Laws of logarithms. Basic logarithmic identity

1.1. Determining the degree for an integer exponent

X 1 = X
X 2 = X * X
X 3 = X * X * X
…
X N \u003d X * X * ... * X - N times

1.2. Zero degree.

By definition, it is customary to assume that the zero power of any number is equal to 1:

1.3. negative degree.

X-N = 1/XN

1.4. Fractional exponent, root.

X 1/N = N-th root of X.

For example: X 1/2 = √X.

1.5. The formula for adding powers.

X (N+M) = X N * X M

1.6. Formula for subtracting degrees.

X (N-M) = X N / X M

1.7. Power multiplication formula.

XN*M = (XN)M

1.8. The formula for raising a fraction to a power.

(X/Y)N = XN /YN

2. Number e.

The value of the number e is equal to the following limit:

E = lim(1+1/N), as N → ∞.

With a precision of 17 digits, the number e is 2.71828182845904512.

3. Euler's equality.

This equality links five numbers that play a special role in mathematics: 0, 1, the number e, the number pi, the imaginary unit.

E(i*pi) + 1 = 0

4. Exponential function exp (x)

exp(x) = e x

5. Derivative of the exponential function

An exponential function has a remarkable property: the derivative of a function is equal to the exponential function itself:

(exp(x))" = exp(x)

6. Logarithm.

6.1. Definition of the logarithm function

If x = b y , then the logarithm is the function

Y = Logb(x).

The logarithm shows to what degree it is necessary to raise a number - the base of the logarithm (b) to get a given number (X). The logarithm function is defined for X greater than zero.

For example: Log 10 (100) = 2.

6.2. Decimal logarithm

This is the logarithm to base 10:

Y = Log 10 (x) .

Denoted Log(x): Log(x) = Log 10 (x).

An example of using the decimal logarithm is decibel.

6.3. Decibel

Item is highlighted on a separate page Decibel

6.4. binary logarithm

This is the base 2 logarithm:

Y = Log2(x).

Denoted by Lg(x): Lg(x) = Log 2 (X)

6.5. natural logarithm

This is the logarithm to base e:

Y = loge(x) .

Denoted by Ln(x): Ln(x) = Log e (X)
The natural logarithm is the inverse of the exponential function exp(X).

6.6. characteristic points

Loga(1) = 0
Log a(a) = 1

6.7. The formula for the logarithm of the product

Log a (x*y) = Log a (x)+Log a (y)

6.8. The formula for the logarithm of the quotient

Log a (x/y) = Log a (x) - Log a (y)

6.9. Power logarithm formula

Log a (x y) = y*Log a (x)

6.10. Formula for converting to a logarithm with a different base

Log b (x) = (Log a (x)) / Log a (b)

Example:

Log 2 (8) = Log 10 (8) / Log 10 (2) =
0.903089986991943552 / 0.301029995663981184 = 3

7. Formulas useful in life

Often there are problems of converting volume into area or length, and the inverse problem is converting area into volume. For example, boards are sold in cubes (cubic meters), and we need to calculate how much wall area can be sheathed with boards contained in a certain volume, see the calculation of boards, how many boards are in a cube. Or, the dimensions of the wall are known, it is necessary to calculate the number of bricks, see brick calculation.

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EXPONENTIAL AND LOGARITHMIC FUNCTIONS VIII

§ 184. Logarithm of degree and root

Theorem 1. The logarithm of the power of a positive number is equal to the product of the exponent of this power by the logarithm of its base.

In other words, if a and X positive and a =/= 1, then for any real number k

log a x k = k log a x . (1)

To prove this formula, it suffices to show that

= a k log a x . (2)

= x k

a k log a x = (a log a x ) k = x k .

This implies the validity of formula (2), and hence also (1).

Note that if the number k is natural ( k = n ), then formula (1) is a particular case of the formula

log a (x 1 x 2 x 3 ... x n ) = log a x 1 + log a x 2 + log a x 3 + ...log a x n .

proven in the previous section. Indeed, assuming in this formula

x 1 = x 2 = ... = x n = x ,

we get:

log a x n = n log a x .

1) log 3 25 = log 3 5 2 = 2 log 3 5;

2) log 3 2 √ 3 = √3 log 3 2.

For negative values X formula (1) loses its meaning. For example, you cannot write log 2 (-4) 2 = 2 log 2 (- 4) because the expression log 2 (-4) is undefined. Note that the expression on the left side of this formula makes sense:

log 2 (-4) 2 = log 2 16 = 4.

In general, if the number X is negative, then the expression log a x 2k = 2k log a x determined because x 2k > 0. The expression is 2 k log a x in this case it doesn't make sense. So write

Log a x 2k = 2k log a x

it is forbidden. However, one can write

log a x 2k = 2k log a | x | (3)

This formula is easily obtained from (1) if we take into account that

x 2k = | x | 2k

For example,

log 3 (-3) 4 = 4 log 3 | -3 | = 4 log 3 3 = 4.

Theorem 2. The logarithm of the root of a positive number is equal to the logarithm of the root expression divided by the exponent of the root.

In other words, if the numbers a and X are positive a =/= 1 and P is a natural number, then

log a n √x = 1 / n log a x

Really, n √x = . Therefore, by Theorem 1

log a n √x = log a = 1 / n log a x .

1) log 3 √ 8 = 1 / 2 log 3 8; 2) log 2 5 √27 = 1/5 log 2 27.

Exercises

1408. How will the logarithm of a number change if, without changing the base:

a) square the number

b) take the square root of a number?

