Prepare well in advance for your exams, and good luck!
Instructions
Write the given logarithmic expression. If the expression uses the logarithm of 10, then its notation is shortened and looks like this: lg b is decimal logarithm. If the logarithm has the number e as its base, then write the expression: ln b – natural logarithm. It is understood that the result of any is the power to which the base number must be raised to obtain the number b.
When finding the sum of two functions, you simply need to differentiate them one by one and add the results: (u+v)" = u"+v";
When finding the derivative of the product of two functions, it is necessary to multiply the derivative of the first function by the second and add the derivative of the second function multiplied by the first function: (u*v)" = u"*v+v"*u;
In order to find the derivative of the quotient of two functions, it is necessary to subtract from the product of the derivative of the dividend multiplied by the divisor function the product of the derivative of the divisor multiplied by the function of the dividend, and divide all this by the divisor function squared. (u/v)" = (u"*v-v"*u)/v^2;
If given complex function, then it is necessary to multiply the derivative of internal function and the derivative of the external one. Let y=u(v(x)), then y"(x)=y"(u)*v"(x).
Using the results obtained above, you can differentiate almost any function. So let's look at a few examples:
y=x^4, y"=4*x^(4-1)=4*x^3;
y=2*x^3*(e^x-x^2+6), y"=2*(3*x^2*(e^x-x^2+6)+x^3*(e^x-2 *x));
There are also problems involving calculating the derivative at a point. Let the function y=e^(x^2+6x+5) be given, you need to find the value of the function at the point x=1.
1) Find the derivative of the function: y"=e^(x^2-6x+5)*(2*x +6).
2) Calculate the value of the function in given point y"(1)=8*e^0=8
Video on the topic
Learn the table of elementary derivatives. This will significantly save time.
Sources:
- derivative of a constant
So, what is the difference between rational equation from the rational? If the unknown variable is under the sign square root, then the equation is considered irrational.
Instructions
The main method for solving such equations is the method of constructing both sides equations into a square. However. this is natural, the first thing you need to do is get rid of the sign. This method is not technically difficult, but sometimes it can lead to trouble. For example, the equation is v(2x-5)=v(4x-7). By squaring both sides you get 2x-5=4x-7. Solving such an equation is not difficult; x=1. But the number 1 will not be given equations. Why? Substitute one into the equation instead of the value of x. And the right and left sides will contain expressions that do not make sense, that is. This value is not valid for a square root. Therefore, 1 is an extraneous root, and therefore this equation has no roots.
So, irrational equation is solved using the method of squaring both its parts. And having solved the equation, it is necessary to cut off extraneous roots. To do this, substitute the found roots into the original equation.
Consider another one.
2х+vх-3=0
Of course, this equation can be solved using the same equation as the previous one. Move Compounds equations, which do not have a square root, in right side and then use the squaring method. solve the resulting rational equation and roots. But also another, more elegant one. Enter a new variable; vх=y. Accordingly, you will receive an equation of the form 2y2+y-3=0. That is, the usual quadratic equation. Find its roots; y1=1 and y2=-3/2. Next, solve two equations vх=1; vх=-3/2. The second equation has no roots; from the first we find that x=1. Don't forget to check the roots.
Solving identities is quite simple. To do this you need to do identity transformations until the goal is achieved. Thus, with the help of simple arithmetic operations, the problem posed will be solved.
You will need
- - paper;
- - pen.
Instructions
The simplest of such transformations are algebraic abbreviated multiplications (such as the square of the sum (difference), difference of squares, sum (difference), cube of the sum (difference)). In addition, there are many and trigonometric formulas, which are essentially the same identities.
Indeed, the square of the sum of two terms is equal to the square of the first plus twice the product of the first by the second and plus the square of the second, that is, (a+b)^2= (a+b)(a+b)=a^2+ab +ba+b ^2=a^2+2ab+b^2.
Simplify both
General principles of the solution
Repeat from a textbook on mathematical analysis or higher mathematics what a definite integral is. As is known, the solution definite integral there is a function whose derivative gives an integrand. This function is called antiderivative. Based on this principle, the main integrals are constructed.Determine by the form of the integrand which of the table integrals fits in in this case. It is not always possible to determine this immediately. Often, the tabular form becomes noticeable only after several transformations to simplify the integrand.
