How to expand the logarithm. Protection of personal information
Problem B7 gives some expression that needs to be simplified. The result should be a regular number that can be written down on your answer sheet. All expressions are conventionally divided into three types:
- Logarithmic,
- Indicative,
- Combined.
Exponential and logarithmic expressions in their pure form are practically never found. However, knowing how they are calculated is absolutely necessary.
In general, problem B7 is solved quite simply and is quite within the capabilities of the average graduate. The lack of clear algorithms is compensated for by its standardization and monotony. You can learn to solve such problems simply by large quantity training.
Logarithmic Expressions
The vast majority of B7 problems involve logarithms in one form or another. This topic is traditionally considered difficult, since its study usually occurs in the 11th grade - the era of mass preparation for final exams. As a result, many graduates have a very vague understanding of logarithms.
But in this task no one requires deep theoretical knowledge. We will encounter only the simplest expressions that require simple reasoning and can easily be mastered independently. Below are the basic formulas you need to know to cope with logarithms:
In addition, you must be able to replace roots and fractions with powers with a rational exponent, otherwise in some expressions there will simply be nothing to take out from under the logarithm sign. Replacement formulas:
Task. Find the meaning of expressions:
log 6 270 − log 6 7.5
log 5 775 − log 5 6.2
The first two expressions are converted as the difference of logarithms:
log 6 270 − log 6 7.5 = log 6 (270: 7.5) = log 6 36 = 2;
log 5 775 − log 5 6.2 = log 5 (775: 6.2) = log 5 125 = 3.
To calculate the third expression, you will have to isolate powers - both in the base and in the argument. First, let's find the internal logarithm:
Then - external:
Constructions of the form log a log b x seem complex and misunderstood to many. Meanwhile, this is just a logarithm of the logarithm, i.e. log a (log b x ). First, the internal logarithm is calculated (put log b x = c), and then the external one: log a c.
Demonstrative Expressions
We will call an exponential expression any construction of the form a k, where the numbers a and k are arbitrary constants, and a > 0. Methods for working with such expressions are quite simple and are discussed in 8th grade algebra lessons.
Below are the basic formulas that you definitely need to know. The application of these formulas in practice, as a rule, does not cause problems.
- a n · a m = a n + m ;
- a n / a m = a n − m ;
- (a n ) m = a n · m ;
- (a · b ) n = a n · b n ;
- (a : b ) n = a n : b n .
If met complex expression with degrees, and it is not clear how to approach it, they use a universal technique - decomposition into prime factors. As a result big numbers in the bases of degrees are replaced by simple and understandable elements. Then all that remains is to apply the above formulas - and the problem will be solved.
Task. Find the values of the expressions: 7 9 · 3 11: 21 8, 24 7: 3 6: 16 5, 30 6: 6 5: 25 2.
Solution. Let's decompose all the bases of powers into simple factors:
7 9 3 11: 21 8 = 7 9 3 11: (7 3) 8 = 7 9 3 11: (7 8 3 8) = 7 9 3 11: 7 8: 3 8 = 7 3 3 = 189.
24 7: 3 6: 16 5 = (3 2 3) 7: 3 6: (2 4) 5 = 3 7 2 21: 3 6: 2 20 = 3 2 = 6.
30 6: 6 5: 25 2 = (5 3 2) 6: (3 2) 5: (5 2) 2 = 5 6 3 6 2 6: 3 5: 2 5: 5 4 = 5 2 3 2 = 150.
Combined tasks
If you know the formulas, then all exponential and logarithmic expressions can be solved literally in one line. However, in Problem B7 powers and logarithms can be combined to form quite strong combinations.
The basic properties of the natural logarithm, graph, domain of definition, set of values, basic formulas, derivative, integral, expansion in power series and representation of the function ln x using complex numbers.
Definition
Natural logarithm is the function y = ln x, the inverse of the exponential, x = e y, and is the logarithm to the base of the number e: ln x = log e x.
The natural logarithm is widely used in mathematics because its derivative has the simplest form: (ln x)′ = 1/ x.
Based definitions, the base of the natural logarithm is the number e:
e ≅ 2.718281828459045...;
.
Graph of the function y = ln x.
