Examples of solutions of logarithms. Examples for logarithms
With this video I begin a long series of lessons about logarithmic equations. Now you have three examples in front of you, on the basis of which we will learn to solve the most simple tasks, which are called so - protozoa.
log 0.5 (3x − 1) = −3
log (x + 3) = 3 + 2 log 5
Let me remind you that the simplest logarithmic equation is the following:
log a f (x) = b
In this case, it is important that the variable x is present only inside the argument, that is, only in the function f (x). And the numbers a and b are just numbers, and in no case are functions containing the variable x.
Basic solution methods
There are many ways to solve such structures. For example, most teachers at school offer this method: Immediately express the function f (x) using the formula f ( x ) = a b . That is, when you come across the simplest construction, you can immediately move on to the solution without additional actions and constructions.
Yes, of course, the decision will be correct. However, the problem with this formula is that most students do not understand, where it comes from and why we raise the letter a to the letter b.
As a result, I often see very annoying mistakes when, for example, these letters are swapped. This formula must either be understood or crammed, and the second method leads to mistakes at the most inopportune and most crucial moments: during exams, tests, etc.
That is why I suggest to all my students to abandon the standard school formula and use the second approach to solve logarithmic equations, which, as you probably guessed from the name, is called canonical form.
The idea of the canonical form is simple. Let's look at our problem again: on the left we have log a, and by the letter a we mean a number, and in no case a function containing the variable x. Consequently, this letter is subject to all the restrictions that are imposed on the base of the logarithm. namely:
1 ≠ a > 0
On the other hand, from the same equation we see that the logarithm must be equal to the number b , and no restrictions are imposed on this letter, because it can take any values - both positive and negative. It all depends on what values the function f(x) takes.
And here we remember our wonderful rule that any number b can be represented as a logarithm to the base a of a to the power of b:
b = log a a b
How to remember this formula? Yes, very simple. Let's write the following construction:
b = b 1 = b log a a
Of course, in this case all the restrictions that we wrote down at the beginning arise. Now let's use the basic property of the logarithm and introduce the multiplier b as the power of a. We get:
b = b 1 = b log a a = log a a b
As a result, the original equation will be rewritten as follows:
log a f (x) = log a a b → f (x) = a b
That's all. The new function no longer contains a logarithm and can be solved using standard algebraic techniques.
Of course, someone will now object: why was it necessary to come up with some kind of canonical formula at all, why perform two additional unnecessary steps if it was possible to immediately move from the original design to the final formula? Yes, if only because most students do not understand where this formula comes from and, as a result, regularly make mistakes when applying it.
But this sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where the final formula comes from. By the way, this entry is called the canonical formula:
log a f (x) = log a a b
The convenience of the canonical form also lies in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest ones that we are considering today.
Examples of solutions
Now let's take a look real examples. So, let's decide:
log 0.5 (3x − 1) = −3
Let's rewrite it like this:
log 0.5 (3x − 1) = log 0.5 0.5 −3
Many students are in a hurry and try to immediately raise the number 0.5 to the power that came to us from the original problem. Indeed, when you are already well trained in solving such problems, you can immediately perform this step.
However, if you are now just starting to study this topic, it is better not to rush anywhere in order to avoid making offensive mistakes. So, we have the canonical form. We have:
3x − 1 = 0.5 −3
This is no longer a logarithmic equation, but linear with respect to the variable x. To solve it, let's first look at the number 0.5 to the power of −3. Note that 0.5 is 1/2.
(1/2) −3 = (2/1) 3 = 8
All decimals convert to ordinary ones when you solve a logarithmic equation.
We rewrite and get:
3x − 1 = 8
3x = 9
x = 3
That's it, we got the answer. The first problem has been solved.
Second task
Let's move on to the second task:
As we see, this equation is no longer the simplest. If only because there is a difference on the left, and not a single logarithm to one base.
Therefore, we need to somehow get rid of this difference. IN in this case everything is very simple. Let's take a closer look at the bases: on the left is the number under the root:
General recommendation: in all logarithmic equations, try to get rid of radicals, i.e., from entries with roots and move on to power functions, simply because the exponents of these powers are easily taken out of the sign of the logarithm and, ultimately, such a notation significantly simplifies and speeds up calculations. Let's write it down like this:
Now let us remember the remarkable property of the logarithm: powers can be derived from the argument, as well as from the base. In the case of grounds, the following happens:
log a k b = 1/k loga b
In other words, the number that was in the base power is brought forward and at the same time inverted, i.e. it becomes reciprocal number. In our case, the base degree was 1/2. Therefore, we can take it out as 2/1. We get:
5 2 log 5 x − log 5 x = 18
10 log 5 x − log 5 x = 18
Please note: under no circumstances should you get rid of logarithms at this step. Remember 4th-5th grade math and the order of operations: multiplication is performed first, and only then addition and subtraction. In this case, we subtract one of the same elements from 10 elements:
9 log 5 x = 18
log 5 x = 2
Now our equation looks as it should. This is the simplest construction, and we solve it using the canonical form:
log 5 x = log 5 5 2
x = 5 2
x = 25
That's all. The second problem has been solved.
