Let's start with properties of the logarithm of unity. 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 straightforward: since a 0 =1 for any a that satisfies the above conditions a>0 and a≠1 , then the proven equality log a 1=0 immediately follows from the definition of the logarithm.
Let's give examples of application of the considered property: log 3 1=0 , lg1=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 the definition of the logarithm log a a=1 .
Examples of using this property of logarithms are 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 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 the 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 , whence the required equality follows by the definition of the logarithm.
Let's show examples of using the property of the logarithm of the product: log 5 (2 3)=log 5 2+log 5 3 and 
.
The product logarithm property 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 is easily proved.
For example, the natural logarithm of a product can be replaced by the sum of three natural logarithms of the 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 quotient logarithm property 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 proved like the formula for the logarithm of the product: since 
, then by the definition of the logarithm .
Here is an example of using this property of the logarithm: 
.
Let's move on to property of the logarithm of degree. 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. We write this property of the logarithm of the degree in the form of a formula: log a b p =p log a |b|, where a>0 , a≠1 , b and p are numbers such that the degree of b p makes sense and b p >0 .
We first prove this property for positive b . The basic 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 power property, is equal to a p log a b . So we arrive at the equality b p =a p log a b , from which, by the definition of the 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|, whence 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 root of the nth degree is equal to the product of the fraction 1/n and the logarithm of the root 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 degree: 
.
Here is an example of using this property: 
.
Now let's prove conversion formula to the new base of the logarithm kind 
. To do this, it suffices 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. Thus, the equality log c b=log a b log c a is proved, which means that the formula for the transition to a new base of the logarithm is also proved.
Let's show a couple of examples of applying 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, it can be used to go to natural or decimal logarithms so that you can calculate the value of the logarithm from the table of logarithms. The formula for the transition to a new base of the logarithm also allows in some cases to find the value of a given logarithm, when the values of some logarithms with other bases are known.
Often used is a special case of the formula for the transition to a new base of the logarithm for c=b of the form 
. This shows that log a b and log b a – . For example, 
.
Also often used is the formula 
, which is useful for finding logarithm values. To confirm our words, we will show how the value of the logarithm of the form is calculated using it. We have 
. To prove the formula 
it is enough to use the transition formula to the new base of the logarithm a: 
.
It remains to prove the comparison properties 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. We confine ourselves to proving its first part, that is, we 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 by a similar principle. Let's use the opposite method. Suppose that for a 1 >1 , a 2 >1 and a 1 1 log a 1 b≤log a 2 b is true. By 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, by 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 be satisfied, that is, a 1 ≥a 2 . Thus, we have arrived at 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: A Textbook for Grades 10-11 of General Educational Institutions.
- Gusev V.A., Mordkovich A.G. Mathematics (a manual for applicants to technical schools).
basic properties.
- logax + logay = log(x y);
- logax − logay = log(x: y).
same grounds
log6 4 + log6 9.
Now let's complicate the task a little.
Examples of solving logarithms
What if there is a degree in the base or argument of the logarithm? Then the exponent of this degree can be taken out of the sign of the logarithm according to the following rules:
Of course, all these rules make sense if the ODZ logarithm is observed: a > 0, a ≠ 1, x >
A task. Find the value of the expression:
Transition to a new foundation
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
A task. Find the value of the expression:
See also:
Basic properties of the logarithm
1.
2.
3.
4.
5.
6.
7.
8.
9.
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11.
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14.
15.
The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Tolstoy.
Basic properties of logarithms
Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.
Examples for logarithms
Take the logarithm of expressions
Example 1
a). x=10ac^2 (a>0, c>0).
By properties 3,5 we calculate
2.
3.
Example 2 Find x if
Example 3. Let the value of logarithms be given
Calculate log(x) if
Basic properties of logarithms
Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.
These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.
Addition and subtraction of logarithms
Consider two logarithms with the same base: logax and logay. Then they can be added and subtracted, and:
- logax + logay = log(x y);
- logax − logay = log(x: y).
