It will be useful for every student who is preparing for the Unified State Exam in mathematics to repeat the topic “Finding an angle between straight lines.” As statistics show, when passing the certification test, tasks on this section Stereometry causes difficulties for large quantity students. At the same time, tasks that require finding the angle between straight lines are found in the Unified State Examination of both basic and profile level. This means that everyone should be able to solve them.
Basic moments
There are 4 types in space relative position straight They can coincide, intersect, be parallel or intersecting. The angle between them can be acute or straight.
To find the angle between lines in the Unified State Exam or, for example, in solving, schoolchildren in Moscow and other cities can use several ways to solve problems in this section of stereometry. You can complete the task using classical constructions. To do this, it is worth learning the basic axioms and theorems of stereometry. The student needs to be able to reason logically and create drawings in order to bring the task to a planimetric problem.
You can also use the vector coordinate method using simple formulas, rules and algorithms. The main thing in this case is to perform all calculations correctly. Hone your skills in solving problems in stereometry and other areas school course will help you educational project"Shkolkovo".
This material is devoted to such a concept as the angle between two intersecting lines. In the first paragraph we will explain what it is and show it in illustrations. Then we will look at how you can find the sine, cosine of this angle and the angle itself (we will separately consider cases with a plane and three-dimensional space), we present necessary formulas and show with examples exactly how they are used in practice.
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In order to understand what the angle formed when two lines intersect is, we need to remember the very definition of angle, perpendicularity and point of intersection.
Definition 1
We call two lines intersecting if they have one common point. This point is called the point of intersection of two lines.
Each straight line is divided by an intersection point into rays. Both straight lines form 4 angles, two of which are vertical, and two are adjacent. If we know the measure of one of them, then we can determine the remaining ones.
Let's say we know that one of the angles is equal to α. In this case, the angle that is vertical with respect to it will also be equal to α. To find the remaining angles, we need to calculate the difference 180 ° - α. If α is equal to 90 degrees, then all angles will be right angles. Lines intersecting at right angles are called perpendicular (a separate article is devoted to the concept of perpendicularity).
Take a look at the picture:
Let's move on to formulating the main definition.
Definition 2
The angle formed by two intersecting lines is the measure of the smaller of the 4 angles that form these two lines.
From the definition we need to make important conclusion: the size of the angle in this case will be expressed by any real number in the interval (0, 90]. If the lines are perpendicular, then the angle between them will in any case be equal to 90 degrees.
The ability to find the measure of the angle between two intersecting lines is useful for solving many practical problems. The solution method can be chosen from several options.
To begin with we can take geometric methods. If we know something about supplementary angles, then we can relate them to the angle we need using the properties of equal or similar figures. For example, if we know the sides of a triangle and need to calculate the angle between the lines on which these sides are located, then the cosine theorem is suitable for our solution. If we have the condition right triangle, then for calculations we will also need knowledge of sine, cosine and tangent of an angle.
The coordinate method is also very convenient for solving problems of this type. Let us explain how to use it correctly.
We have a rectangular (Cartesian) coordinate system O x y, in which two straight lines are given. Let's denote them by letters a and b. The straight lines can be described using some equations. The original lines have an intersection point M. How to determine the required angle (let's denote it α) between these straight lines?
Let's start by formulating the basic principle of finding an angle under given conditions.
We know that the concept of a straight line is closely related to such concepts as a direction vector and a normal vector. If we have an equation of a certain line, we can take the coordinates of these vectors from it. We can do this for two intersecting lines at once.
The angle subtended by two intersecting lines can be found using:
- angle between direction vectors;
- angle between normal vectors;
- the angle between the normal vector of one line and the direction vector of the other.
Now let's look at each method separately.
1. Let us assume that we have a line a with a direction vector a → = (a x, a y) and a line b with a direction vector b → (b x, b y). Now let's plot two vectors a → and b → from the intersection point. After this we will see that they will each be located on their own straight line. Then we have four options for their relative arrangement. See illustration:
If the angle between two vectors is not obtuse, then it will be the angle we need between the intersecting lines a and b. If it is obtuse, then the desired angle will be equal to the angle adjacent to the angle a →, b → ^. Thus, α = a → , b → ^ if a → , b → ^ ≤ 90 ° , and α = 180 ° - a → , b → ^ if a → , b → ^ > 90 ° .
Based on the fact that the cosines equal angles are equal, we can rewrite the resulting equalities as follows: cos α = cos a → , b → ^ , if a → , b → ^ ≤ 90 ° ; cos α = cos 180 ° - a →, b → ^ = - cos a →, b → ^, if a →, b → ^ > 90 °.
In the second case, reduction formulas were used. Thus,
cos α cos a → , b → ^ , cos a → , b → ^ ≥ 0 - cos a → , b → ^ , cos a → , b → ^< 0 ⇔ cos α = cos a → , b → ^
Let's write the last formula in words:
Definition 3
The cosine of the angle formed by two intersecting lines will be equal to modulus cosine of the angle between its direction vectors.
The general form of the formula for the cosine of the angle between two vectors a → = (a x , a y) and b → = (b x , b y) looks like this:
cos a → , b → ^ = a → , b → ^ a → b → = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2
From it we can derive the formula for the cosine of the angle between two given straight lines:
cos α = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2 = a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2
Then the angle itself can be found by the following formula:
α = a r c cos a x b x + a y + b y a x 2 + a y 2 b x 2 + b y 2
Here a → = (a x , a y) and b → = (b x , b y) are the direction vectors of the given lines.
Let's give an example of solving the problem.
Example 1
In a rectangular coordinate system on a plane, two intersecting lines a and b are given. They can be described by the parametric equations x = 1 + 4 · λ y = 2 + λ λ ∈ R and x 5 = y - 6 - 3. Calculate the angle between these lines.