Graph of natural logarithm (functions y = ln x) is obtained from the exponential graph by mirror reflection relative to the straight line y = x.
The natural logarithm is defined at positive values variable x. It increases monotonically in its domain of definition.
At x → 0 the limit of the natural logarithm is minus infinity (-∞).
As x → + ∞, the limit of the natural logarithm is plus infinity (+ ∞). For large x, the logarithm increases quite slowly. Any power function x a with a positive exponent a grows faster than the logarithm.
Properties of the natural logarithm
Domain of definition, set of values, extrema, increase, decrease
The natural logarithm is a monotonically increasing function, so it has no extrema. The main properties of the natural logarithm are presented in the table.
ln x values
ln 1 = 0
Basic formulas for natural logarithms
Formulas following from the definition of the inverse function:
The main property of logarithms and its consequences
Base replacement formula
Any logarithm can be expressed in terms of natural logarithms using the base substitution formula:
Proofs of these formulas are presented in the section "Logarithm".
Inverse function
The inverse of the natural logarithm is the exponent.
If , then
If, then.
Derivative ln x
Derivative of the natural logarithm:
.
Derivative of the natural logarithm of modulus x:
.
Derivative of nth order:
.
Deriving formulas > > >
Integral
The integral is calculated by integration by parts:
.
So,
Expressions using complex numbers
Consider the function of the complex variable z:
.
Let's express the complex variable z via module r and argument φ
:
.
Using the properties of the logarithm, we have:
.
Or
.
The argument φ is not uniquely defined. If you put
, where n is an integer,
it will be the same number for different n.
Therefore, the natural logarithm, as a function of a complex variable, is not a single-valued function.
Power series expansion
When the expansion takes place:
References:
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.
Tasks whose solution is transformation logarithmic expressions , are quite common on the Unified State Examination.
In order to successfully cope with them with minimal time, in addition to the basic logarithmic identities, you need to know and correctly use some more formulas.
This is: a log a b = b, where a, b > 0, a ≠ 1 (It follows directly from the definition of the logarithm).
log a b = log c b / log c a or log a b = 1/log b a
where a, b, c > 0; a, c ≠ 1.
log a m b n = (m/n) log |a| |b|
where a, b > 0, a ≠ 1, m, n Є R, n ≠ 0.
a log c b = b log c a
where a, b, c > 0 and a, b, c ≠ 1
To show the validity of the fourth equality, let us take the logarithm of the left and right side based on a. We get log a (a log with b) = log a (b log with a) or log with b = log with a · log a b; log c b = log c a · (log c b / log c a); log with b = log with b.
We have proven the equality of logarithms, which means that the expressions under the logarithms are also equal. Formula 4 has been proven.
Example 1.
Calculate 81 log 27 5 log 5 4 .
Solution.
81 = 3 4 , 27 = 3 3 .
log 27 5 = 1/3 log 3 5, log 5 4 = log 3 4 / log 3 5. Therefore,
log 27 5 log 5 4 = 1/3 log 3 5 (log 3 4 / log 3 5) = 1/3 log 3 4.
Then 81 log 27 5 log 5 4 = (3 4) 1/3 log 3 4 = (3 log 3 4) 4/3 = (4) 4/3 = 4 3 √4.
You can complete the following task yourself.
Calculate (8 log 2 3 + 3 1/ log 2 3) - log 0.2 5.
As a hint, 0.2 = 1/5 = 5 -1 ; log 0.2 5 = -1.
Answer: 5.
Example 2.
Calculate (√11) log √3 9- log 121 81 .
Solution.
Let's change the expressions: 9 = 3 2, √3 = 3 1/2, log √3 9 = 4,
121 = 11 2, 81 = 3 4, log 121 81 = 2 log 11 3 (formula 3 was used).
Then (√11) log √3 9- log 121 81 = (11 1/2) 4-2 log 11 3 = (11) 2- log 11 3 = 11 2 / (11) log 11 3 = 11 2 / ( 11 log 11 3) = 121/3.
Example 3.
Calculate log 2 24 / log 96 2 - log 2 192 / log 12 2.
Solution.