Variable Replacement Method
If the integrand function is trigonometric function, whose argument contains some polynomial, then try using the variable replacement method. In order to do this, replace the polynomial in the argument of the integrand with some new variable. Based on the relationship between the new and old variables, determine the new limits of integration. By differentiating this expression, find the new differential in . So you will get the new kind of the previous integral, close to or even corresponding to any tabular one.Solving integrals of the second kind
If the integral is an integral of the second kind, a vector form of the integrand, then you will need to use the rules for the transition from these integrals to scalar ones. One such rule is the Ostrogradsky-Gauss relation. This law allows you to move from the rotor flow to some vector function to the triple integral over the divergence of a given vector field.Substitution of integration limits
After finding the antiderivative, it is necessary to substitute the limits of integration. First, substitute the value of the upper limit into the expression for the antiderivative. You will get some number. Next, subtract from the resulting number another number obtained from the lower limit into the antiderivative. If one of the limits of integration is infinity, then when substituting it into antiderivative function it is necessary to go to the limit and find what the expression strives for.If the integral is two-dimensional or three-dimensional, then you will have to represent the limits of integration geometrically to understand how to evaluate the integral. Indeed, in the case of, say, a three-dimensional integral, the limits of integration can be entire planes that limit the volume being integrated.
The final videos in a long series of lessons about solving logarithmic equations. This time we will work primarily with the ODZ of the logarithm - it is precisely because of incorrect consideration (or even ignoring) of the domain of definition that most errors arise when solving such problems.
In this short video lesson, we will look at the use of formulas for adding and subtracting logarithms, and also deal with fractional rational equations, which many students also have problems with.
What will we talk about? The main formula I would like to understand looks like this:
log a (f g ) = log a f + log a g
This is a standard transition from the product to the sum of logarithms and back. You probably know this formula from the very beginning of studying logarithms. However, there is one hitch.
As long as the variables a, f and g are ordinary numbers, no problems arise. This formula works great.
However, as soon as functions appear instead of f and g, the problem of expanding or narrowing the domain of definition arises depending on which direction to transform. Judge for yourself: in the logarithm written on the left, the domain of definition is as follows:
fg > 0
But in the amount written on the right, the domain of definition is already somewhat different:
f > 0
g > 0
This set of requirements is more stringent than the original one. In the first case, we will be satisfied with option f< 0, g < 0 (ведь их произведение положительное, поэтому неравенство fg >0 is executed).
So, when moving from the left construction to the right one, a narrowing of the domain of definition occurs. If at first we had a sum, and we rewrite it in the form of a product, then the domain of definition expands.
In other words, in the first case we could lose roots, and in the second we could get extra ones. This must be taken into account when solving real logarithmic equations.
So, the first task:
[Caption for the picture] On the left we see the sum of logarithms using the same base. Therefore, these logarithms can be added:
[Caption for the picture] As you can see, on the right we replaced the zero using the formula:
a = log b b a
Let's rearrange our equation a little more:
log 4 (x − 5) 2 = log 4 1
Before us is the canonical form of the logarithmic equation; we can cross out the log sign and equate the arguments:
(x − 5) 2 = 1
|x − 5| = 1
Please note: where did the module come from? Let me remind you that the root of an exact square is equal to the modulus:
[Caption for the picture] Then we solve the classical equation with modulus:
|f | = g (g > 0) ⇒f = ±g
x − 5 = ±1 ⇒x 1 = 5 − 1 = 4; x 2 = 5 + 1 = 6
Here are two candidate answers. Are they a solution to the original logarithmic equation? No way!
We have no right to leave everything just like that and write down the answer. Take a look at the step where we replace the sum of logarithms with one logarithm of the product of the arguments. The problem is that in the original expressions we have functions. Therefore, you should require:
x(x − 5) > 0; (x − 5)/x > 0.
When we transformed the product, obtaining an exact square, the requirements changed:
(x − 5) 2 > 0
When is this requirement met? Yes, almost always! Except for the case when x − 5 = 0. That is the inequality will be reduced to one punctured point:
x − 5 ≠ 0 ⇒ x ≠ 5
As you can see, the scope of definition has expanded, which is what we talked about at the very beginning of the lesson. Consequently, extra roots may appear.