Third example
Let's move on to the third task:
log (x + 3) = 3 + 2 log 5
Let me remind you of the following formula:
log b = log 10 b
If for some reason you are confused by the notation log b , then when performing all the calculations you can simply write log 10 b . You can work with decimal logarithms in the same way as with others: take powers, add and represent any numbers in the form lg 10.
It is these properties that we will now use to solve the problem, since it is not the simplest one that we wrote down at the very beginning of our lesson.
First, note that the factor 2 in front of lg 5 can be added and becomes a power of base 5. In addition, the free term 3 can also be represented as a logarithm - this is very easy to observe from our notation.
Judge for yourself: any number can be represented as log to base 10:
3 = log 10 10 3 = log 10 3
Let's rewrite the original problem taking into account the obtained changes:
log (x − 3) = log 1000 + log 25
log (x − 3) = log 1000 25
log (x − 3) = log 25,000
We have before us the canonical form again, and we got it without going through the transformation stage, i.e. the simplest logarithmic equation did not appear anywhere.
This is exactly what I talked about at the very beginning of the lesson. The canonical form allows you to solve a wider class of problems than the standard school formula that most school teachers give.
That's it, let's get rid of the sign decimal logarithm, and we get a simple linear construction:
x + 3 = 25,000
x = 24,997
All! The problem is solved.
A note on scope
Here I would like to bring important note regarding the scope of definition. Surely now there will be students and teachers who will say: “When we solve expressions with logarithms, we must remember that the argument f (x) must be greater than zero!” In this regard, a logical question arises: why did we not require this inequality to be satisfied in any of the problems considered?
Do not worry. In these cases, no extra roots will appear. And this is another great trick that allows you to speed up the solution. Just know that if in the problem the variable x occurs only in one place (or rather, in one single argument of a single logarithm), and nowhere else in our case does the variable x appear, then write down the domain of definition no need, because it will be executed automatically.
Judge for yourself: in the first equation we got that 3x − 1, i.e. the argument should be equal to 8. This automatically means that 3x − 1 will be greater than zero.
With the same success we can write that in the second case x should be equal to 5 2, i.e. it is certainly greater than zero. And in the third case, where x + 3 = 25,000, i.e., again, obviously greater than zero. In other words, the scope is satisfied automatically, but only if x occurs only in the argument of only one logarithm.
That's all you need to know to solve the simplest problems. This rule alone, together with the transformation rules, will allow you to solve a very wide class of problems.
But let's be honest: in order to finally understand this technique, to learn how to apply the canonical form of the logarithmic equation, it is not enough to just watch one video lesson. Therefore, right now, download the options for independent solutions that are attached to this video lesson and start solving at least one of these two independent works.
It will take you literally a few minutes. But the effect of such training will be much higher than if you simply watched this video lesson.
I hope this lesson will help you understand logarithmic equations. Use the canonical form, simplify expressions using the rules for working with logarithms - and you won’t be afraid of any problems. That's all I have for today.
Taking into account the domain of definition
Now let's talk about the domain of definition logarithmic function, as well as how this affects the solution of logarithmic equations. Consider a construction of the form
log a f (x) = b
Such an expression is called the simplest - it contains only one function, and the numbers a and b are just numbers, and in no case a function that depends on the variable x. It can be solved very simply. You just need to use the formula:
b = log a a b
This formula is one of the key properties of the logarithm, and when substituting into our original expression we get the following:
log a f (x) = log a a b
f (x) = a b
This is a familiar formula from school textbooks. Many students will probably have a question: since in the original expression the function f (x) is under the log sign, the following restrictions are imposed on it:
f(x) > 0
This limitation applies because the logarithm of negative numbers does not exist. So, perhaps, as a result of this limitation, a check on answers should be introduced? Perhaps they need to be inserted into the source?
No, in the simplest logarithmic equations additional checking is unnecessary. And that's why. Take a look at our final formula:
f (x) = a b
The fact is that the number a is in any case greater than 0 - this requirement is also imposed by the logarithm. The number a is the base. In this case, no restrictions are imposed on the number b. But this doesn’t matter, because no matter what power we raise a positive number to, we will still get a positive number at the output. Thus, the requirement f (x) > 0 is satisfied automatically.
What's really worth checking is the domain of the function under the log sign. There may be quite complex structures, and you definitely need to keep an eye on them during the solution process. Let's get a look.
First task:
First step: convert the fraction on the right. We get:
We get rid of the logarithm sign and get the usual irrational equation:
Of the obtained roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks are required to ensure that the expression under the logarithm sign is greater than 0, because it is not just greater than 0, but according to the condition of the equation it is equal to 2. Therefore, the requirement “greater than zero” is satisfied automatically.
