Logarithm of number examples. Solving logarithmic equations
The final videos in a long series of lessons about solving logarithmic equations. This time we will work primarily with the ODZ of the logarithm - it is precisely because of incorrect consideration (or even ignoring) of the domain of definition that most errors arise when solving such problems.
In this short video lesson, we will look at the use of formulas for adding and subtracting logarithms, and also deal with fractional rational equations, which many students also have problems with.
What will we talk about? The main formula I would like to understand looks like this:
log a (f g ) = log a f + log a g
This is a standard transition from the product to the sum of logarithms and back. You probably know this formula from the very beginning of studying logarithms. However, there is one hitch.
As long as the variables a, f and g are ordinary numbers, no problems arise. This formula works great.
However, as soon as functions appear instead of f and g, the problem of expanding or narrowing the domain of definition arises depending on which direction to transform. Judge for yourself: in the logarithm written on the left, the domain of definition is as follows:
fg > 0
But in the amount written on the right, the domain of definition is already somewhat different:
f > 0
g > 0
This set of requirements is more stringent than the original one. In the first case, we will be satisfied with option f< 0, g < 0 (ведь их произведение положительное, поэтому неравенство fg >0 is executed).
So, when moving from the left construction to the right one, a narrowing of the domain of definition occurs. If at first we had a sum, and we rewrite it in the form of a product, then the domain of definition expands.
In other words, in the first case we could lose roots, and in the second we could get extra ones. This must be taken into account when solving real logarithmic equations.
So, the first task:
[Caption for the picture] On the left we see the sum of logarithms using the same base. Therefore, these logarithms can be added:
[Caption for the picture] As you can see, on the right we replaced the zero using the formula:
a = log b b a
Let's rearrange our equation a little more:
log 4 (x − 5) 2 = log 4 1
Before us is the canonical form of the logarithmic equation; we can cross out the log sign and equate the arguments:
(x − 5) 2 = 1
|x − 5| = 1
Please note: where did the module come from? Let me remind you that the root of an exact square is equal to the modulus:
[Caption for the picture] Then we solve the classical equation with modulus:
|f | = g (g > 0) ⇒f = ±g
x − 5 = ±1 ⇒x 1 = 5 − 1 = 4; x 2 = 5 + 1 = 6
Here are two candidate answers. Are they a solution to the original logarithmic equation? No way!
We have no right to leave everything just like that and write down the answer. Take a look at the step where we replace the sum of logarithms with one logarithm of the product of the arguments. The problem is that in the original expressions we have functions. Therefore, you should require:
x(x − 5) > 0; (x − 5)/x > 0.
When we transformed the product, obtaining an exact square, the requirements changed:
(x − 5) 2 > 0
When is this requirement met? Yes, almost always! Except for the case when x − 5 = 0. That is the inequality will be reduced to one punctured point:
x − 5 ≠ 0 ⇒ x ≠ 5
As you can see, the scope of definition has expanded, which is what we talked about at the very beginning of the lesson. Consequently, extra roots may appear.
How can you prevent these extra roots from appearing? It’s very simple: we look at our obtained roots and compare them with the domain of definition of the original equation. Let's count:
x (x − 5) > 0
We will solve using the interval method:
x (x − 5) = 0 ⇒ x = 0; x = 5
We mark the resulting numbers on the line. All points are missing because the inequality is strict. Take any number greater than 5 and substitute:
[Caption for the picture] We are interested in the intervals (−∞; 0) ∪ (5; ∞). If we mark our roots on the segment, we will see that x = 4 does not suit us, because this root lies outside the domain of definition of the original logarithmic equation.
We return to the totality, cross out the root x = 4 and write down the answer: x = 6. This is the final answer to the original logarithmic equation. That's it, problem solved.
Let's move on to the second logarithmic equation:
[Caption for the picture] Let's solve it. Note that the first term is a fraction, and the second is the same fraction, but inverted. Don't be scared by the expression lgx - it's simple decimal logarithm, we can write:
lgx = log 10 x
Since we have two inverted fractions, I propose introducing a new variable:
[Caption for the picture] Therefore, our equation can be rewritten as follows:
t + 1/t = 2;
t + 1/t − 2 = 0;
(t 2 − 2t + 1)/t = 0;
(t − 1) 2 /t = 0.