How can you prevent these extra roots from appearing? It’s very simple: we look at our obtained roots and compare them with the domain of definition of the original equation. Let's count:
x (x − 5) > 0
We will solve using the interval method:
x (x − 5) = 0 ⇒ x = 0; x = 5
We mark the resulting numbers on the line. All points are missing because the inequality is strict. Take any number greater than 5 and substitute:
[Caption for the picture] We are interested in the intervals (−∞; 0) ∪ (5; ∞). If we mark our roots on the segment, we will see that x = 4 does not suit us, because this root lies outside the domain of definition of the original logarithmic equation.
We return to the totality, cross out the root x = 4 and write down the answer: x = 6. This is the final answer to the original logarithmic equation. That's it, problem solved.
Let's move on to the second logarithmic equation:
[Caption for the picture] Let's solve it. Note that the first term is a fraction, and the second is the same fraction, but inverted. Don't be scared by the expression lgx - it's just a decimal logarithm, we can write it:
lgx = log 10 x
Since we have two inverted fractions, I propose introducing a new variable:
[Caption for the picture] Therefore, our equation can be rewritten as follows:
t + 1/t = 2;
t + 1/t − 2 = 0;
(t 2 − 2t + 1)/t = 0;
(t − 1) 2 /t = 0.
As you can see, the numerator of the fraction is an exact square. A fraction is equal to zero when its numerator is zero and its denominator is non-zero:
(t − 1) 2 = 0; t ≠ 0
Let's solve the first equation:
t − 1 = 0;
t = 1.
This value satisfies the second requirement. Therefore, we can say that we have completely solved our equation, but only with respect to the variable t. Now let’s remember what t is:
[Caption for the picture]We got the proportion:
logx = 2 logx + 1
2 logx − logx = −1
logx = −1
We bring this equation to its canonical form:
logx = log 10 −1
x = 10 −1 = 0.1
As a result, we received a single root, which, in theory, is the solution to the original equation. However, let’s still play it safe and write out the domain of definition of the original equation:
[Caption for the picture] Therefore, our root satisfies all the requirements. We have found a solution to the original logarithmic equation. Answer: x = 0.1. The problem is solved.
There is only one key point in today's lesson: when using the formula for moving from a product to a sum and back, be sure to take into account that the scope of definition can narrow or expand depending on which direction the transition is made.
How to understand what is happening: contraction or expansion? Very simple. If earlier the functions were together, but now they are separate, then the scope of definition has narrowed (because there are more requirements). If at first the functions stood separately, and now they are together, then the domain of definition is expanded (fewer requirements are imposed on the product than on individual factors).
Taking into account this remark, I would like to note that the second logarithmic equation does not require these transformations at all, that is, we do not add or multiply the arguments anywhere. However, here I would like to draw your attention to another wonderful technique that can significantly simplify the solution. It's about replacing a variable.
However, remember that no substitutions free us from the scope of definition. That is why after all the roots were found, we were not lazy and returned to the original equation to find its ODZ.
Often, when replacing a variable, an annoying error occurs when students find the value of t and think that the solution is complete. No way!
Once you have found the value of t, you need to go back to the original equation and see what exactly we meant with this letter. As a result, we have to solve one more equation, which, however, will be much simpler than the original one.
This is precisely the point of introducing a new variable. We split the original equation into two intermediate ones, each of which has a much simpler solution.
How to solve "nested" logarithmic equations
Today we continue to study logarithmic equations and will analyze constructions when one logarithm is under the sign of another logarithm. We will solve both equations using the canonical form.
Today we continue to study logarithmic equations and will analyze constructions when one logarithm is under the sign of another. We will solve both equations using the canonical form. Let me remind you that if we have the simplest logarithmic equation of the form log a f (x) = b, then to solve such an equation we perform the following steps. First of all, we need to replace the number b :
b = log a a b
Note: a b is an argument. Similarly, in the original equation, the argument is the function f(x). Then we rewrite the equation and get this construction:
log a f (x) = log a a b
Then we can perform the third step - get rid of the logarithm sign and simply write:
f (x) = a b
As a result, we get a new equation. In this case, no restrictions are imposed on the function f (x). For example, in its place there may also be logarithmic function. And then we will again obtain a logarithmic equation, which we will again reduce to its simplest form and solve through the canonical form.