Let's move on to the second task:
Everything is the same here. We rewrite the construction, replacing the triple:
We get rid of the logarithm signs and get an irrational equation:
We square both sides taking into account the restrictions and get:
4 − 6x − x 2 = (x − 4) 2
4 − 6x − x 2 = x 2 + 8x + 16
x 2 + 8x + 16 −4 + 6x + x 2 = 0
2x 2 + 14x + 12 = 0 |:2
x 2 + 7x + 6 = 0
We solve the resulting equation through the discriminant:
D = 49 − 24 = 25
x 1 = −1
x 2 = −6
But x = −6 does not suit us, because if we substitute this number into our inequality, we get:
−6 + 4 = −2 < 0
In our case, it is required that it be greater than 0 or, in extreme cases, equal. But x = −1 suits us:
−1 + 4 = 3 > 0
The only answer in our case will be x = −1. That's the solution. Let's go back to the very beginning of our calculations.
The main takeaway from this lesson is that you don't need to check constraints on a function in simple logarithmic equations. Because during the solution process all constraints are satisfied automatically.
However, this in no way means that you can forget about checking altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own restrictions and requirements for the right side, which we have seen today in two different examples.
Feel free to solve such problems and be especially careful if there is a root in the argument.
Logarithmic equations with different bases
We continue to study logarithmic equations and look at two more quite interesting techniques with which it is fashionable to solve more complex constructions. But first, let’s remember how the simplest problems are solved:
log a f (x) = b
In this entry, a and b are numbers, and in the function f (x) the variable x must be present, and only there, that is, x must only be in the argument. We will transform such logarithmic equations using the canonical form. To do this, note that
b = log a a b
Moreover, a b is precisely an argument. Let's rewrite this expression as follows:
log a f (x) = log a a b
This is exactly what we are trying to achieve, so that there is a logarithm to base a on both the left and the right. In this case, we can, figuratively speaking, cross out the log signs, and from a mathematical point of view we can say that we are simply equating the arguments:
f (x) = a b
As a result, we will get a new expression that will be much easier to solve. Let's apply this rule to our problems today.
So, the first design:
First of all, I note that on the right is a fraction whose denominator is log. When you see an expression like this, it’s a good idea to remember a wonderful property of logarithms:
Translated into Russian, this means that any logarithm can be represented as the quotient of two logarithms with any base c. Of course 0< с ≠ 1.
So: this formula has one wonderful special case, when the variable c is equal to the variable b. In this case we get a construction like:
This is exactly the construction we see from the sign on the right in our equation. Let's replace this construction with log a b , we get:
In other words, in comparison with the original task, we swapped the argument and the base of the logarithm. Instead, we had to reverse the fraction.
We recall that any degree can be derived from the base according to the following rule:
In other words, the coefficient k, which is the power of the base, is expressed as an inverted fraction. Let's render it as an inverted fraction:
The fractional factor cannot be left in front, because in this case we will not be able to represent this notation as a canonical form (after all, in the canonical form there is no additional factor before the second logarithm). Therefore, let's add the fraction 1/4 to the argument as a power:
Now we equate arguments whose bases are the same (and our bases are really the same), and write:
x + 5 = 1
x = −4
That's all. We got the answer to the first logarithmic equation. Please note: in the original problem, the variable x appears in only one log, and it appears in its argument. Therefore, there is no need to check the domain, and our number x = −4 is indeed the answer.
Now let's move on to the second expression:
log 56 = log 2 log 2 7 − 3log (x + 4)
Here, in addition to the usual logarithms, we will have to work with log f (x). How to solve such an equation? To an unprepared student it may seem like this is some kind of tough task, but in fact everything can be solved in an elementary way.
Take a close look at the term lg 2 log 2 7. What can we say about it? The bases and arguments of log and lg are the same, and this should give some ideas. Let's remember once again how powers are taken out from under the sign of the logarithm:
log a b n = nlog a b
In other words, what was a power of b in the argument becomes a factor in front of log itself. Let's apply this formula to the expression lg 2 log 2 7. Don't be scared by lg 2 - this is the most common expression. You can rewrite it as follows:
All the rules that apply to any other logarithm are valid for it. In particular, the factor in front can be added to the degree of the argument. Let's write it down:
Very often, students do not see this action directly, because it is not good to enter one log under the sign of another. In fact, there is nothing criminal about this. Moreover, we get a formula that is easy to calculate if you remember an important rule:
This formula can be considered both as a definition and as one of its properties. In any case, if you are converting a logarithmic equation, you should know this formula just like you would know the log representation of any number.
Let's return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will be simply equal to lg 7. We have:
lg 56 = lg 7 − 3lg (x + 4)
Let's move lg 7 to the left, we get:
lg 56 − lg 7 = −3lg (x + 4)
We subtract the expressions on the left because they have the same base:
lg (56/7) = −3lg (x + 4)
Now let's take a closer look at the equation we got. It is practically the canonical form, but there is a factor −3 on the right. Let's add it to the right lg argument:
log 8 = log (x + 4) −3
Before us is the canonical form of the logarithmic equation, so we cross out the lg signs and equate the arguments:
(x + 4) −3 = 8
x + 4 = 0.5
That's all! We solved the second logarithmic equation. In this case, no additional checks are required, because in the original problem x was present in only one argument.
Let me list the key points of this lesson again.