We replace the logarithms contained in the example with logarithms with base 2.
log 96 2 = 1/log 2 96 = 1/log 2 (2 5 3) = 1/(log 2 2 5 + log 2 3) = 1/(5 + log 2 3);
log 2 192 = log 2 (2 6 3) = (log 2 2 6 + log 2 3) = (6 + log 2 3);
log 2 24 = log 2 (2 3 3) = (log 2 2 3 + log 2 3) = (3 + log 2 3);
log 12 2 = 1/log 2 12 = 1/log 2 (2 2 3) = 1/(log 2 2 2 + log 2 3) = 1/(2 + log 2 3).
Then log 2 24 / log 96 2 – log 2 192 / log 12 2 = (3 + log 2 3) / (1/(5 + log 2 3)) – ((6 + log 2 3) / (1/( 2 + log 2 3)) =
= (3 + log 2 3) · (5 + log 2 3) – (6 + log 2 3)(2 + log 2 3).
After opening the parentheses and bringing similar terms, we get the number 3. (When simplifying the expression, we can denote log 2 3 by n and simplify the expression
(3 + n) · (5 + n) – (6 + n)(2 + n)).
Answer: 3.
You can complete the following task yourself:
Calculate (log 3 4 + log 4 3 + 2) log 3 16 log 2 144 3.
Here it is necessary to make the transition to base 3 logarithms and factorization of large numbers into prime factors.
Answer:1/2
Example 4.
Given three numbers A = 1/(log 3 0.5), B = 1/(log 0.5 3), C = log 0.5 12 – log 0.5 3. Arrange them in ascending order.
Solution.
Let's transform the numbers A = 1/(log 3 0.5) = log 0.5 3; C = log 0.5 12 – log 0.5 3 = log 0.5 12/3 = log 0.5 4 = -2.
Let's compare them
log 0.5 3 > log 0.5 4 = -2 and log 0.5 3< -1 = log 0,5 2, так как функция у = log 0,5 х – убывающая.
Or 2< log 0,5 3 < -1. Тогда -1 < 1/(log 0,5 3) < -1/2.
Answer. Therefore, the order of placing the numbers is: C; A; IN.
Example 5.
How many integers are in the interval (log 3 1 / 16 ; log 2 6 48).
Solution.
Let us determine between which powers of the number 3 the number 1/16 is located. We get 1/27< 1 / 16 < 1 / 9 .
Since the function y = log 3 x is increasing, then log 3 (1 / 27)< log 3 (1 / 16) < log 3 (1 / 9); -3 < log 3 (1 / 16) < -2.
log 6 48 = log 6 (36 4 / 3) = log 6 36 + log 6 (4 / 3) = 2 + log 6 (4 / 3). Let's compare log 6 (4 / 3) and 1 / 5. And for this we compare the numbers 4/3 and 6 1/5. Let's raise both numbers to the 5th power. We get (4 / 3) 5 = 1024 / 243 = 4 52 / 243< 6. Следовательно,
log 6 (4 / 3)< 1 / 5 . 2 < log 6 48 < 2 1 / 5 . Числа, входящие в двойное неравенство, положительные. Их можно возводить в квадрат. Знаки неравенства при этом не изменятся. Тогда 4 < log 6 2 48 < 4 21 / 25.
Therefore, the interval (log 3 1 / 16 ; log 6 48) includes the interval [-2; 4] and the integers -2 are placed on it; -1; 0; 1; 2; 3; 4.
Answer: 7 integers.
Example 6.
Calculate 3 lglg 2/ lg 3 - lg20.
Solution.
3 lg lg 2/ lg 3 = (3 1/ lg3) lg lg 2 = (3 lо g 3 10) lg lg 2 = 10 lg lg 2 = lg2.
Then 3 lglg2/lg3 - lg 20 = lg 2 – lg 20 = lg 0.1 = -1.
Answer: -1.
Example 7.
It is known that log 2 (√3 + 1) + log 2 (√6 – 2) = A. Find log 2 (√3 –1) + log 2 (√6 + 2).
Solution.
Numbers (√3 + 1) and (√3 – 1); (√6 – 2) and (√6 + 2) are conjugate.
Let us carry out the following transformation of expressions
√3 – 1 = (√3 – 1) · (√3 + 1)) / (√3 + 1) = 2/(√3 + 1);
√6 + 2 = (√6 + 2) · (√6 – 2)) / (√6 – 2) = 2/(√6 – 2).