As you can see, the numerator of the fraction is an exact square. A fraction is equal to zero when its numerator is zero and its denominator is non-zero:
(t − 1) 2 = 0; t ≠ 0
Let's solve the first equation:
t − 1 = 0;
t = 1.
This value satisfies the second requirement. Therefore, we can say that we have completely solved our equation, but only with respect to the variable t. Now let’s remember what t is:
[Caption for the picture]We got the proportion:
lgx = 2 lgx + 1
2 logx − logx = −1
logx = −1
We bring this equation to its canonical form:
logx = log 10 −1
x = 10 −1 = 0.1
As a result, we received a single root, which, in theory, is the solution to the original equation. However, let’s still play it safe and write out the domain of definition of the original equation:
[Caption for the picture] Therefore, our root satisfies all the requirements. We have found a solution to the original logarithmic equation. Answer: x = 0.1. The problem is solved.
There is only one key point in today's lesson: when using the formula for moving from a product to a sum and back, be sure to take into account that the scope of definition can narrow or expand depending on which direction the transition is made.
How to understand what is happening: contraction or expansion? Very simple. If earlier the functions were together, but now they are separate, then the scope of definition has narrowed (because there are more requirements). If at first the functions stood separately, and now they are together, then the domain of definition is expanded (fewer requirements are imposed on the product than on individual factors).
Taking into account this remark, I would like to note that the second logarithmic equation does not require these transformations at all, that is, we do not add or multiply the arguments anywhere. However, here I would like to draw your attention to another wonderful technique that can significantly simplify the solution. It's about replacing a variable.
However, remember that no substitutions free us from the scope of definition. That is why after all the roots were found, we were not lazy and returned to the original equation to find its ODZ.
Often, when replacing a variable, an annoying error occurs when students find the value of t and think that the solution is complete. No way!
Once you have found the value of t, you need to go back to the original equation and see what exactly we meant with this letter. As a result, we have to solve one more equation, which, however, will be much simpler than the original one.
This is precisely the point of introducing a new variable. We split the original equation into two intermediate ones, each of which has a much simpler solution.
How to solve "nested" logarithmic equations
Today we continue to study logarithmic equations and will analyze constructions when one logarithm is under the sign of another logarithm. We will solve both equations using the canonical form.
Today we continue to study logarithmic equations and will analyze constructions when one logarithm is under the sign of another. We will solve both equations using the canonical form. Let me remind you that if we have the simplest logarithmic equation of the form log a f (x) = b, then to solve such an equation we perform the following steps. First of all, we need to replace the number b :
b = log a a b
Note: a b is an argument. Similarly, in the original equation, the argument is the function f(x). Then we rewrite the equation and get this construction:
log a f (x) = log a a b
Then we can perform the third step - get rid of the logarithm sign and simply write:
f (x) = a b
As a result, we get a new equation. In this case, no restrictions are imposed on the function f (x). For example, a logarithmic function can also take its place. And then we will again obtain a logarithmic equation, which we will again reduce to its simplest form and solve through the canonical form.
However, enough of the lyrics. Let's solve the real problem. So, task number 1:
log 2 (1 + 3 log 2 x ) = 2
As you can see, we have a simple logarithmic equation. The role of f (x) is the construction 1 + 3 log 2 x, and the role of the number b is the number 2 (the role of a is also played by two). Let's rewrite this two as follows:
It is important to understand that the first two twos came to us from the base of the logarithm, i.e. if there were 5 in the original equation, then we would get that 2 = log 5 5 2. In general, the base depends solely on the logarithm that was originally given in the problem. And in our case this is the number 2.
So, we rewrite our logarithmic equation taking into account the fact that the two on the right is actually also a logarithm. We get:
log 2 (1 + 3 log 2 x ) = log 2 4
Let's move on to the last step of our scheme - getting rid of the canonical form. You could say, we simply cross out the signs of log. However, from a mathematical point of view, it is impossible to “cross out log” - it would be more correct to say that we simply equate the arguments:
1 + 3 log 2 x = 4
From here we can easily find 3 log 2 x:
3 log 2 x = 3
log 2 x = 1
We have again obtained the simplest logarithmic equation, let's bring it back to the canonical form. To do this we need to make the following changes:
1 = log 2 2 1 = log 2 2
Why is there a two at the base? Because in our canonical equation on the left there is a logarithm precisely to base 2. We rewrite the problem taking into account this fact:
log 2 x = log 2 2
Again we get rid of the logarithm sign, i.e. we simply equate the arguments. We have the right to do this because the bases are the same, and no more additional actions were performed either on the right or on the left:
That's all! The problem is solved. We have found a solution to the logarithmic equation.