The main formula that is taught in all the lessons on this page dedicated to solving logarithmic equations is the canonical form. And don’t be scared by the fact that most school textbooks teach you to solve such problems differently. This tool works very effectively and allows you to solve a much wider class of problems than the simplest ones that we studied at the very beginning of our lesson.
In addition, to solve logarithmic equations it will be useful to know the basic properties. Namely:
- The formula for moving to one base and the special case when we reverse log (this was very useful to us in the first problem);
- Formula for adding and subtracting powers from the logarithm sign. Here, many students get stuck and do not see that the degree taken out and introduced can itself contain log f (x). Nothing wrong with that. We can introduce one log according to the sign of the other and at the same time significantly simplify the solution of the problem, which is what we observe in the second case.
In conclusion, I would like to add that it is not necessary to check the domain of definition in each of these cases, because everywhere the variable x is present in only one sign of log, and at the same time is in its argument. As a consequence, all requirements of the scope are fulfilled automatically.
Problems with variable base
Today we will look at logarithmic equations, which for many students seem non-standard, if not completely unsolvable. It's about about expressions based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely through the canonical form.
First, let's remember how the simplest problems are solved, based on ordinary numbers. So, the simplest construction is called
log a f (x) = b
To solve such problems we can use the following formula:
b = log a a b
We rewrite our original expression and get:
log a f (x) = log a a b
Then we equate the arguments, i.e. we write:
f (x) = a b
Thus, we get rid of the log sign and solve the usual problem. In this case, the roots obtained from the solution will be the roots of the original logarithmic equation. In addition, a record when both the left and the right are in the same logarithm with the same base is precisely called the canonical form. It is to such a record that we will try to reduce today's designs. So, let's go.
First task:
log x − 2 (2x 2 − 13x + 18) = 1
Replace 1 with log x − 2 (x − 2) 1 . The degree that we observe in the argument is actually the number b that stood to the right of the equal sign. Thus, let's rewrite our expression. We get:
log x − 2 (2x 2 − 13x + 18) = log x − 2 (x − 2)
What do we see? Before us is the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:
2x 2 − 13x + 18 = x − 2
But the solution doesn't end there, because given equation not equivalent to the original one. After all, the resulting construction consists of functions that are defined on the entire number line, and our original logarithms are not defined everywhere and not always.
Therefore, we must write down the domain of definition separately. Let's not split hairs and first write down all the requirements:
First, the argument of each of the logarithms must be greater than 0:
2x 2 − 13x + 18 > 0
x − 2 > 0
Secondly, the base must not only be greater than 0, but also different from 1:
x − 2 ≠ 1
As a result, we get the system:
But don’t be alarmed: when processing logarithmic equations, such a system can be significantly simplified.
Judge for yourself: on the one hand, we are required that the quadratic function be greater than zero, and on the other hand, this quadratic function is equated to a certain linear expression, which is also required that it be greater than zero.
In this case, if we require that x − 2 > 0, then the requirement 2x 2 − 13x + 18 > 0 will automatically be satisfied. Therefore, we can safely cross out the inequality containing the quadratic function. Thus, the number of expressions contained in our system will be reduced to three.
Of course, we could just as well cross out linear inequality, that is, cross out x − 2 > 0 and demand that 2x 2 − 13x + 18 > 0. But you must agree that solving the simplest linear inequality is much faster and easier than quadratic, even if as a result of solving the entire this system we will get the same roots.
In general, try to optimize calculations whenever possible. And in the case of logarithmic equations, cross out the most difficult inequalities.
Let's rewrite our system:
Here is a system of three expressions, two of which we, in fact, have already dealt with. Let's write it down separately quadratic equation and let's solve it:
2x 2 − 14x + 20 = 0
x 2 − 7x + 10 = 0
Before us is a reduced quadratic trinomial and, therefore, we can use Vieta’s formulas. We get:
(x − 5)(x − 2) = 0
x 1 = 5
x 2 = 2
Now we return to our system and find that x = 2 does not suit us, because we are required that x be strictly greater than 2.
But x = 5 suits us perfectly: the number 5 is greater than 2, and at the same time 5 is not equal to 3. Therefore, the only solution to this system will be x = 5.
That's it, the problem is solved, including taking into account the ODZ. Let's move on to the second equation. More interesting and informative calculations await us here:
The first step: like last time, we bring this whole matter to canonical form. To do this, we can write the number 9 as follows:
You don’t have to touch the base with the root, but it’s better to transform the argument. Let's move from the root to the power with a rational exponent. Let's write down:
Let me not rewrite our entire large logarithmic equation, but just immediately equate the arguments:
x 3 + 10x 2 + 31x + 30 = x 3 + 9x 2 + 27x + 27
x 2 + 4x + 3 = 0
Before us is a newly reduced quadratic trinomial, let’s use Vieta’s formulas and write:
(x + 3)(x + 1) = 0
x 1 = −3
x 2 = −1
So, we got the roots, but no one guaranteed us that they would fit the original logarithmic equation. After all, the log signs impose additional restrictions (here we should have written down the system, but due to the cumbersome nature of the entire structure, I decided to calculate the domain of definition separately).