However, enough of the lyrics. Let's solve the real problem. So, task number 1:
log 2 (1 + 3 log 2 x ) = 2
As you can see, we have a simple logarithmic equation. The role of f (x) is the construction 1 + 3 log 2 x, and the role of the number b is the number 2 (the role of a is also played by two). Let's rewrite this two as follows:
It is important to understand that the first two twos came to us from the base of the logarithm, i.e. if there were 5 in the original equation, then we would get that 2 = log 5 5 2. In general, the base depends solely on the logarithm that was originally given in the problem. And in our case this is the number 2.
So, we rewrite our logarithmic equation taking into account the fact that the two on the right is actually also a logarithm. We get:
log 2 (1 + 3 log 2 x ) = log 2 4
Let's move on to the last step of our scheme - getting rid of the canonical form. You could say, we simply cross out the signs of log. However, from a mathematical point of view, it is impossible to “cross out log” - it would be more correct to say that we simply equate the arguments:
1 + 3 log 2 x = 4
From here we can easily find 3 log 2 x:
3 log 2 x = 3
log 2 x = 1
We have again obtained the simplest logarithmic equation, let's bring it back to the canonical form. To do this we need to make the following changes:
1 = log 2 2 1 = log 2 2
Why is there a two at the base? Because in our canonical equation on the left there is a logarithm precisely to base 2. We rewrite the problem taking into account this fact:
log 2 x = log 2 2
Again we get rid of the logarithm sign, i.e. we simply equate the arguments. We have the right to do this because the bases are the same, and no more additional actions were performed either on the right or on the left:
That's all! The problem is solved. We have found a solution to the logarithmic equation.
Note! Although the variable x appears in the argument (i.e., there are requirements for the domain of definition), we will not make any additional requirements.
As I said above, this check is redundant if the variable appears in only one argument of only one logarithm. In our case, x really appears only in the argument and only under one log sign. Therefore, no additional checks are required.
However, if you don't trust this method, then you can easily verify that x = 2 is indeed a root. It is enough to substitute this number into the original equation.
Let's move on to the second equation, it's a little more interesting:
log 2 (log 1/2 (2x − 1) + log 2 4) = 1
If we denote the expression inside the large logarithm with the function f (x), we get the simplest logarithmic equation with which we started today's video lesson. Therefore, we can apply the canonical form, for which we will have to represent the unit in the form log 2 2 1 = log 2 2.
Let's rewrite our big equation:
log 2 (log 1/2 (2x − 1) + log 2 4) = log 2 2
Let's get away from the sign of the logarithm, equating the arguments. We have the right to do this, because both on the left and on the right the bases are the same. Additionally, note that log 2 4 = 2:
log 1/2 (2x − 1) + 2 = 2
log 1/2 (2x − 1) = 0
Before us again is the simplest logarithmic equation of the form log a f (x) = b. Let's move on to the canonical form, that is, we represent zero in the form log 1/2 (1/2)0 = log 1/2 1.
We rewrite our equation and get rid of the log sign, equating the arguments:
log 1/2 (2x − 1) = log 1/2 1
2x − 1 = 1
Again, we received an answer immediately. No additional checks are required because in the original equation only one logarithm contains the function as an argument.
Therefore, no additional checks are required. We can safely say that x = 1 is the only root of this equation.
But if in the second logarithm there was some function of x instead of four (or 2x was not in the argument, but in the base) - then it would be necessary to check the domain of definition. Otherwise, there is a high chance of running into extra roots.
Where do these extra roots come from? This point must be understood very clearly. Take a look at the original equations: everywhere the function x is under the logarithm sign. Consequently, since we wrote down log 2 x, we automatically set the requirement x > 0. Otherwise, this entry simply does not make sense.
However, as we solve the logarithmic equation, we get rid of all the log signs and get simple constructions. There are no restrictions set here anymore, because linear function defined for any value of x.
It is this problem, when the final function is defined everywhere and always, but the original one is not defined everywhere and not always, that is the reason why extra roots very often arise in solving logarithmic equations.
But I repeat once again: this only happens in a situation where the function is either in several logarithms or at the base of one of them. In the problems that we are considering today, there are, in principle, no problems with expanding the domain of definition.