Then log 2 (√3 – 1) + log 2 (√6 + 2) = log 2 (2/(√3 + 1)) + log 2 (2/(√6 – 2)) =
Log 2 2 – log 2 (√3 + 1) + log 2 2 – log 2 (√6 – 2) = 1 – log 2 (√3 + 1) + 1 – log 2 (√6 – 2) =
2 – log 2 (√3 + 1) – log 2 (√6 – 2) = 2 – A.
Answer: 2 – A.
Example 8.
Simplify and find the approximate value of the expression (log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9.
Solution.
We reduce all logarithms to common ground 10.
(log 3 2 log 4 3 log 5 4 log 6 5 ... log 10 9 = (lg 2 / lg 3) (lg 3 / lg 4) (lg 4 / lg 5) (lg 5 / lg 6) · … · (lg 8 / lg 9) · lg 9 = lg 2 ≈ 0.3010 (The approximate value of lg 2 can be found using a table, slide rule or calculator).
Answer: 0.3010.
Example 9.
Calculate log a 2 b 3 √(a 11 b -3) if log √ a b 3 = 1. (In this example, a 2 b 3 is the base of the logarithm).
Solution.
If log √ a b 3 = 1, then 3/(0.5 log a b = 1. And log a b = 1/6.
Then log a 2 b 3√(a 11 b -3) = 1/2 log a 2 b 3 (a 11 b -3) = log a (a 11 b -3) / (2log a (a 2 b 3) ) = (log a a 11 + log a b -3) / (2(log a a 2 + log a b 3)) = (11 – 3log a b) / (2(2 + 3log a b)) Considering that that log a b = 1/6 we get (11 – 3 1 / 6) / (2(2 + 3 1 / 6)) = 10.5/5 = 2.1.
Answer: 2.1.
You can complete the following task yourself:
Calculate log √3 6 √2.1 if log 0.7 27 = a.
Answer: (3 + a) / (3a).
Example 10.
Calculate 6.5 4/ log 3 169 · 3 1/ log 4 13 + log125.
Solution.
6.5 4/ log 3 169 · 3 1/ log 4 13 + log 125 = (13/2) 4/2 log 3 13 · 3 2/ log 2 13 + 2log 5 5 3 = (13/2) 2 log 13 3 3 2 log 13 2 + 6 = (13 log 13 3 / 2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3/2 log 13 3) 2 (3 log 13 2) 2 + 6 = (3 2 /(2 log 13 3) 2) · (2 log 13 3) 2 + 6.
(2 log 13 3 = 3 log 13 2 (formula 4))
We get 9 + 6 = 15.
Answer: 15.
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Let's start with properties of the logarithm of one. Its formulation is as follows: the logarithm of unity is equal to zero, that is, log a 1=0 for any a>0, a≠1. The proof is not difficult: since a 0 =1 for any a satisfying the above conditions a>0 and a≠1, then the equality log a 1=0 to be proved follows immediately from the definition of the logarithm.
Let us give examples of the application of the considered property: log 3 1=0, log1=0 and .
Let's move on to the next property: the logarithm of a number equal to the base is equal to one, that is, log a a=1 for a>0, a≠1. Indeed, since a 1 =a for any a, then by definition of the logarithm log a a=1.
Examples of using this property of logarithms are the equalities log 5 5=1, log 5.6 5.6 and lne=1.
For example, log 2 2 7 =7, log10 -4 =-4 and 
.
Logarithm of the product of two positive numbers x and y is equal to the product of the logarithms of these numbers: log a (x y)=log a x+log a y, a>0 , a≠1 . Let us prove the property of the logarithm of a product. Due to the properties of the degree a log a x+log a y =a log a x ·a log a y, and since by the main logarithmic identity a log a x =x and a log a y =y, then a log a x ·a log a y =x·y. Thus, a log a x+log a y =x·y, from which, by the definition of a logarithm, the equality being proved follows.
Let's show examples of using the property of the logarithm of a product: log 5 (2 3)=log 5 2+log 5 3 and 
.