Note! Although the variable x appears in the argument (i.e., there are requirements for the domain of definition), we will not make any additional requirements.
As I said above, this check is redundant if the variable appears in only one argument of only one logarithm. In our case, x really appears only in the argument and only under one log sign. Therefore, no additional checks are required.
However, if you don't trust this method, you can easily verify that x = 2 is indeed a root. It is enough to substitute this number into the original equation.
Let's move on to the second equation, it's a little more interesting:
log 2 (log 1/2 (2x − 1) + log 2 4) = 1
If we denote the expression inside the large logarithm with the function f (x), we get the simplest logarithmic equation with which we started today's video lesson. Therefore, we can apply the canonical form, for which we will have to represent the unit in the form log 2 2 1 = log 2 2.
Let's rewrite our big equation:
log 2 (log 1/2 (2x − 1) + log 2 4) = log 2 2
Let's get away from the sign of the logarithm, equating the arguments. We have the right to do this, because both on the left and on the right the bases are the same. Additionally, note that log 2 4 = 2:
log 1/2 (2x − 1) + 2 = 2
log 1/2 (2x − 1) = 0
Before us again is the simplest logarithmic equation of the form log a f (x) = b. Let's move on to the canonical form, that is, we represent zero in the form log 1/2 (1/2)0 = log 1/2 1.
We rewrite our equation and get rid of the log sign, equating the arguments:
log 1/2 (2x − 1) = log 1/2 1
2x − 1 = 1
Again, we received an answer immediately. No additional checks are required because in the original equation only one logarithm contains the function as an argument.
Therefore, no additional checks are required. We can safely say that x = 1 is the only root of this equation.
But if in the second logarithm there was some function of x instead of four (or 2x was not in the argument, but in the base) - then it would be necessary to check the domain of definition. Otherwise, there is a high chance of running into extra roots.
Where do these extra roots come from? This point must be understood very clearly. Take a look at the original equations: everywhere the function x is under the logarithm sign. Consequently, since we wrote down log 2 x, we automatically set the requirement x > 0. Otherwise, this entry simply does not make sense.
However, as we solve the logarithmic equation, we get rid of all the log signs and get simple constructions. There are no restrictions set here, because the linear function is defined for any value of x.
It is this problem, when the final function is defined everywhere and always, but the original one is not defined everywhere and not always, that is the reason why extra roots very often arise in solving logarithmic equations.
But I repeat once again: this only happens in a situation where the function is either in several logarithms or at the base of one of them. In the problems that we are considering today, there are, in principle, no problems with expanding the domain of definition.
Cases of different grounds
This lesson is dedicated to more complex structures. Logarithms in today's equations will no longer be solved straight away; some transformations will need to be done first.
We begin solving logarithmic equations with completely different bases, which are not exact powers of each other. Don’t let such problems scare you - they are no more difficult to solve than the simplest designs that we discussed above.
But before moving directly to the problems, let me remind you of the formula for solving the simplest logarithmic equations using the canonical form. Consider a problem like this:
log a f(x) = b
It is important that the function f (x) is just a function, and the role of numbers a and b should be numbers (without any variables x). Of course, literally in a minute we will look at such cases when instead of variables a and b there are functions, but that’s not about that now.
As we remember, the number b must be replaced by a logarithm to the same base a, which is on the left. This is done very simply:
b = log a a b
Of course, the words “any number b” and “any number a” mean values that satisfy the scope of definition. In particular, in this equation we are talking only about the base a > 0 and a ≠ 1.