First of all, remember that the arguments must be greater than 0, namely:
These are the requirements imposed by the scope of definition.
Let us immediately note that since we equate the first two expressions of the system to each other, we can cross out any of them. Let's cross out the first one because it looks more threatening than the second one.
In addition, note that the solution to the second and third inequalities will be the same sets (the cube of some number is greater than zero, if this number itself is greater than zero; similarly, with a root of the third degree - these inequalities are completely analogous, so we can cross it out).
But with the third inequality this will not work. Let's get rid of the radical sign on the left by raising both parts to a cube. We get:
So we get the following requirements:
− 2 ≠ x > −3
Which of our roots: x 1 = −3 or x 2 = −1 meets these requirements? Obviously, only x = −1, because x = −3 does not satisfy the first inequality (since our inequality is strict). So, returning to our problem, we get one root: x = −1. That's it, problem solved.
Once again, the key points of this task:
- Feel free to apply and solve logarithmic equations using canonical form. Students who make such a notation, rather than moving directly from the original problem to a construction like log a f (x) = b, make much fewer errors than those who rush somewhere, skipping intermediate steps of calculations;
- As soon as the logarithm appears variable base, the task ceases to be the simplest. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but they also must not be equal to 1.
The final requirements can be applied to the final answers in different ways. For example, you can solve an entire system containing all the requirements for the domain of definition. On the other hand, you can first solve the problem itself, and then remember the domain of definition, separately work it out in the form of a system and apply it to the obtained roots.
Which method to choose when solving a particular logarithmic equation is up to you. In any case, the answer will be the same.
We continue to study logarithms. In this article we will talk about calculating logarithms, this process is called logarithm. First we will understand the calculation of logarithms by definition. Next, let's look at how the values of logarithms are found using their properties. After this, we will focus on calculating logarithms through the initially specified values of other logarithms. Finally, let's learn how to use logarithm tables. The entire theory is provided with examples with detailed solutions.
Page navigation.
Calculating logarithms by definition
In the simplest cases it is possible to perform quite quickly and easily finding the logarithm by definition. Let's take a closer look at how this process happens.
Its essence is to represent the number b in the form a c, from which, by the definition of a logarithm, the number c is the value of the logarithm. That is, by definition, the following chain of equalities corresponds to finding the logarithm: log a b=log a a c =c.
So, calculating a logarithm by definition comes down to finding a number c such that a c = b, and the number c itself is the desired value of the logarithm.
Taking into account the information in the previous paragraphs, when the number under the logarithm sign is given by a certain power of the logarithm base, you can immediately indicate what the logarithm is equal to - it is equal to the exponent. Let's show solutions to examples.
Example.
Find log 2 2 −3, and also calculate the natural logarithm of the number e 5,3.
Solution.
The definition of the logarithm allows us to immediately say that log 2 2 −3 =−3. Indeed, the number under the logarithm sign is equal to base 2 to the −3 power.
Similarly, we find the second logarithm: lne 5.3 =5.3.
Answer:
log 2 2 −3 =−3 and lne 5,3 =5,3.
If the number b under the logarithm sign is not specified as a power of the base of the logarithm, then you need to carefully look to see if it is possible to come up with a representation of the number b in the form a c . Often this representation is quite obvious, especially when the number under the logarithm sign is equal to the base to the power of 1, or 2, or 3, ...
Example.
Calculate the logarithms log 5 25 , and .
Solution.
It is easy to see that 25=5 2, this allows you to calculate the first logarithm: log 5 25=log 5 5 2 =2.
Let's move on to calculating the second logarithm. The number can be represented as a power of 7: 
(see if necessary). Hence, 
.
Let's rewrite the third logarithm in the following form. Now you can see that 
, from which we conclude that 
. Therefore, by the definition of logarithm 
.
Briefly, the solution could be written as follows: .
Answer:
log 5 25=2 , 
And .
When under the logarithm sign there is a sufficiently large natural number, then it wouldn’t hurt to decompose it into prime factors. It often helps to represent such a number as some power of the base of the logarithm, and therefore calculate this logarithm by definition.
Example.
Find the value of the logarithm.
Solution.
Some properties of logarithms allow you to immediately specify the value of logarithms. These properties include the property of the logarithm of one and the property of the logarithm of a number equal to the base: log 1 1=log a a 0 =0 and log a a=log a a 1 =1. That is, when under the sign of the logarithm there is a number 1 or a number a equal to the base of the logarithm, then in these cases the logarithms are equal to 0 and 1, respectively.
Example.
What are logarithms and log10 equal to?
Solution.
Since , then from the definition of logarithm it follows 
.
In the second example, the number 10 under the logarithm sign coincides with its base, so the decimal logarithm of ten is equal to one, that is, lg10=lg10 1 =1.
Answer:
AND lg10=1 .
Note that the calculation of logarithms by definition (which we discussed in the previous paragraph) implies the use of the equality log a a p =p, which is one of the properties of logarithms.