So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note: the key point here is - same grounds. If the bases are different, these rules do not work!
These formulas will help calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples and see:
Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
A task. Find the value of the expression: log2 48 − log2 3.
The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.
A task. Find the value of the expression: log3 135 − log3 5.
Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.
As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.
Removing the exponent from the logarithm
It is easy to see that the last rule follows their 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 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 sign of the logarithm into the logarithm itself. This is what is most often required.
A task. Find the value of the expression: log7 496.
Let's get rid of the degree in the argument according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12
A task. Find the value of the expression:
Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:
I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator.
Formulas of logarithms. Logarithms are examples of solutions.
They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator have the same number: log2 7. Since log2 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 was done. The result is 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 bases are different? What if they are not exact powers of the same number?
Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
In particular, if we put c = x, we get:
It follows from the second formula that the base and the argument of the logarithm can be interchanged, but the whole expression is “turned over”, i.e. the logarithm is 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 tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:
A task. Find the value of the expression: log5 16 log2 25.
Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;
Now let's flip the second logarithm:
Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.
A task. Find the value of the expression: log9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Basic logarithmic identity
Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.
The second formula is actually a paraphrased definition. It's called like this:
Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.
Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.
A task. Find the value of the expression:
Note that log25 64 = log5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:
If someone is not in the know, this was a real task from the Unified State Examination 🙂
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.
- logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 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.
See also:
The logarithm of the number b to the base a denotes the expression. To calculate the logarithm means to find such a power x () at which the equality is true
Basic properties of the logarithm
The above properties need to be known, since, on their basis, almost all problems and examples are solved based on logarithms. The remaining exotic properties can be derived by mathematical manipulations with these formulas
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When calculating the formulas for the sum and difference of logarithms (3.4) are encountered quite often. The rest are somewhat complex, but in a number of tasks they are indispensable for simplifying complex expressions and calculating their values.
Common cases of logarithms
Some of the common logarithms are those in which the base is even ten, exponential or deuce.
The base ten logarithm is usually called the base ten logarithm and is simply denoted lg(x).
It can be seen from the record that the basics are not written in the record. For example
The natural logarithm is the logarithm whose basis is the exponent (denoted ln(x)).
The exponent is 2.718281828…. To remember the exponent, you can study the rule: the exponent is 2.7 and twice the year of birth of Leo Tolstoy. Knowing this rule, you will know both the exact value of the exponent and the date of birth of Leo Tolstoy.
And another important base two logarithm is
The derivative of the logarithm of the function is equal to one divided by the variable
The integral or antiderivative logarithm is determined by the dependence
The above material is enough for you to solve a wide class of problems related to logarithms and logarithms. To assimilate the material, I will give only a few common examples from the school curriculum and universities.
Examples for logarithms
Take the logarithm of expressions
Example 1
a). x=10ac^2 (a>0, c>0).
By properties 3,5 we calculate
2.
By the difference property of logarithms, we have
3.
Using properties 3.5 we find
A seemingly complex expression using a series of rules is simplified to the form
Finding Logarithm Values
Example 2 Find x if
Solution. For the calculation, we apply properties 5 and 13 up to the last term
Substitute in the record and mourn
Since the bases are equal, we equate the expressions
Logarithms. First level.
Let the value of the logarithms be given
Calculate log(x) if
Solution: Take the logarithm of the variable to write the logarithm through the sum of the terms
This is just the beginning of acquaintance with logarithms and their properties. Practice calculations, enrich your practical skills - you will soon need the acquired knowledge to solve logarithmic equations. Having studied the basic methods for solving such equations, we will expand your knowledge for another equally important topic - logarithmic inequalities ...
Basic properties of logarithms
Logarithms, like any number, can be added, subtracted and converted in every possible way. But since logarithms are not quite ordinary numbers, there are rules here, which are called basic properties.
These rules must be known - no serious logarithmic problem can be solved without them. In addition, there are very few of them - everything can be learned in one day. So let's get started.