1409. How the difference log 2 will change a - log 2 b if numbers a and b replace accordingly with:

a) a 3 and b 3; b) 3 a and 3 b ?

1410. Knowing that log 10 2 ≈ 0.3010, log 10 3 ≈ 0.4771, find the logarithms to the base of 10 numbers:

8; 9; 3 √2 ; 3 √6 ; 0,5; 1 / 9

1411. Prove that the logarithms of successive members of a geometric progression form an arithmetic progression.

1412. Are the functions different from each other

at = log 3 X 2 and at = 2 log 3 X

Construct graphs of these functions.

1413. Find an error in the following transformations:

log 2 1 / 3 = log 2 1 / 3

2log 2 1 / 3 > log 2 1 / 3 ;

log 2 (1 / 3) 2 > log 2 1 / 3

(1 / 3) 2 > 1 / 3 ;

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 not defined. Also, the base of the logarithm must be a positive number, not equal to 1. For example, if we square -2, we get the number 4, but this does not mean that the base -2 logarithm of 4 is 2.

Basic logarithmic identity

a log a b = b (a > 0, a ≠ 1) (2)

It is important that the domains of definition of the right and left parts of this formula are different. The left side is defined only for b>0, a>0 and a ≠ 1. The right side is defined for any b, and does not depend on a at all. Thus, the application of the basic logarithmic "identity" in solving equations and inequalities can lead to a change in the DPV.

Two obvious consequences of the definition of the 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 zero power, we get one.

The logarithm of the product and the 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 the thoughtless use of these formulas when solving logarithmic equations and inequalities. When they are used "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 restrict ourselves only to the case when f(x)>0 and g(x)>0. There is a narrowing of the range of admissible values, and this is categorically unacceptable, since it can lead to the 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! Taking the power out of the logarithm, we again narrow the ODZ. The reverse procedure leads to an expansion of the range of admissible values. All these remarks apply not only to the power of 2, but also to any even power.

Formula for moving to a new base

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 conversion. If you have chosen the base c wisely (positive and not equal to 1), the formula for moving to a new base is perfectly safe.

If we choose the number b as a new base c, we obtain an important particular case of formula (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: lg2 + lg50.
Solution. lg2 + lg50 = lg100 = 2. We used the formula for the sum of logarithms (5) and the definition of the decimal logarithm.

Example 2 Calculate: lg125/lg5.
Solution. lg125/lg5 = log 5 125 = 3. We used the new base transition formula (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)

What is a logarithm?

Attention!
There are additional
material in Special Section 555.
For those who strongly "not very..."
And for those who "very much...")

What is a logarithm? How to solve logarithms? These questions confuse many graduates. Traditionally, the topic of logarithms is considered complex, incomprehensible and scary. Especially - equations with logarithms.

This is absolutely not true. Absolutely! Don't believe? Good. Now, for some 10 - 20 minutes you:

1. Understand what is a logarithm.

2. Learn to solve a whole class of exponential equations. Even if you haven't heard of them.

3. Learn to calculate simple logarithms.

Moreover, for this you will only need to know the multiplication table, and how a number is raised to a power ...

I feel you doubt ... Well, keep time! Go!

First, solve the following equation in your mind:

If you like this site...

By the way, I have a couple more interesting sites for you.)

You can practice solving examples and find out your level. Testing with instant verification. Learning - with interest!)

you can get acquainted with functions and derivatives.

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:

Basic properties of the logarithm

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
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

4. where .

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.

You must know these rules - 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 it is possible to interchange the base and the argument of the logarithm, but in this case 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.

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

1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.

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

4. where .

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

You must know these rules - 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).

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.

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.

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.

Transition to a new foundation

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:

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:

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

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.

Portal for the student. Self-training

1.1. Determining the degree for an integer exponent

1.2. Zero degree.

1.3. negative degree.

1.4. Fractional exponent, root.

1.5. The formula for adding powers.

1.6. Formula for subtracting degrees.

1.7. Power multiplication formula.

1.8. The formula for raising a fraction to a power.

2. Number e.

3. Euler's equality.

4. Exponential function exp (x)

5. Derivative of the exponential function

6. Logarithm.

6.1. Definition of the logarithm function

6.2. Decimal logarithm

6.3. Decibel

6.4. binary logarithm

6.5. natural logarithm

6.6. characteristic points

6.7. The formula for the logarithm of the product

6.8. The formula for the logarithm of the quotient

6.9. Power logarithm formula

6.10. Formula for converting to a logarithm with a different base

Example:

7. Formulas useful in life

Basic logarithmic identity

Two obvious consequences of the definition of the logarithm

The logarithm of the product and the logarithm of the quotient

The degree can be taken out of the sign of the logarithm

Formula for moving to a new base

Some simple examples with logarithms

Table of formulas related to logarithms

What is a logarithm?

If you like this site...

Examples of solving logarithms

Transition to a new foundation

Basic properties of the logarithm

Basic properties of logarithms

Examples for logarithms

Take the logarithm of expressions

Basic properties of logarithms

Addition and subtraction of logarithms

Removing the exponent from the logarithm

Formulas of logarithms. Logarithms are examples of solutions.

Transition to a new foundation

Basic logarithmic identity

Logarithmic unit and logarithmic zero

Basic properties of the logarithm

Common cases of logarithms

Examples for logarithms

Take the logarithm of expressions

Finding Logarithm Values

Logarithms. First level.

Basic properties of logarithms

Addition and subtraction of logarithms

Removing the exponent from the logarithm

How to solve logarithms

Transition to a new foundation

Basic logarithmic identity

Logarithmic unit and logarithmic zero

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