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 logarithmic problem can be solved. 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. Note: key moment Here - 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 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. Taking into account the rules for multiplying powers with the same base, 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 direct consequence from 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.
Logarithm of a number N based on A called exponent X , to which you need to build A to get the number N
Provided that 
,
,
From the definition of logarithm it follows that 
, i.e.
- this equality is the basic logarithmic identity.
Logarithms to base 10 are called decimal logarithms. Instead of 
write 
.
Logarithms to the base e
are called natural and are designated 
.
Basic properties of logarithms.
The logarithm of one is equal to zero for any base.
Logarithm of the product equal to the sum logarithms of factors.
3) The logarithm of the quotient is equal to the difference of the logarithms
Factor 
called the modulus of transition from logarithms to the base a
to logarithms at the base b
.
Using properties 2-5, it is often possible to reduce the logarithm of a complex expression to the result of simple arithmetic operations on logarithms.
For example, 
Such transformations of a logarithm are called logarithms. Transformations inverse to logarithms are called potentiation.
Chapter 2. Elements of higher mathematics.
1. Limits
Limit of the function 
is a finite number A if, as xx
0
for each predetermined 
, there is such a number 
that as soon as 
, That 
.
A function that has a limit differs from it by an infinitesimal amount: 
, where- b.m.v., i.e. 
.
Example. Consider the function 
.
When striving 
, function y
tends to zero: 
1.1. Basic theorems about limits.
The limit of a constant value is equal to this constant value
.
The limit of the sum (difference) of a finite number of functions is equal to the sum (difference) of the limits of these functions.
The limit of the product of a finite number of functions is equal to the product of the limits of these functions.
The limit of the quotient of two functions is equal to the quotient of the limits of these functions if the limit of the denominator is not zero.
Wonderful Limits
,
, Where 
1.2. Limit Calculation Examples
However, not all limits are calculated so easily. More often, calculating the limit comes down to revealing an uncertainty of the type: 
or .
.
2. Derivative of a function
Let us have a function 
, continuous on the segment 
.
Argument 
got some increase 
. Then the function will receive an increment 
.
Argument value 
corresponds to the function value 
.
Argument value 
corresponds to the function value.
Hence, .
Let us find the limit of this ratio at 
. If this limit exists, then it is called the derivative of the given function.
Definition 3 Derivative of a given function 
by argument 
is called the limit of the ratio of the increment of a function to the increment of the argument, when the increment of the argument arbitrarily tends to zero.
Derivative of a function 
can be designated as follows:
;
;
;
.
Definition 4The operation of finding the derivative of a function is called differentiation.
2.1. Mechanical meaning of derivative.
Let's consider the rectilinear motion of some rigid body or material point.
Let at some point in time 
moving point 
was at a distance 
from the starting position 
.
After some period of time 
she moved a distance 
. Attitude 
=
- average speed of a material point 
. Let us find the limit of this ratio, taking into account that 
.
Therefore, the definition instantaneous speed motion of a material point comes down to finding the derivative of the path with respect to time.
2.2. Geometric value of the derivative
Let us have a graphically defined function 
.
Rice. 1. Geometric meaning of derivative
If 
, then point 
, will move along the curve, approaching the point 
.
Hence 
, i.e. the value of the derivative for a given value of the argument 
numerically equal to the tangent of the angle formed by the tangent at a given point with the positive direction of the axis 
.
2.3. Table of basic differentiation formulas.
Power function
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Exponential function
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Logarithmic function
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Trigonometric function
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Inverse trigonometric function
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2.4. Rules of differentiation.
Derivative of 
Derivative of the sum (difference) of functions
Derivative of the product of two functions
Derivative of the quotient of two functions
2.5. Derivative of a complex function.
Let the function be given 
such that it can be represented in the form
And 
, where the variable 
is an intermediate argument, then 
The derivative of a complex function is equal to the product of the derivative of the given function with respect to the intermediate argument and the derivative of the intermediate argument with respect to x.
Example 1. 
Example 2. 
3. Differential function.
Let there be 
, differentiable on some interval 
let it go at
this function has a derivative
,
then we can write
(1),
Where 
- an infinitesimal quantity,
since when 
Multiplying all terms of equality (1) by 
we have:
Where 
- b.m.v. higher order.
Magnitude 
called the differential of the function 
and is designated 
.
3.1. Geometric value of the differential.
Let the function be given 
.
Fig.2. Geometric meaning of differential.
.
Obviously, the differential of the function 
is equal to the increment of the ordinate of the tangent at a given point.
3.2. Derivatives and differentials of various orders.
If there 
, Then 
is called the first derivative.
The derivative of the first derivative is called the second-order derivative and is written 
.
Derivative of the nth order of the function 
is called the (n-1)th order derivative and is written:
.
The differential of the differential of a function is called the second differential or second order differential.
.
.
3.3 Solving biological problems using differentiation.
Task 1. Studies have shown that the growth of a colony of microorganisms obeys the law 
, Where N
– number of microorganisms (in thousands), t
– time (days).
b) Will the population of the colony increase or decrease during this period?
Answer. The size of the colony will increase.
Task 2. The water in the lake is periodically tested to monitor the content of pathogenic bacteria. Through t days after testing, the concentration of bacteria is determined by the ratio
.
When will the lake have a minimum concentration of bacteria and will it be possible to swim in it?