Addition and subtraction of logarithms
Consider two logarithms with the same base: logax and logay. Then they can be added and subtracted, and:
- logax + logay = log(x y);
- logax − logay = log(x: y).
So, the sum of the logarithms is equal to the logarithm of the product, and the difference is the logarithm of the quotient. Please note: the key point here is - same grounds. If the bases are different, these rules do not work!
These formulas will help calculate the logarithmic expression even when its individual parts are not considered (see the lesson "What is a logarithm"). Take a look at the examples and see:
A task. Find the value of the expression: log6 4 + log6 9.
Since the bases of logarithms are the same, we use the sum formula:
log6 4 + log6 9 = log6 (4 9) = log6 36 = 2.
A task. Find the value of the expression: log2 48 − log2 3.
The bases are the same, we use the difference formula:
log2 48 − log2 3 = log2 (48: 3) = log2 16 = 4.
A task. Find the value of the expression: log3 135 − log3 5.
Again, the bases are the same, so we have:
log3 135 − log3 5 = log3 (135: 5) = log3 27 = 3.
As you can see, the original expressions are made up of "bad" logarithms, which are not considered separately. But after transformations quite normal numbers turn out. Many tests are based on this fact. Yes, control - similar expressions in all seriousness (sometimes - with virtually no changes) are offered at the exam.
Removing the exponent from the logarithm
Now let's complicate the task a little. What if there is a degree in the base or argument of the logarithm? 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 their 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 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 sign of the logarithm into the logarithm itself.
How to solve logarithms
This is what is most often required.
A task. Find the value of the expression: log7 496.
Let's get rid of the degree in the argument according to the first formula:
log7 496 = 6 log7 49 = 6 2 = 12
A task. Find the value of the expression:
Note that the denominator is a logarithm whose base and argument are exact powers: 16 = 24; 49 = 72. We have:
I think the last example needs clarification. Where have logarithms gone? Until the very last moment, we work only with the denominator. They presented the base and the argument of the logarithm standing there in the form of degrees and took out the indicators - they got a “three-story” fraction.
Now let's look at the main fraction. The numerator and denominator have the same number: log2 7. Since log2 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 was done. The result is 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 bases are different? What if they are not exact powers of the same number?
Formulas for transition to a new base come to the rescue. We formulate them in the form of a theorem:
Let the logarithm logax be given. Then for any number c such that c > 0 and c ≠ 1, the equality is true:
In particular, if we put c = x, we get:
It follows from the second formula that the base and the argument of the logarithm can be interchanged, but the whole expression is “turned over”, i.e. the logarithm is 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 tasks that cannot be solved at all except by moving to a new foundation. Let's consider a couple of these:
A task. Find the value of the expression: log5 16 log2 25.
Note that the arguments of both logarithms are exact exponents. Let's take out the indicators: log5 16 = log5 24 = 4log5 2; log2 25 = log2 52 = 2log2 5;
Now let's flip the second logarithm:
Since the product does not change from permutation of factors, we calmly multiplied four and two, and then figured out the logarithms.
A task. Find the value of the expression: log9 100 lg 3.
The base and argument of the first logarithm are exact powers. Let's write it down and get rid of the indicators:
Now let's get rid of the decimal logarithm by moving to a new base:
Basic logarithmic identity
Often in the process of solving it is required to represent a number as a logarithm to a given base. In this case, the formulas will help us:
In the first case, the number n becomes the exponent in the argument. The number n can be absolutely anything, because it's just the value of the logarithm.
The second formula is actually a paraphrased definition. It's called like this:
Indeed, what will happen if the number b is raised to such a degree that the number b in this degree gives the number a? That's right: this is the same number a. Read this paragraph carefully again - many people “hang” on it.
Like the new base conversion formulas, the basic logarithmic identity is sometimes the only possible solution.