However, this requirement is satisfied automatically, because the original problem already contains a logarithm to base a - it will certainly be greater than 0 and not equal to 1. Therefore, we continue solving the logarithmic equation:
log a f (x) = log a a b
Such a notation is called canonical form. Its convenience lies in the fact that we can immediately get rid of the log sign by equating the arguments:
f (x) = a b
It is this technique that we will now use to solve logarithmic equations with a variable base. So, let's go!
log 2 (x 2 + 4x + 11) = log 0.5 0.125
What's next? Someone will now say that you need to calculate the right logarithm, or reduce them to the same base, or something else. And indeed, now we need to bring both bases to the same form - either 2 or 0.5. But let's learn the following rule once and for all:
If a logarithmic equation contains decimals, be sure to convert these fractions from decimal notation to ordinary ones. This transformation can greatly simplify the solution.
Such a transition must be performed immediately, even before performing any actions or transformations. Let's get a look:
log 2 (x 2 + 4x + 11) = log 1 /2 1/8
What does such a record give us? We can represent 1/2 and 1/8 as powers with a negative exponent:
[Caption for the picture] Before us is the canonical form. We equate the arguments and get the classic quadratic equation:
x 2 + 4x + 11 = 8
x 2 + 4x + 3 = 0
We have before us the following quadratic equation, which can be easily solved using Vieta’s formulas. In high school, you should see similar displays literally orally:
(x + 3)(x + 1) = 0
x 1 = −3
x 2 = −1
That's all! The original logarithmic equation has been solved. We got two roots.
Let me remind you that in this case it is not necessary to determine the domain of definition, since the function with the variable x is present in only one argument. Therefore, the definition scope is performed automatically.
So, the first equation is solved. Let's move on to the second:
log 0.5 (5x 2 + 9x + 2) = log 3 1/9
log 1/2 (5x 2 + 9x + 2) = log 3 9 −1
Now note that the argument of the first logarithm can also be written as a power with a negative exponent: 1/2 = 2 −1. Then you can take out the powers on both sides of the equation and divide everything by −1:
[Caption for the picture] And now we have completed a very important step in solving the logarithmic equation. Perhaps someone didn't notice something, so let me explain.
Look at our equation: both on the left and on the right there is a log sign, but on the left there is a logarithm to base 2, and on the right there is a logarithm to base 3. Three is not an integer power of two and, conversely, you cannot write that 2 is 3 in an integer degrees.
Consequently, these are logarithms with different bases that cannot be reduced to each other by simply adding powers. The only way to solve such problems is to get rid of one of these logarithms. In this case, since we are still considering fairly simple problems, the logarithm on the right was simply calculated, and we got the simplest equation - exactly the one we talked about at the very beginning of today's lesson.
Let's represent the number 2, which is on the right, as log 2 2 2 = log 2 4. And then we get rid of the logarithm sign, after which we are simply left with a quadratic equation:
log 2 (5x 2 + 9x + 2) = log 2 4
5x 2 + 9x + 2 = 4
5x 2 + 9x − 2 = 0
We have before us an ordinary quadratic equation, but it is not reduced because the coefficient of x 2 is different from unity. Therefore, we will solve it using a discriminant:
D = 81 − 4 5 (−2) = 81 + 40 = 121
x 1 = (−9 + 11)/10 = 2/10 = 1/5
x 2 = (−9 − 11)/10 = −2
That's all! We have found both roots, which means we have obtained a solution to the original logarithmic equation. Indeed, in the original problem, the function with variable x is present in only one argument. Consequently, no additional checks on the domain of definition are required - both roots that we found certainly meet all possible restrictions.
This could be the end of today’s video lesson, but in conclusion I would like to say again: be sure to convert all decimal fractions to ordinary fractions when solving logarithmic equations. In most cases, this greatly simplifies their solution.
Rarely, very rarely, do you come across problems in which getting rid of decimal fractions only complicates the calculations. However, in such equations, as a rule, it is initially clear that there is no need to get rid of decimal fractions.
In most other cases (especially if you are just starting to practice solving logarithmic equations), feel free to get rid of the decimals and convert them to ordinary ones. Because practice shows that in this way you will significantly simplify the subsequent solution and calculations.
Subtleties and tricks of the solution
Today we move on to more complex tasks and we will solve a logarithmic equation, the basis of which is not a number, but a function.
And even if this function is linear, small changes will have to be made to the solution scheme, the meaning of which boils down to additional requirements imposed on the domain of definition of the logarithm.