The property of the logarithm of a product can be generalized to the product of a finite number n of positive numbers x 1 , x 2 , …, x n as log a (x 1 ·x 2 ·…·x n)= log a x 1 +log a x 2 +…+log a x n . This equality can be proven without problems.
For example, the natural logarithm of a product can be replaced by the sum of three natural logarithms numbers 4 , e , and .
Logarithm of the quotient of two positive numbers x and y is equal to the difference between the logarithms of these numbers. The property of the logarithm of a quotient corresponds to a formula of the form , where a>0, a≠1, x and y are some positive numbers. The validity of this formula is proven as well as the formula for the logarithm of a product: since 
, then by definition of a logarithm.
Here is an example of using this property of the logarithm: 
.
Let's move on to property of the logarithm of the power. The logarithm of a degree is equal to the product of the exponent and the logarithm of the modulus of the base of this degree. Let us write this property of the logarithm of a power as a formula: log a b p =p·log a |b|, where a>0, a≠1, b and p are numbers such that the degree b p makes sense and b p >0.
First we prove this property for positive b. Basics logarithmic identity allows us to represent the number b as a log a b , then b p =(a log a b) p , and the resulting expression, due to the property of power, is equal to a p·log a b . So we come to the equality b p =a p·log a b, from which, by the definition of a logarithm, we conclude that log a b p =p·log a b.
It remains to prove this property for negative b. Here we note that the expression log a b p for negative b makes sense only for even exponents p (since the value of the degree b p must be greater than zero, otherwise the logarithm will not make sense), and in this case b p =|b| p. Then b p =|b| p =(a log a |b|) p =a p·log a |b|, from where log a b p =p·log a |b| .
For example, 
and ln(-3) 4 =4·ln|-3|=4·ln3 .
It follows from the previous property property of the logarithm from the root: the logarithm of the nth root is equal to the product of the fraction 1/n by the logarithm of the radical expression, that is, 
, where a>0, a≠1, n – natural number, greater than one, b>0.
The proof is based on the equality (see), which is valid for any positive b, and the property of the logarithm of the power: 
.
Here is an example of using this property: 
.
Now let's prove formula for moving to a new logarithm base kind 
. To do this, it is enough to prove the validity of the equality log c b=log a b·log c a. The basic logarithmic identity allows us to represent the number b as a log a b , then log c b=log c a log a b . It remains to use the property of the logarithm of the degree: log c a log a b =log a b log c a. This proves the equality log c b=log a b·log c a, which means that the formula for transition to a new base of the logarithm has also been proven.
Let's show a couple of examples of using this property of logarithms: and 
.
The formula for moving to a new base allows you to move on to working with logarithms that have a “convenient” base. For example, with its help you can switch to natural or decimal logarithms so that you can calculate the value of the logarithm from the table of logarithms. The formula for moving to a new logarithm base also allows, in some cases, to find the value of a given logarithm when the values of some logarithms with other bases are known.
A special case of the formula for transition to a new logarithm base for c=b of the form is often used 
. This shows that log a b and log b a – . Eg, 
.
The formula is also often used 
, which is convenient for finding logarithm values. To confirm our words, we will show how it can be used to calculate the value of a logarithm of the form . We have 
. To prove the formula 
it is enough to use the formula for transition to a new base of the logarithm a: 
.
It remains to prove the properties of comparison of logarithms.
Let us prove that for any positive numbers b 1 and b 2, b 1 log a b 2 , and for a>1 – the inequality log a b 1 Finally, it remains to prove the last of the listed properties of logarithms. Let us limit ourselves to the proof of its first part, that is, we will prove that if a 1 >1, a 2 >1 and a 1 1 is true log a 1 b>log a 2 b . The remaining statements of this property of logarithms are proved according to a similar principle. Let's use the opposite method. Suppose that for a 1 >1, a 2 >1 and a 1 1 is true log a 1 b≤log a 2 b . Based on the properties of logarithms, these inequalities can be rewritten as 
And 
respectively, and from them it follows that log b a 1 ≤log b a 2 and log b a 1 ≥log b a 2, respectively. Then, according to the properties of powers with the same bases, the equalities b log b a 1 ≥b log b a 2 and b log b a 1 ≥b log b a 2 must hold, that is, a 1 ≥a 2 . So we came to a contradiction to the condition a 1
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).