In practice, when a number under the logarithm sign and the base of the logarithm are easily represented as a power of a certain number, it is very convenient to use the formula 
, which corresponds to one of the properties of logarithms. Let's look at an example of finding a logarithm that illustrates the use of this formula.
Example.
Calculate the logarithm.
Solution.
Answer:
.
Properties of logarithms not mentioned above are also used in calculations, but we will talk about this in the following paragraphs.
Finding logarithms through other known logarithms
The information in this paragraph continues the topic of using the properties of logarithms when calculating them. But here the main difference is that the properties of logarithms are used to express the original logarithm in terms of another logarithm, the value of which is known. Let's give an example for clarification. Let's say we know that log 2 3≈1.584963, then we can find, for example, log 2 6 by doing a little transformation using the properties of the logarithm: log 2 6=log 2 (2 3)=log 2 2+log 2 3≈ 1+1,584963=2,584963 .
In the above example, it was enough for us to use the property of the logarithm of a product. However, much more often it is necessary to use a wider arsenal of properties of logarithms in order to calculate the original logarithm through the given ones.
Example.
Calculate the logarithm of 27 to base 60 if you know that log 60 2=a and log 60 5=b.
Solution.
So we need to find log 60 27 . It is easy to see that 27 = 3 3 , and the original logarithm, due to the property of the logarithm of the power, can be rewritten as 3·log 60 3 .
Now let's see how to express log 60 3 in terms of known logarithms. The property of the logarithm of a number equal to the base allows us to write the equality log 60 60=1. On the other hand, log 60 60=log60(2 2 3 5)= log 60 2 2 +log 60 3+log 60 5= 2·log 60 2+log 60 3+log 60 5 . Thus, 2 log 60 2+log 60 3+log 60 5=1. Hence, log 60 3=1−2·log 60 2−log 60 5=1−2·a−b.
Finally, we calculate the original logarithm: log 60 27=3 log 60 3= 3·(1−2·a−b)=3−6·a−3·b.
Answer:
log 60 27=3·(1−2·a−b)=3−6·a−3·b.
Separately, it is worth mentioning the meaning of the formula for transition to a new base of the logarithm of the form 
. It allows you to move from logarithms with any base to logarithms with a specific base, the values of which are known or it is possible to find them. Usually, from the original logarithm, using the transition formula, they move to logarithms in one of the bases 2, e or 10, since for these bases there are tables of logarithms that allow their values to be calculated with a certain degree of accuracy. In the next paragraph we will show how this is done.
Logarithm tables and their uses
For approximate calculation of logarithm values can be used logarithm tables. The most commonly used base 2 logarithm table is the table natural logarithms and a table of decimal logarithms. When working in decimal system For calculus, it is convenient to use a table of logarithms based on base ten. With its help we will learn to find the values of logarithms.
The presented table allows you to find the values of the decimal logarithms of numbers from 1,000 to 9,999 (with three decimal places) with an accuracy of one ten-thousandth. We will analyze the principle of finding the value of a logarithm using a table of decimal logarithms into specific example– it’s clearer that way. Let's find log1.256.
In the left column of the table of decimal logarithms we find the first two digits of the number 1.256, that is, we find 1.2 (this number is circled in blue for clarity). The third digit of the number 1.256 (digit 5) is found in the first or last line to the left of the double line (this number is circled in red). The fourth digit of the original number 1.256 (digit 6) is found in the first or last line to the right of the double line (this number is circled with a green line). Now we find the numbers in the cells of the table of logarithms at the intersection of the marked row and marked columns (these numbers are highlighted orange). The sum of the marked numbers gives the desired value of the decimal logarithm accurate to the fourth decimal place, that is, log1.236≈0.0969+0.0021=0.0990.
Is it possible, using the table above, to find the values of decimal logarithms of numbers that have more than three digits after the decimal point, as well as those that go beyond the range from 1 to 9.999? Yes, you can. Let's show how this is done with an example.
Let's calculate lg102.76332. First you need to write down number in standard form : 102.76332=1.0276332·10 2. After this, the mantissa should be rounded to the third decimal place, we have 1.0276332 10 2 ≈1.028 10 2, while the original decimal logarithm is approximately equal to the logarithm of the resulting number, that is, we take log102.76332≈lg1.028·10 2. Now we apply the properties of the logarithm: lg1.028·10 2 =lg1.028+lg10 2 =lg1.028+2. Finally, we find the value of the logarithm lg1.028 from the table of decimal logarithms lg1.028≈0.0086+0.0034=0.012. As a result, the entire process of calculating the logarithm looks like this: log102.76332=log1.0276332 10 2 ≈lg1.028 10 2 = log1.028+lg10 2 =log1.028+2≈0.012+2=2.012.
In conclusion, it is worth noting that using a table of decimal logarithms you can calculate the approximate value of any logarithm. To do this, it is enough to use the transition formula to go to decimal logarithms, find their values in the table, and perform the remaining calculations.
For example, let's calculate log 2 3 . According to the formula for transition to a new base of the logarithm, we have . From the table of decimal logarithms we find log3≈0.4771 and log2≈0.3010. Thus, 
.