Complex tasks
This tutorial will be quite long. In it we will analyze two rather serious logarithmic equations, when solving which many students make mistakes. During my practice as a math tutor, I constantly encountered two types of errors:
- The appearance of extra roots due to the expansion of the domain of definition of logarithms. To avoid such offensive mistakes, just carefully monitor each transformation;
- Loss of roots due to the fact that the student forgot to consider some “subtle” cases - these are the situations we will focus on today.
This is the last lesson on logarithmic equations. It will be long, we will analyze complex logarithmic equations. Make yourself comfortable, make yourself some tea, and let's get started.
The first equation looks quite standard:
log x + 1 (x − 0.5) = log x − 0.5 (x + 1)
Let's immediately note that both logarithms are inverted copies of each other. Let's remember the wonderful formula:
log a b = 1/log b a
However, this formula has a number of limitations that arise if instead of the numbers a and b there are functions of the variable x:
b > 0
1 ≠ a > 0
These requirements apply to the base of the logarithm. On the other hand, in a fraction we are required to have 1 ≠ a > 0, since not only is the variable a in the argument of the logarithm (hence a > 0), but the logarithm itself is in the denominator of the fraction. But log b 1 = 0, and the denominator must be non-zero, so a ≠ 1.
So, the restrictions on the variable a remain. But what happens to the variable b? On the one hand, the base implies b > 0, on the other hand, the variable b ≠ 1, because the base of the logarithm must be different from 1. In total, from the right side of the formula it follows that 1 ≠ b > 0.
But here’s the problem: the second requirement (b ≠ 1) is missing from the first inequality, which deals with the left logarithm. In other words, when performing this transformation we must check separately, that the argument b is different from one!
So let's check it out. Let's apply our formula:
[Caption for the picture] 1 ≠ x − 0.5 > 0; 1 ≠ x + 1 > 0
So we got that already from the original logarithmic equation it follows that both a and b must be greater than 0 and not equal to 1. This means that we can easily invert the logarithmic equation:
I suggest introducing a new variable:
log x + 1 (x − 0.5) = t
In this case, our construction will be rewritten as follows:
(t 2 − 1)/t = 0
Note that in the numerator we have the difference of squares. We reveal the difference of squares using the abbreviated multiplication formula:
(t − 1)(t + 1)/t = 0
A fraction is equal to zero when its numerator is zero and its denominator is non-zero. But the numerator contains a product, so we equate each factor to zero:
t 1 = 1;
t 2 = −1;
t ≠ 0.
As we can see, both values of the variable t suit us. However, the solution does not end there, because we need to find not t, but the value of x. We return to the logarithm and get:
log x + 1 (x − 0.5) = 1;
log x + 1 (x − 0.5) = −1.
Let's put each of these equations in canonical form:
log x + 1 (x − 0.5) = log x + 1 (x + 1) 1
log x + 1 (x − 0.5) = log x + 1 (x + 1) −1
We get rid of the logarithm sign in the first case and equate the arguments:
x − 0.5 = x + 1;
x − x = 1 + 0.5;
Such an equation has no roots, therefore the first logarithmic equation also has no roots. But with the second equation everything is much more interesting:
(x − 0.5)/1 = 1/(x + 1)
Solving the proportion, we get:
(x − 0.5)(x + 1) = 1
Let me remind you that when solving logarithmic equations it is much more convenient to use all decimal fractions as ordinary ones, so let's rewrite our equation as follows:
(x − 1/2)(x + 1) = 1;
x 2 + x − 1/2x − 1/2 − 1 = 0;
x 2 + 1/2x − 3/2 = 0.
We have before us the quadratic equation below, it can be easily solved using Vieta’s formulas:
(x + 3/2) (x − 1) = 0;
x 1 = −1.5;
x 2 = 1.
We got two roots - they are candidates for solving the original logarithmic equation. In order to understand what roots will really go into the answer, let's return to the original problem. Now we will check each of our roots to see if they fit within the domain of definition:
1.5 ≠ x > 0.5; 0 ≠ x > −1.
These requirements are tantamount to a double inequality:
1 ≠ x > 0.5
From here we immediately see that the root x = −1.5 does not suit us, but x = 1 suits us quite well. Therefore x = 1 is the final solution to the logarithmic equation.