As you know, when multiplying expressions with powers, their exponents always add up (a b *a c = a b+c). This mathematical law was derived by Archimedes, and later, in the 8th century, the mathematician Virasen created a table of integer exponents. It was they who served for the further discovery of logarithms. Examples of using this function can be found almost everywhere where you need to simplify cumbersome multiplication by simple addition. If you spend 10 minutes reading this article, we will explain to you what logarithms are and how to work with them. In simple and accessible language.
Definition in mathematics
A logarithm is an expression of the following form: log a b=c, that is, the logarithm of any non-negative number (that is, any positive) “b” to its base “a” is considered to be the power “c” to which the base “a” must be raised in order to ultimately get the value "b". Let's analyze the logarithm using examples, let's say there is an expression log 2 8. How to find the answer? It’s very simple, you need to find a power such that from 2 to the required power you get 8. After doing some calculations in your head, we get the number 3! And that’s true, because 2 to the power of 3 gives the answer as 8.
Types of logarithms
For many pupils and students, this topic seems complicated and incomprehensible, but in fact logarithms are not so scary, the main thing is to understand their general meaning and remember their properties and some rules. There are three separate types of logarithmic expressions:
- Natural logarithm ln a, where the base is the Euler number (e = 2.7).
- Decimal a, where the base is 10.
- Logarithm of any number b to base a>1.
Each of them is solved in a standard way, including simplification, reduction and subsequent reduction to a single logarithm using logarithmic theorems. To obtain the correct values of logarithms, you should remember their properties and the sequence of actions when solving them.
Rules and some restrictions
In mathematics, there are several rules-constraints that are accepted as an axiom, that is, they are not subject to discussion and are the truth. For example, it is impossible to divide numbers by zero, and it is also impossible to extract the even root of negative numbers. Logarithms also have their own rules, following which you can easily learn to work even with long and capacious logarithmic expressions:
- The base “a” must always be greater than zero, and not equal to 1, otherwise the expression will lose its meaning, because “1” and “0” to any degree are always equal to their values;
- if a > 0, then a b >0, it turns out that “c” must also be greater than zero.
How to solve logarithms?
For example, the task is given to find the answer to the equation 10 x = 100. This is very easy, you need to choose a power by raising the number ten to which we get 100. This, of course, is 10 2 = 100.
Now let's represent this expression in logarithmic form. We get log 10 100 = 2. When solving logarithms, all actions practically converge to find the power to which it is necessary to enter the base of the logarithm in order to obtain a given number.
To accurately determine the value of an unknown degree, you need to learn how to work with a table of degrees. It looks like this:
As you can see, some exponents can be guessed intuitively if you have a technical mind and knowledge of the multiplication table. However, for larger values you will need a power table. It can be used even by those who know nothing at all about complex mathematical topics. The left column contains numbers (base a), the top row of numbers is the value of the power c to which the number a is raised. At the intersection, the cells contain the number values that are the answer (a c =b). Let's take, for example, the very first cell with the number 10 and square it, we get the value 100, which is indicated at the intersection of our two cells. Everything is so simple and easy that even the most true humanist will understand!
Equations and inequalities
It turns out that under certain conditions the exponent is the logarithm. Therefore, any mathematical numerical expressions can be written as a logarithmic equality. For example, 3 4 =81 can be written as the base 3 logarithm of 81 equal to four (log 3 81 = 4). For negative powers the rules are the same: 2 -5 = 1/32 we write it as a logarithm, we get log 2 (1/32) = -5. One of the most fascinating sections of mathematics is the topic of “logarithms”. We will look at examples and solutions of equations below, immediately after studying their properties. Now let's look at what inequalities look like and how to distinguish them from equations.
The following expression is given: log 2 (x-1) > 3 - it is a logarithmic inequality, since the unknown value “x” is under the logarithmic sign. And also in the expression two quantities are compared: the logarithm of the desired number to base two is greater than the number three.
The most important difference between logarithmic equations and inequalities is that equations with logarithms (for example, the logarithm 2 x = √9) imply one or more specific numerical values in the answer, while when solving an inequality, both the range of acceptable values and the points are determined breaking this function. As a consequence, the answer is not a simple set of individual numbers, as in the answer to an equation, but a continuous series or set of numbers.