Cases of different grounds
This lesson is devoted to more complex structures. Logarithms in today's equations will no longer be solved straight away; some transformations will need to be done first.
We begin solving logarithmic equations with completely different bases, which are not exact powers of each other. Don’t let such problems scare you - they are no more difficult to solve than the simplest designs that we discussed above.
But before moving directly to the problems, let me remind you of the formula for solving the simplest logarithmic equations using the canonical form. Consider a problem like this:
log a f (x) = b
It is important that the function f (x) is just a function, and the role of numbers a and b should be numbers (without any variables x). Of course, literally in a minute we will look at such cases when instead of variables a and b there are functions, but that’s not about that now.
As we remember, the number b must be replaced by a logarithm to the same base a, which is on the left. This is done very simply:
b = log a a b
Of course, the words “any number b” and “any number a” mean values that satisfy the scope of definition. In particular, in given equation we're talking about only the base a > 0 and a ≠ 1.
However, this requirement is satisfied automatically, because the original problem already contains a logarithm to base a - it will certainly be greater than 0 and not equal to 1. Therefore, we continue solving the logarithmic equation:
log a f (x) = log a a b
Such a notation is called canonical form. Its convenience lies in the fact that we can immediately get rid of the log sign by equating the arguments:
f (x) = a b
It is this technique that we will now use to solve logarithmic equations with variable base. So, let's go!
log 2 (x 2 + 4x + 11) = log 0.5 0.125
What's next? Someone will now say that you need to calculate the right logarithm, or reduce them to the same base, or something else. And indeed, now we need to bring both bases to the same form - either 2 or 0.5. But let's learn the following rule once and for all:
If there are decimals in a logarithmic equation, be sure to convert those fractions from decimal notation to normal. This transformation can greatly simplify the solution.
Such a transition must be performed immediately, even before performing any actions or transformations. Let's get a look:
log 2 (x 2 + 4x + 11) = log 1 /2 1/8
What does such a record give us? We can represent 1/2 and 1/8 as powers with a negative exponent:
[Caption for the picture] Before us is the canonical form. We equate the arguments and get the classic quadratic equation:
x 2 + 4x + 11 = 8
x 2 + 4x + 3 = 0
We have before us the following quadratic equation, which can be easily solved using Vieta’s formulas. In high school, you should see similar displays literally orally:
(x + 3)(x + 1) = 0
x 1 = −3
x 2 = −1
That's all! The original logarithmic equation has been solved. We got two roots.
Let me remind you that in this case it is not necessary to determine the domain of definition, since the function with the variable x is present in only one argument. Therefore, the definition scope is performed automatically.
So, the first equation is solved. Let's move on to the second:
log 0.5 (5x 2 + 9x + 2) = log 3 1/9
log 1/2 (5x 2 + 9x + 2) = log 3 9 −1
Now note that the argument of the first logarithm can also be written as a power with a negative exponent: 1/2 = 2 −1. Then you can take out the powers on both sides of the equation and divide everything by −1:
[Caption for the picture] And now we have completed a very important step in solving the logarithmic equation. Perhaps someone didn't notice something, so let me explain.
Look at our equation: both the sign on the left and right is log, but on the left there is a logarithm to the base 2, and on the right there is a logarithm to the base 3. Three is not whole degree twos and vice versa: it is impossible to write that 2 is 3 to the integer power.
Consequently, these are logarithms with different bases that cannot be reduced to each other by simply adding powers. The only way to solve such problems is to get rid of one of these logarithms. In this case, since we are still considering quite simple tasks, the logarithm on the right was simply calculated, and we got the simplest equation - exactly the one we talked about at the very beginning of today's lesson.
Let's represent the number 2, which is on the right, as log 2 2 2 = log 2 4. And then we get rid of the logarithm sign, after which we are simply left with a quadratic equation:
log 2 (5x 2 + 9x + 2) = log 2 4
5x 2 + 9x + 2 = 4
5x 2 + 9x − 2 = 0
We have before us an ordinary quadratic equation, but it is not reduced because the coefficient of x 2 is different from unity. Therefore, we will solve it using the discriminant:
D = 81 − 4 5 (−2) = 81 + 40 = 121
x 1 = (−9 + 11)/10 = 2/10 = 1/5
x 2 = (−9 − 11)/10 = −2
That's all! We have found both roots, which means we have obtained a solution to the original logarithmic equation. Indeed, in the original problem, the function with variable x is present in only one argument. Consequently, no additional checks on the domain of definition are required - both roots that we found certainly meet all possible restrictions.