A task. Find the value of the expression:
Note that log25 64 = log5 8 - just took out the square from the base and the argument of the logarithm. Given the rules for multiplying powers with the same base, we get:
If someone is not in the know, this was a real task from the Unified State Examination 🙂
Logarithmic unit and logarithmic zero
In conclusion, I will give two identities that are difficult to call properties - rather, these are consequences from the definition of the logarithm. They are constantly found in problems and, surprisingly, create problems even for "advanced" students.
- logaa = 1 is. Remember once and for all: the logarithm to any base a from that base itself is equal to one.
- loga 1 = 0 is. The base a can be anything, but if the argument is one, the logarithm is zero! Because a0 = 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.
derived from its definition. And so the logarithm of the number b by reason a defined as the exponent to which a number must be raised a to get the number b(the logarithm exists only for positive numbers).
From this formulation it follows that the calculation x=log a b, is equivalent to solving the equation ax=b. For example, log 2 8 = 3 because 8 = 2 3 . The formulation of the logarithm makes it possible to justify that if b=a c, then the logarithm of the number b by reason a equals With. It is also clear that the topic of the logarithm is closely related to the topic of the power of a number.
With logarithms, as with any numbers, you can perform operations of addition, subtraction and transform in every possible way. But in view of the fact that logarithms are not quite ordinary numbers, their own special rules apply here, which are called basic properties.
Addition and subtraction of logarithms.
Take two logarithms with the same base: log x and log a y. Then remove it is possible to perform addition and subtraction operations:
log a x+ log a y= log a (x y);
log a x - log a y = log a (x:y).
log a(x 1 . x 2 . x 3 ... x k) = log x 1 + log x 2 + log x 3 + ... + log a x k.
From quotient logarithm theorems one more property of the logarithm can be obtained. It is well known that log a 1= 0, therefore,
log a 1 /b= log a 1 - log a b= -log a b.
So there is an equality:
log a 1 / b = - log a b.
Logarithms of two mutually reciprocal numbers on the same basis will differ from each other only in sign. So:
Log 3 9= - log 3 1 / 9 ; log 5 1 / 125 = -log 5 125.
Acceptable range (ODZ) of the logarithm
Now let's talk about restrictions (ODZ - the area of admissible values of variables).
We remember that, for example, the square root cannot be taken from negative numbers; or if we have a fraction, then the denominator cannot be equal to zero. There are similar restrictions for logarithms:
That is, both the argument and the base must be greater than zero, and the base cannot be equal.
Why is that?
Let's start simple: let's say that. Then, for example, the number does not exist, since no matter what degree we raise, it always turns out. Moreover, it does not exist for any. But at the same time it can be equal to anything (for the same reason - it is equal to any degree). Therefore, the object is of no interest, and it was simply thrown out of mathematics.
We have a similar problem in the case: in any positive degree - this, but it cannot be raised to a negative power at all, since division by zero will result (I remind you that).
When we are faced with the problem of erecting in fractional degree(which is represented as a root: . For example, (that is), but does not exist.
Therefore, negative reasons are easier to throw away than to mess with them.
Well, since the base a is only positive for us, then no matter what degree we raise it, we will always get a strictly positive number. So the argument must be positive. For example, does not exist, since in no way will negative number(and even zero, so it doesn't exist either).
In problems with logarithms, the first step is to write down the ODZ. I'll give an example:
Let's solve the equation.
Recall the definition: the logarithm is the power to which the base must be raised in order to obtain an argument. And by the condition, this degree is equal to: .
We get the usual quadratic equation: . We solve it using the Vieta theorem: the sum of the roots is equal, and the product. Easy to pick up, these are numbers and.
But if you immediately take and write down both of these numbers in the answer, you can get 0 points for the task. Why? Let's think about what happens if we substitute these roots into the initial equation?
This is clearly false, since the base cannot be negative, that is, the root is "third-party".
To avoid such unpleasant tricks, you need to write down the ODZ even before starting to solve the equation:
Then, having received the roots and, we immediately discard the root, and write the correct answer.