Let's move on to the second task:
log x 25 + log 125 x 5 = log 25 x 625
At first glance, it may seem that all logarithms have different bases and different arguments. What to do with such structures? First of all, note that the numbers 25, 5 and 625 are powers of 5:
25 = 5 2 ; 625 = 5 4
Now let's take advantage of the wonderful property of the logarithm. The point is that you can extract powers from an argument in the form of factors:
log a b n = n ∙ log a b
This transformation is also subject to restrictions in the case where b is replaced by a function. But for us, b is just a number, and no additional restrictions arise. Let's rewrite our equation:
2 ∙ log x 5 + log 125 x 5 = 4 ∙ log 25 x 5
We have obtained an equation with three terms containing the log sign. Moreover, the arguments of all three logarithms are equal.
It's time to reverse the logarithms to bring them to the same base - 5. Since the variable b is a constant, no changes in the domain of definition occur. We just rewrite:
[Caption for the picture] As expected, the same logarithms appeared in the denominator. I suggest replacing the variable:
log 5 x = t
In this case, our equation will be rewritten as follows:
Let's write out the numerator and open the brackets:
2 (t + 3) (t + 2) + t (t + 2) − 4t (t + 3) = 2 (t 2 + 5t + 6) + t 2 + 2t − 4t 2 − 12t = 2t 2 + 10t + 12 + t 2 + 2t − 4t 2 − 12t = −t 2 + 12
Let's return to our fraction. The numerator must be zero:
[Caption for the picture] And the denominator is different from zero:
t ≠ 0; t ≠ −3; t ≠ −2
The last requirements are fulfilled automatically, since they are all “tied” to integers, and all answers are irrational.
So, fractional rational equation solved, the values of the variable t are found. Let's return to solving the logarithmic equation and remember what t is:
[Caption for the picture] We reduce this equation to canonical form and obtain a number with an irrational degree. Don’t let this confuse you - even such arguments can be equated:
[Caption for the picture] We got two roots. More precisely, two candidate answers - let's check them for compliance with the domain of definition. Since the base of the logarithm is the variable x, we require the following:
1 ≠ x > 0;
With the same success we assert that x ≠ 1/125, otherwise the base of the second logarithm will turn to unity. Finally, x ≠ 1/25 for the third logarithm.
In total, we received four restrictions:
1 ≠ x > 0; x ≠ 1/125; x ≠ 1/25
Now the question is: do our roots satisfy these requirements? Of course they satisfy! Because 5 to any power will be greater than zero, and the requirement x > 0 is satisfied automatically.
On the other hand, 1 = 5 0, 1/25 = 5 −2, 1/125 = 5 −3, which means that these restrictions for our roots (which, let me remind you, have an irrational number in the exponent) are also satisfied, and both answers are solutions to the problem.
So, we have the final answer. Key Points There are two in this problem:
- Be careful when flipping a logarithm when the argument and base are swapped. Such transformations impose unnecessary restrictions on the scope of definition.
- Don't be afraid to transform logarithms: they can not only be reversed, but also expanded using the sum formula and generally changed using any formulas that you studied when solving logarithmic expressions. However, always remember: some transformations expand the scope of definition, and some narrow them.
One of the elements of primitive level algebra is the logarithm. The name comes from the Greek language from the word “number” or “power” and means the power to which the number in the base must be raised to find the final number.
Types of logarithms
- log a b – logarithm of the number b to base a (a > 0, a ≠ 1, b > 0);
- log b – decimal logarithm (logarithm to base 10, a = 10);
- ln b – natural logarithm (logarithm to base e, a = e).
How to solve logarithms?
The logarithm of b to base a is an exponent that requires b to be raised to base a. The result obtained is pronounced like this: “logarithm of b to base a.” The solution to logarithmic problems is that you need to determine a given power in numbers by the indicated numbers. There are some basic rules to determine or solve the logarithm, as well as convert the notation itself. Using them, logarithmic equations are solved, derivatives are found, integrals are solved, and many other operations are carried out. Basically, the solution to the logarithm itself is its simplified notation. Below are the basic formulas and properties:
For any a ; a > 0; a ≠ 1 and for any x ; y > 0.