This could be the end of today’s video lesson, but in conclusion I would like to say again: be sure to convert all decimal fractions to ordinary fractions when solving logarithmic equations. In most cases, this greatly simplifies their solution.
Rarely, very rarely, do you come across problems in which getting rid of decimal fractions only complicates the calculations. However, in such equations, as a rule, it is initially clear that there is no need to get rid of decimal fractions.
In most other cases (especially if you are just starting to practice solving logarithmic equations), feel free to get rid of the decimals and convert them to ordinary ones. Because practice shows that in this way you will significantly simplify the subsequent solution and calculations.
Subtleties and tricks of the solution
Today we move on to more complex problems and will solve a logarithmic equation, which is based not on a number, but on a function.
And even if this function is linear, you will have to add minor changes, the meaning of which boils down to additional requirements imposed on the domain of definition of the logarithm.
Complex tasks
This tutorial will be quite long. In it we will analyze two rather serious logarithmic equations, when solving which many students make mistakes. During my practice as a math tutor, I constantly encountered two types of errors:
- The appearance of extra roots due to the expansion of the domain of definition of logarithms. To avoid such offensive mistakes, just carefully monitor each transformation;
- Loss of roots due to the fact that the student forgot to consider some “subtle” cases - these are the situations we will focus on today.
This is the last lesson on logarithmic equations. It will be long, we will analyze complex logarithmic equations. Make yourself comfortable, make yourself some tea, and let's get started.
The first equation looks quite standard:
log x + 1 (x − 0.5) = log x − 0.5 (x + 1)
Let us immediately note that both logarithms are inverted copies of each other. Let's remember the wonderful formula:
log a b = 1/log b a
However, this formula has a number of limitations that arise if instead of the numbers a and b there are functions of the variable x:
b > 0
1 ≠ a > 0
These requirements apply to the base of the logarithm. On the other hand, in a fraction we are required to have 1 ≠ a > 0, since not only is the variable a in the argument of the logarithm (hence a > 0), but the logarithm itself is in the denominator of the fraction. But log b 1 = 0, and the denominator must be non-zero, so a ≠ 1.
So, the restrictions on the variable a remain. But what happens to the variable b? On the one hand, the base implies b > 0, on the other hand, the variable b ≠ 1, because the base of the logarithm must be different from 1. In total, from the right side of the formula it follows that 1 ≠ b > 0.
But here’s the problem: the second requirement (b ≠ 1) is missing from the first inequality, which deals with the left logarithm. In other words, when performing this transformation we must check separately, that the argument b is different from one!
So let's check it out. Let's apply our formula:
[Caption for the picture] 1 ≠ x − 0.5 > 0; 1 ≠ x + 1 > 0
So we got that already from the original logarithmic equation it follows that both a and b must be greater than 0 and not equal to 1. This means that we can easily invert the logarithmic equation:
I suggest introducing a new variable:
log x + 1 (x − 0.5) = t
In this case, our construction will be rewritten as follows:
(t 2 − 1)/t = 0
Note that in the numerator we have the difference of squares. We reveal the difference of squares using the abbreviated multiplication formula:
(t − 1)(t + 1)/t = 0
A fraction is equal to zero when its numerator is zero and its denominator is non-zero. But the numerator contains a product, so we equate each factor to zero:
t 1 = 1;
t 2 = −1;
t ≠ 0.
As we can see, both values of the variable t suit us. However, the solution does not end there, because we need to find not t, but the value of x. We return to the logarithm and get:
log x + 1 (x − 0.5) = 1;
log x + 1 (x − 0.5) = −1.
Let's put each of these equations in canonical form:
log x + 1 (x − 0.5) = log x + 1 (x + 1) 1
log x + 1 (x − 0.5) = log x + 1 (x + 1) −1
We get rid of the logarithm sign in the first case and equate the arguments:
x − 0.5 = x + 1;
x − x = 1 + 0.5;
Such an equation has no roots, therefore the first logarithmic equation also has no roots. But with the second equation everything is much more interesting:
(x − 0.5)/1 = 1/(x + 1)
Solving the proportion, we get:
(x − 0.5)(x + 1) = 1
Let me remind you that when solving logarithmic equations it is much more convenient to use all decimal fractions as ordinary ones, so let's rewrite our equation as follows:
(x − 1/2)(x + 1) = 1;
x 2 + x − 1/2x − 1/2 − 1 = 0;
x 2 + 1/2x − 3/2 = 0.