Basic theorems about logarithms
When solving primitive tasks of finding the values of the logarithm, its properties may not be known. However, when it comes to logarithmic equations or inequalities, first of all, it is necessary to clearly understand and apply in practice all the basic properties of logarithms. We will look at examples of equations later; let's first look at each property in more detail.
- The main identity looks like this: a logaB =B. It applies only when a is greater than 0, not equal to one, and B is greater than zero.
- The logarithm of the product can be represented in the following formula: log d (s 1 * s 2) = log d s 1 + log d s 2. In this case, the mandatory condition is: d, s 1 and s 2 > 0; a≠1. You can give a proof for this logarithmic formula, with examples and solution. Let log a s 1 = f 1 and log a s 2 = f 2, then a f1 = s 1, a f2 = s 2. We obtain that s 1 * s 2 = a f1 *a f2 = a f1+f2 (properties of degrees ), and then by definition: log a (s 1 * s 2) = f 1 + f 2 = log a s1 + log a s 2, which is what needed to be proven.
- The logarithm of the quotient looks like this: log a (s 1/ s 2) = log a s 1 - log a s 2.
- The theorem in the form of a formula takes the following form: log a q b n = n/q log a b.
This formula is called the “property of the degree of logarithm.” It resembles the properties of ordinary degrees, and it is not surprising, because all mathematics is based on natural postulates. Let's look at the proof.
Let log a b = t, it turns out a t =b. If we raise both parts to the power m: a tn = b n ;
but since a tn = (a q) nt/q = b n, therefore log a q b n = (n*t)/t, then log a q b n = n/q log a b. The theorem has been proven.
Examples of problems and inequalities
The most common types of problems on logarithms are examples of equations and inequalities. They are found in almost all problem books, and are also a required part of mathematics exams. To enter a university or pass entrance examinations in mathematics, you need to know how to correctly solve such tasks.
Unfortunately, there is no single plan or scheme for solving and determining the unknown value of the logarithm, but certain rules can be applied to each mathematical inequality or logarithmic equation. First of all, you should find out whether the expression can be simplified or reduced to a general form. You can simplify long logarithmic expressions if you use their properties correctly. Let's get to know them quickly.
When solving logarithmic equations, we must determine what type of logarithm we have: an example expression may contain a natural logarithm or a decimal one.
Here are examples ln100, ln1026. Their solution boils down to the fact that they need to determine the power to which the base 10 will be equal to 100 and 1026, respectively. To solve natural logarithms, you need to apply logarithmic identities or their properties. Let's look at examples of solving logarithmic problems of various types.
How to Use Logarithm Formulas: With Examples and Solutions
So, let's look at examples of using the basic theorems about logarithms.
- The property of the logarithm of a product can be used in tasks where it is necessary to decompose a large value of the number b into simpler factors. For example, log 2 4 + log 2 128 = log 2 (4*128) = log 2 512. The answer is 9.
- log 4 8 = log 2 2 2 3 = 3/2 log 2 2 = 1.5 - as you can see, using the fourth property of the logarithm power, we managed to solve a seemingly complex and unsolvable expression. You just need to factor the base and then take the exponent values out of the sign of the logarithm.
Assignments from the Unified State Exam
Logarithms are often found in entrance exams, especially many logarithmic problems in the Unified State Exam (state exam for all school graduates). Typically, these tasks are present not only in part A (the easiest test part of the exam), but also in part C (the most complex and voluminous tasks). The exam requires accurate and perfect knowledge of the topic “Natural logarithms”.
Examples and solutions to problems are taken from the official versions of the Unified State Exam. Let's see how such tasks are solved.
Given log 2 (2x-1) = 4. Solution:
let's rewrite the expression, simplifying it a little log 2 (2x-1) = 2 2, by the definition of the logarithm we get that 2x-1 = 2 4, therefore 2x = 17; x = 8.5.
- It is best to reduce all logarithms to the same base so that the solution is not cumbersome and confusing.
- All expressions under the logarithm sign are indicated as positive, therefore, when the exponent of an expression that is under the logarithm sign and as its base is taken out as a multiplier, the expression remaining under the logarithm must be positive.