- a log a b = b – basic logarithmic identity
- log a 1 = 0
- loga a = 1
- log a (x y) = log a x + log a y
- log a x/ y = log a x – log a y
- log a 1/x = -log a x
- log a x p = p log a x
- log a k x = 1/k log a x , for k ≠ 0
- log a x = log a c x c
- log a x = log b x/ log b a – formula for moving to a new base
- log a x = 1/log x a
How to solve logarithms - step-by-step instructions for solving
- First, write down the required equation.
Please note: if the base logarithm is 10, then the entry is shortened, resulting in a decimal logarithm. If it's worth natural number e, then we write it down, abbreviating it to natural logarithm. This means that the result of all logarithms is the power to which the base number is raised to obtain the number b.
Directly, the solution lies in calculating this degree. Before solving an expression with a logarithm, it must be simplified according to the rule, that is, using formulas. You can find the main identities by going back a little in the article.
Adding and subtracting logarithms with two different numbers, but with on the same grounds, replace with one logarithm with the product or division of the numbers b and c, respectively. In this case, you can apply the formula for moving to another base (see above).
If you use expressions to simplify a logarithm, there are some limitations to consider. And that is: the base of the logarithm a is only a positive number, but not equal to one. The number b, like a, must be greater than zero.
There are cases where, by simplifying an expression, you will not be able to calculate the logarithm numerically. It happens that such an expression does not make sense, because many powers are irrational numbers. Under this condition, leave the power of the number as a logarithm.
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Today we will talk about logarithmic formulas and we will give indicative solution examples.
They themselves imply solution patterns according to the basic properties of logarithms. Before applying logarithm formulas to solve, let us remind you of all the properties:
Now, based on these formulas (properties), we will show examples of solving logarithms.
Examples of solving logarithms based on formulas.
Logarithm a positive number b to base a (denoted by log a b) is an exponent to which a must be raised to get b, with b > 0, a > 0, and 1.
According to the definition, log a b = x, which is equivalent to a x = b, therefore log a a x = x.
Logarithms, examples:
log 2 8 = 3, because 2 3 = 8
log 7 49 = 2, because 7 2 = 49
log 5 1/5 = -1, because 5 -1 = 1/5
Decimal logarithm- this is an ordinary logarithm, the base of which is 10. It is denoted as lg.
log 10 100 = 2, because 10 2 = 100
Natural logarithm- also an ordinary logarithm, a logarithm, but with the base e (e = 2.71828... - an irrational number). Denoted as ln.
It is advisable to memorize the formulas or properties of logarithms, because we will need them later when solving logarithms, logarithmic equations and inequalities. Let's work through each formula again with examples.
- Basic logarithmic identity
a log a b = b8 2log 8 3 = (8 2log 8 3) 2 = 3 2 = 9
- Logarithm of the product equal to the sum logarithms
log a (bc) = log a b + log a clog 3 8.1 + log 3 10 = log 3 (8.1*10) = log 3 81 = 4
- The logarithm of the quotient is equal to the difference of the logarithms
log a (b/c) = log a b - log a c9 log 5 50 /9 log 5 2 = 9 log 5 50- log 5 2 = 9 log 5 25 = 9 2 = 81
- Properties of the power of a logarithmic number and the base of the logarithm
Exponent of the logarithmic number log a b m = mlog a b
Exponent of the base of the logarithm log a n b =1/n*log a b
log a n b m = m/n*log a b,
if m = n, we get log a n b n = log a b
log 4 9 = log 2 2 3 2 = log 2 3
- Transition to a new foundation
log a b = log c b/log c a,if c = b, we get log b b = 1
then log a b = 1/log b a
log 0.8 3*log 3 1.25 = log 0.8 3*log 0.8 1.25/log 0.8 3 = log 0.8 1.25 = log 4/5 5/4 = -1
As you can see, the formulas for logarithms are not as complicated as they seem. Now, having looked at examples of solving logarithms, we can move on to logarithmic equations. We will look at examples of solving logarithmic equations in more detail in the article: "". Do not miss!
If you still have questions about the solution, write them in the comments to the article.
Note: we decided to get a different class of education and study abroad as an option.
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.