Bibliography.
- Kolmogorov A.N., Abramov A.M., Dudnitsyn Yu.P. and others. Algebra and the beginnings of analysis: Textbook for grades 10 - 11 of general education institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for those entering technical schools).
Logarithmic expressions, solving examples. In this article we will look at problems related to solving logarithms. The tasks ask the question of finding the meaning of an expression. It should be noted that the concept of logarithm is used in many tasks and understanding its meaning is extremely important. As for the Unified State Exam, the logarithm is used when solving equations, in applied problems, and also in tasks related to the study of functions.
Let us give examples to understand the very meaning of the logarithm:
Basics logarithmic identity:
Properties of logarithms that must always be remembered:
*The logarithm of the product is equal to the sum of the logarithms of the factors.
* * *
*The logarithm of a quotient (fraction) is equal to the difference between the logarithms of the factors.
* * *
*The logarithm of an exponent is equal to the product of the exponent and the logarithm of its base.
* * *
*Transition to a new foundation
* * *
More properties:
* * *
The calculation of logarithms is closely related to the use of properties of exponents.
Let's list some of them:
The essence of this property is that when the numerator is transferred to the denominator and vice versa, the sign of the exponent changes to the opposite. For example:
A corollary from this property:
* * *
When raising a power to a power, the base remains the same, but the exponents are multiplied.
* * *
As you have seen, the concept of a logarithm itself is simple. The main thing is what is needed good practice, which gives a certain skill. Of course, knowledge of formulas is required. If the skill in converting elementary logarithms has not been developed, then when solving simple tasks It's easy to make a mistake.
Practice, solve the simplest examples from the mathematics course first, then move on to more complex ones. In the future, I will definitely show how “scary” logarithms are solved; they won’t appear on the Unified State Examination, but they are of interest, don’t miss them!
That's all! Good luck to you!
Sincerely, Alexander Krutitskikh
P.S: I would be grateful if you tell me about the site on social networks.
Logarithms, like any numbers, can be added, subtracted and transformed in every way. But since logarithms are not exactly ordinary numbers, there are rules here, which are called main properties.
You definitely need to know these rules - without them not a single serious problem can be solved. logarithmic problem. In addition, there are very few of them - you can learn everything in one day. So let's get started.
Adding and subtracting logarithms
Consider two logarithms with the same bases: log a x and log a y. Then they can be added and subtracted, and:
- log a x+ log a y=log a (x · y);
- log a x− log a y=log a (x : y).
So, the sum of logarithms is equal to the logarithm of the product, and the difference is equal to the logarithm of the quotient. Please note: the key point here is identical grounds. If the reasons are different, these rules do not work!
These formulas will help you calculate logarithmic expression even when its individual parts are not counted (see lesson “What is a logarithm”). Take a look at the examples and see:
Log 6 4 + log 6 9.
Since logarithms have the same bases, we use the sum formula:
log 6 4 + log 6 9 = log 6 (4 9) = log 6 36 = 2.
Task. Find the value of the expression: log 2 48 − log 2 3.
The bases are the same, we use the difference formula:
log 2 48 − log 2 3 = log 2 (48: 3) = log 2 16 = 4.
Task. Find the value of the expression: log 3 135 − log 3 5.
Again the bases are the same, so we have:
log 3 135 − log 3 5 = log 3 (135: 5) = log 3 27 = 3.
As you can see, the original expressions are made up of “bad” logarithms, which are not calculated separately. But after the transformations, completely normal numbers are obtained. Many are built on this fact test papers. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.
Extracting the exponent from the logarithm
Now let's complicate the task a little. What if the base or argument of a logarithm is a power? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
It is easy to see that the last rule follows the first two. But it’s better to remember it anyway - in some cases it will significantly reduce the amount of calculations.
Of course, all these rules make sense if the ODZ of the logarithm is observed: a > 0, a ≠ 1, x> 0. And one more thing: learn to apply all formulas not only from left to right, but also vice versa, i.e. You can enter the numbers before the logarithm sign into the logarithm itself. This is what is most often required.
Task. Find the value of the expression: log 7 49 6 .
Let's get rid of the degree in the argument using the first formula:
log 7 49 6 = 6 log 7 49 = 6 2 = 12
Task. Find the meaning of the expression:
[Caption for the picture]
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2. We have:
[Caption for the picture] I think the last example requires some clarification. Where have logarithms gone? Until the very last moment we work only with the denominator. We presented the base and argument of the logarithm standing there in the form of powers and took out the exponents - we got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator contain the same number: log 2 7. Since log 2 7 ≠ 0, we can reduce the fraction - 2/4 will remain in the denominator. According to the rules of arithmetic, the four can be transferred to the numerator, which is what was done. The result was the answer: 2.
Transition to a new foundation
Speaking about the rules for adding and subtracting logarithms, I specifically emphasized that they only work with the same bases. What if the reasons are different? What if they are not exact powers of the same number?