We have before us the quadratic equation below, it can be easily solved using Vieta’s formulas:
(x + 3/2) (x − 1) = 0;
x 1 = −1.5;
x 2 = 1.
We got two roots - they are candidates for solving the original logarithmic equation. In order to understand what roots will actually go into the answer, let's return to the original problem. Now we will check each of our roots to see if they fit within the domain of definition:
1.5 ≠ x > 0.5; 0 ≠ x > −1.
These requirements are tantamount to a double inequality:
1 ≠ x > 0.5
From here we immediately see that the root x = −1.5 does not suit us, but x = 1 suits us quite well. Therefore x = 1 is the final solution to the logarithmic equation.
Let's move on to the second task:
log x 25 + log 125 x 5 = log 25 x 625
At first glance it may seem that all logarithms different reasons and different arguments. What to do with such structures? First of all, note that the numbers 25, 5 and 625 are powers of 5:
25 = 5 2 ; 625 = 5 4
Now let's take advantage of the wonderful property of the logarithm. The point is that you can extract powers from an argument in the form of factors:
log a b n = n ∙ log a b
This transformation is also subject to restrictions in the case where b is replaced by a function. But for us, b is just a number, and no additional restrictions arise. Let's rewrite our equation:
2 ∙ log x 5 + log 125 x 5 = 4 ∙ log 25 x 5
We have obtained an equation with three terms containing the log sign. Moreover, the arguments of all three logarithms are equal.
It's time to reverse the logarithms to bring them to the same base - 5. Since the variable b is a constant, no changes in the domain of definition occur. We just rewrite:
[Caption for the picture] As expected, the same logarithms appeared in the denominator. I suggest replacing the variable:
log 5 x = t
In this case, our equation will be rewritten as follows:
Let's write out the numerator and open the brackets:
2 (t + 3) (t + 2) + t (t + 2) − 4t (t + 3) = 2 (t 2 + 5t + 6) + t 2 + 2t − 4t 2 − 12t = 2t 2 + 10t + 12 + t 2 + 2t − 4t 2 − 12t = −t 2 + 12
Let's return to our fraction. The numerator must be zero:
[Caption for the picture] And the denominator is different from zero:
t ≠ 0; t ≠ −3; t ≠ −2
The last requirements are fulfilled automatically, since they are all “tied” to integers, and all answers are irrational.
So, fractional rational equation solved, the values of the variable t are found. Let's return to solving the logarithmic equation and remember what t is:
[Caption for the picture] We reduce this equation to canonical form and obtain a number with an irrational degree. Don’t let this confuse you - even such arguments can be equated:
[Caption for the picture] We got two roots. More precisely, two candidate answers - let's check them for compliance with the domain of definition. Since the base of the logarithm is the variable x, we require the following:
1 ≠ x > 0;
With the same success we assert that x ≠ 1/125, otherwise the base of the second logarithm will turn to unity. Finally, x ≠ 1/25 for the third logarithm.
In total, we received four restrictions:
1 ≠ x > 0; x ≠ 1/125; x ≠ 1/25
Now the question is: do our roots satisfy these requirements? Of course they satisfy! Because 5 to any power will be greater than zero, and the requirement x > 0 is satisfied automatically.
On the other hand, 1 = 5 0, 1/25 = 5 −2, 1/125 = 5 −3, which means that these restrictions for our roots (which, let me remind you, have irrational number) are also satisfied, and both answers are solutions to the problem.
So, we have the final answer. There are two key points in this task:
- Be careful when flipping a logarithm when the argument and base are swapped. Such transformations impose unnecessary restrictions on the scope of definition.
- Don’t be afraid to transform logarithms: you can not only turn them over, but also open them using the sum formula and generally change them using any formulas that you studied when solving logarithmic expressions. However, always remember: some transformations expand the scope of definition, and some narrow them.