Example 1(try to solve it yourself) :
Find the root of the equation. If there are several roots, indicate the smaller one in your answer.
Solution:
First of all, let's write the ODZ:
Now we remember what a logarithm is: to what power do you need to raise the base to get an argument? In the second. That is:
It would seem that the smaller root is equal. But this is not so: according to the ODZ, the root is third-party, that is, it is not the root of this equation at all. Thus, the equation has only one root: .
Answer: .
Basic logarithmic identity
Recall the definition of a logarithm in general terms:
Substitute in the second equality instead of the logarithm:
This equality is called basic logarithmic identity. Although in essence this equality is just written differently definition of the logarithm:
This is the power to which you need to raise in order to get.
For example:
Solve the following examples:
Example 2
Find the value of the expression.
Solution:
Recall the rule from the section:, that is, when raising a degree to a power, the indicators are multiplied. Let's apply it:
Example 3
Prove that.
Solution:
Properties of logarithms
Unfortunately, the tasks are not always so simple - often you first need to simplify the expression, bring it to the usual form, and only then it will be possible to calculate the value. It's easiest to do this knowing properties of logarithms. So let's learn the basic properties of logarithms. I will prove each of them, because any rule is easier to remember if you know where it comes from.
All these properties must be remembered; without them, most problems with logarithms cannot be solved.
And now about all the properties of logarithms in more detail.
Property 1:
Proof:
Let, then.
We have: , h.t.d.
Property 2: Sum of logarithms
The sum of logarithms with the same base is equal to the logarithm of the product: .
Proof:
Let, then. Let, then.
Example: Find the value of the expression: .
Solution: .
The formula you just learned helps to simplify the sum of the logarithms, not the difference, so that these logarithms cannot be combined right away. But you can do the opposite - "break" the first logarithm into two: And here is the promised simplification:
.
Why is this needed? Well, for example: what does it matter?
Now it's obvious that.
Now make it easy for yourself:
Tasks:
Answers:
Property 3: Difference of logarithms:
Proof:
Everything is exactly the same as in paragraph 2:
Let, then.
Let, then. We have:
The example from the last point is now even simpler:
More complicated example: . Guess yourself how to decide?
Here it should be noted that we do not have a single formula about logarithms squared. This is something akin to an expression - this cannot be simplified right away.
Therefore, let's digress from the formulas about logarithms, and think about what formulas we generally use in mathematics most often? Ever since 7th grade!
It - . You have to get used to the fact that they are everywhere! And in exponential, and in trigonometric, and in irrational problems, they are found. Therefore, they must be remembered.
If you look closely at the first two terms, it becomes clear that this is difference of squares:
Answer to check:
Simplify yourself.
Examples
Answers.
Property 4: Derivation of the exponent from the argument of the logarithm:
Proof: And here we also use the definition of the logarithm: let, then. We have: , h.t.d.
You can understand this rule like this:
That is, the degree of the argument is taken forward of the logarithm, as a coefficient.
Example: Find the value of the expression.
Solution: .
Decide for yourself:
Examples:
Answers:
Property 5: Derivation of the exponent from the base of the logarithm:
Proof: Let, then.
We have: , h.t.d.
Remember: from grounds degree is rendered as reverse number, unlike the previous case!
Property 6: Derivation of the exponent from the base and the argument of the logarithm:
Or if the degrees are the same: .
Property 7: Transition to new base:
Proof: Let, then.
We have: , h.t.d.
Property 8: Swapping the base and the argument of the logarithm:
Proof: This is a special case of formula 7: if we substitute, we get: , p.t.d.
Let's look at a few more examples.
Example 4
Find the value of the expression.
We use the property of logarithms No. 2 - the sum of logarithms with the same base is equal to the logarithm of the product:
Example 5
Find the value of the expression.
Solution:
We use the property of logarithms No. 3 and No. 4:
Example 6
Find the value of the expression.
Solution:
Using property number 7 - go to base 2:
Example 7
Find the value of the expression.
Solution:
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