Formulas for transition to a new foundation come to the rescue. Let us formulate them in the form of a theorem:
Let the logarithm log be given a x. Then for any number c such that c> 0 and c≠ 1, the equality is true:
[Caption for the picture]
In particular, if we put c = x, we get:
[Caption for the picture]
From the second formula it follows that the base and argument of the logarithm can be swapped, but in this case the entire expression is “turned over”, i.e. the logarithm appears in the denominator.
These formulas are rarely found in ordinary numerical expressions. It is possible to evaluate how convenient they are only when solving logarithmic equations and inequalities.
However, there are problems that cannot be solved at all except by moving to a new foundation. Let's look at a couple of these:
Task. Find the value of the expression: log 5 16 log 2 25.
Note that the arguments of both logarithms contain exact powers. Let's take out the indicators: log 5 16 = log 5 2 4 = 4log 5 2; log 2 25 = log 2 5 2 = 2log 2 5;
Now let’s “reverse” the second logarithm:
[Caption for the picture]Since the product does not change when rearranging factors, we calmly multiplied four and two, and then dealt with logarithms.
Task. Find the value of the expression: log 9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write this down and get rid of the indicators:
[Caption for the picture]Now let's get rid of the decimal logarithm by moving to a new base:
[Caption for the picture]Basic logarithmic identity
Often in the solution process it is necessary to represent a number as a logarithm to a given base. In this case, the following formulas will help us:
In the first case, the number n becomes an indicator of the degree standing in the argument. Number n can be absolutely anything, because it’s just a logarithm value.
The second formula is actually a paraphrased definition. That’s what it’s called: the basic logarithmic identity.
In fact, what will happen if the number b raise to such a power that the number b to this power gives the number a? That's right: you get this same number a. Read this paragraph carefully again - many people get stuck on it.
Like formulas for moving to a new base, the basic logarithmic identity is sometimes the only possible solution.
Task. Find the meaning of the expression:
[Caption for the picture]
Note that log 25 64 = log 5 8 - simply took the square from the base and argument of the logarithm. Considering the rules for multiplying powers with the same basis, we get:
[Caption for the picture]If anyone doesn't know, this was a real task from the Unified State Exam :)
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that can hardly be called properties - rather, they are consequences of the definition of the logarithm. They constantly appear in problems and, surprisingly, create problems even for “advanced” students.
- log a a= 1 is a logarithmic unit. Remember once and for all: logarithm to any base a from this very base is equal to one.
- log a 1 = 0 is logarithmic zero. Base a can be anything, but if the argument contains one, the logarithm is equal to zero! Because a 0 = 1 is a direct consequence of the definition.
That's all the properties. Be sure to practice putting them into practice! Download the cheat sheet at the beginning of the lesson, print it out, and solve the problems.
Today we will talk about logarithmic formulas and we will give indicative solution examples.
They themselves imply solution patterns according to the basic properties of logarithms. Before applying logarithm formulas to solve, let us remind you of all the properties:
Now, based on these formulas (properties), we will show examples of solving logarithms.
Examples of solving logarithms based on formulas.
Logarithm positive number b to base a (denoted by log a b) is an exponent to which a must be raised to obtain b, with b > 0, a > 0, and 1.
According to the definition, log a b = x, which is equivalent to a x = b, therefore log a a x = x.
Logarithms, examples:
log 2 8 = 3, because 2 3 = 8
log 7 49 = 2, because 7 2 = 49
log 5 1/5 = -1, because 5 -1 = 1/5
Decimal logarithm- this is an ordinary logarithm, the base of which is 10. It is denoted as lg.
log 10 100 = 2, because 10 2 = 100
Natural logarithm- also the usual logarithm logarithm, but with the base e (e = 2.71828... - irrational number). Denoted as ln.
It is advisable to memorize the formulas or properties of logarithms, because we will need them later when solving logarithms, logarithmic equations and inequalities. Let's work through each formula again with examples.
- Basic logarithmic identity
a log a b = b8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9
- The logarithm of the product is equal to the sum of the logarithms
log a (bc) = log a b + log a clog 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4
- The logarithm of the quotient is equal to the difference of the logarithms
log a (b/c) = log a b - log a c9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81
- Properties of the power of a logarithmic number and the base of the logarithm
Exponent of the logarithmic number log a b m = mlog a b
Exponent of the base of the logarithm log a n b =1/n*log a b
log a n b m = m/n*log a b,
if m = n, we get log a n b n = log a b
log 4 9 = log 2 2 3 2 = log 2 3
- Transition to a new foundation
log a b = log c b/log c a,if c = b, we get log b b = 1
then log a b = 1/log b a
log 0.8 3*log 3 1.25 = log 0.8 3*log 0.8 1.25/log 0.8 3 = log 0.8 1.25 = log 4/5 5/4 = -1
As you can see, the formulas for logarithms are not as complicated as they seem. Now, having looked at examples of solving logarithms, we can move on to logarithmic equations. We will look at examples of solving logarithmic equations in more detail in the article: "". Do not miss!
If you still have questions about the solution, write them in the comments to the article.
Note: we decided to get a different class of education and study abroad as an option.