Solution: A function reaches max or min when its derivative is zero.
,
Let's determine the max or min will be in 6 days. To do this, let's take the second derivative.
Answer: After 6 days there will be a minimum concentration of bacteria.
The logarithm of a positive number b to base a (a>0, a is not equal to 1) is a number c such that a c = b: log a b = c ⇔ a c = b (a > 0, a ≠ 1, b > 0)
Note that the logarithm of a non-positive number is undefined. In addition, the base of the logarithm must be positive number, not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the logarithm to the base -2 of 4 is equal to 2.
Basic logarithmic identity
a log a b = b (a > 0, a ≠ 1) (2)It is important that the scope of definition of the right and left sides of this formula is different. Left side is defined only for b>0, a>0 and a ≠ 1. The right-hand side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic “identity” when solving equations and inequalities can lead to a change in the OD.
Two obvious consequences of the definition of logarithm
log a a = 1 (a > 0, a ≠ 1) (3)log a 1 = 0 (a > 0, a ≠ 1) (4)
Indeed, when raising the number a to the first power, we get the same number, and when raising it to the first power zero degree- one.
Logarithm of the product and logarithm of the quotient
log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) (5)Log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) (6)
I would like to warn schoolchildren against thoughtlessly using these formulas when solving logarithmic equations and inequalities. When using them “from left to right,” the ODZ narrows, and when moving from the sum or difference of logarithms to the logarithm of the product or quotient, the ODZ expands.
Indeed, the expression log a (f (x) g (x)) is defined in two cases: when both functions are strictly positive or when f(x) and g(x) are both less than zero.
Transforming this expression into the sum log a f (x) + log a g (x), we are forced to limit ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of acceptable values, and this is categorically unacceptable, since it can lead to a loss of solutions. A similar problem exists for formula (6).
The degree can be taken out of the sign of the logarithm
log a b p = p log a b (a > 0, a ≠ 1, b > 0) (7)And again I would like to call for accuracy. Consider the following example:
Log a (f (x) 2 = 2 log a f (x)
The left side of the equality is obviously defined for all values of f(x) except zero. The right side is only for f(x)>0! By taking the degree out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of acceptable values. All these remarks apply not only to power 2, but also to any even power.
Formula for moving to a new foundation
log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) (8)That rare case, when the ODZ does not change during the transformation. If you have chosen base c wisely (positive and not equal to 1), the formula for moving to a new base is completely safe.
If we choose the number b as the new base c, we get an important special case formulas (8):
Log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) (9)
Some simple examples with logarithms
Example 1. Calculate: log2 + log50.
Solution. log2 + log50 = log100 = 2. We used the sum of logarithms formula (5) and the definition of the decimal logarithm.
Example 2. Calculate: lg125/lg5.
Solution. log125/log5 = log 5 125 = 3. We used the formula for moving to a new base (8).
Table of formulas related to logarithms
| a log a b = b (a > 0, a ≠ 1) |
| log a a = 1 (a > 0, a ≠ 1) |
| log a 1 = 0 (a > 0, a ≠ 1) |
| log a (b c) = log a b + log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b c = log a b − log a c (a > 0, a ≠ 1, b > 0, c > 0) |
| log a b p = p log a b (a > 0, a ≠ 1, b > 0) |
| log a b = log c b log c a (a > 0, a ≠ 1, b > 0, c > 0, c ≠ 1) |
| log a b = 1 log b a (a > 0, a ≠ 1, b > 0, b ≠ 1) |
Follows from its definition. And so the logarithm of the number b based on A is defined as the exponent to which a number must be raised a to get the number b(logarithm exists only for positive numbers).
From this formulation it follows that the calculation x=log a b, is equivalent to solving the equation a x =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 based on a equals With. It is also clear that the topic of logarithms is closely related to the topic of powers of a number.
With logarithms, as with any numbers, you can do operations of addition, subtraction and transform in every possible way. But due to the fact that logarithms are not entirely ordinary numbers, their own special rules apply here, which are called main properties.
Adding and subtracting logarithms.
Let's take two logarithms with the same bases: log a x And log a y. Then 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 a x 1 + log a x 2 + log a x 3 + ... + log a x k.
From logarithm quotient theorem One more property of the logarithm can be obtained. It is common knowledge that log a 1= 0, therefore
log a 1 /b=log a 1 - log a b= -log a b.
This means there is an equality:
log a 1 / b = - log a b.
Logarithms of two reciprocal numbers for the same reason will differ from each other solely by sign. So:
Log 3 9= - log 3 1 / 9 ; log 5 1 / 125 = -log 5 125.
Adding and subtracting logarithms
Basic properties of logarithms
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 logarithmic problem can be solved. In addition, there are very few of them - you can learn everything in one day. So let's get started.
Consider two logarithms with the same bases: log a x and log a y. Then they can be added and subtracted, and:
1. log a x + log a y = log a (x y);
2. 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 a 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:
Find the meaning of the expression: 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.
Find the meaning 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.
Find the meaning 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 tests are based on this fact. Yes, test-like expressions are offered in all seriousness (sometimes with virtually no changes) on the Unified State Examination.
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:
1. log a x n = n log a x;
3. 
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.
Find the meaning 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
Find the meaning of the expression:
Note that the denominator contains a logarithm, the base and argument of which are exact powers: 16 = 2 4 ; 49 = 7 2 . We have:
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.
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