Solution
We have in our condition parametric equation, which means that for this line we can immediately write down the coordinates of its direction vector. To do this, we need to take the values of the coefficients for the parameter, i.e. the straight line x = 1 + 4 · λ y = 2 + λ λ ∈ R will have a direction vector a → = (4, 1).
The second straight line is described using canonical equation x 5 = y - 6 - 3 . Here we can take the coordinates from the denominators. Thus, this line has a direction vector b → = (5 , - 3) .
Next, we move directly to finding the angle. To do this, simply substitute the existing coordinates of the two vectors into the above formula α = a r c cos a x · b x + a y + b y a x 2 + a y 2 · b x 2 + b y 2 . We get the following:
α = a r c cos 4 5 + 1 (- 3) 4 2 + 1 2 5 2 + (- 3) 2 = a r c cos 17 17 34 = a r c cos 1 2 = 45 °
Answer: These straight lines form an angle of 45 degrees.
We can solve a similar problem by finding the angle between normal vectors. If we have a line a with a normal vector n a → = (n a x , n a y) and a line b with a normal vector n b → = (n b x , n b y), then the angle between them will be equal to the angle between n a → and n b → or the angle that will be adjacent to n a →, n b → ^. This method is shown in the picture:
Formulas for calculating the cosine of the angle between intersecting lines and this angle itself using the coordinates of normal vectors look like this:
cos α = cos n a → , n b → ^ = n a x n b x + n a y + n b y n a x 2 + n a y 2 n b x 2 + n b y 2 α = a r c cos n a x n b x + n a y + n b y n a x 2 + n a y 2 n b x 2 + n b y 2
Here n a → and n b → denote the normal vectors of two given lines.
Example 2
In a rectangular coordinate system, two straight lines are given using the equations 3 x + 5 y - 30 = 0 and x + 4 y - 17 = 0. Find the sine and cosine of the angle between them and the magnitude of this angle itself.
Solution
The original lines are specified using normal equations straight line of the form A x + B y + C = 0. We denote the normal vector as n → = (A, B). Let's find the coordinates of the first normal vector for one line and write them: n a → = (3, 5) . For the second line x + 4 y - 17 = 0, the normal vector will have coordinates n b → = (1, 4). Now let’s add the obtained values to the formula and calculate the total:
cos α = cos n a → , n b → ^ = 3 1 + 5 4 3 2 + 5 2 1 2 + 4 2 = 23 34 17 = 23 2 34
If we know the cosine of an angle, then we can calculate its sine using the basic trigonometric identity. Since the angle α formed by straight lines is not obtuse, then sin α = 1 - cos 2 α = 1 - 23 2 34 2 = 7 2 34.
In this case, α = a r c cos 23 2 34 = a r c sin 7 2 34.
Answer: cos α = 23 2 34, sin α = 7 2 34, α = a r c cos 23 2 34 = a r c sin 7 2 34
Let's sort it out last case– finding the angle between straight lines if we know the coordinates of the direction vector of one straight line and the normal vector of the other.
Let us assume that straight line a has a direction vector a → = (a x , a y) , and straight line b has a normal vector n b → = (n b x , n b y) . We need to set these vectors aside from the intersection point and consider all options for their relative positions. See in the picture:
If the angle between the given vectors is no more than 90 degrees, it turns out that it will complement the angle between a and b to a right angle.
a → , n b → ^ = 90 ° - α if a → , n b → ^ ≤ 90 ° .
If it is less than 90 degrees, then we get the following:
a → , n b → ^ > 90 ° , then a → , n b → ^ = 90 ° + α
Using the rule of equality of cosines of equal angles, we write:
cos a → , n b → ^ = cos (90 ° - α) = sin α for a → , n b → ^ ≤ 90 ° .
cos a → , n b → ^ = cos 90 ° + α = - sin α for a → , n b → ^ > 90 ° .
Thus,
sin α = cos a → , n b → ^ , a → , n b → ^ ≤ 90 ° - cos a → , n b → ^ , a → , n b → ^ > 90 ° ⇔ sin α = cos a → , n b → ^ , a → , n b → ^ > 0 - cos a → , n b → ^ , a → , n b → ^< 0 ⇔ ⇔ sin α = cos a → , n b → ^
Let us formulate a conclusion.
Definition 4
To find the sine of the angle between two lines intersecting on a plane, you need to calculate the modulus of the cosine of the angle between the direction vector of the first line and the normal vector of the second.
Let's write it down necessary formulas. Finding the sine of an angle:
sin α = cos a → , n b → ^ = a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2
Finding the angle itself:
α = a r c sin = a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2
Here a → is the direction vector of the first line, and n b → is the normal vector of the second.
Example 3
Two intersecting lines are given by the equations x - 5 = y - 6 3 and x + 4 y - 17 = 0. Find the angle of intersection.
Solution
We take the coordinates of the guide and normal vector from the given equations. It turns out a → = (- 5, 3) and n → b = (1, 4). We take the formula α = a r c sin = a x n b x + a y n b y a x 2 + a y 2 n b x 2 + n b y 2 and calculate:
α = a r c sin = - 5 1 + 3 4 (- 5) 2 + 3 2 1 2 + 4 2 = a r c sin 7 2 34
Note that we took the equations from previous task and got exactly the same result, but in a different way.
Answer:α = a r c sin 7 2 34
Let us present another way to find the desired angle using the angular coefficients of given straight lines.
We have a line a, which is defined in a rectangular coordinate system using the equation y = k 1 x + b 1, and a line b, defined as y = k 2 x + b 2. These are equations of lines with slopes. To find the angle of intersection, we use the formula:
α = a r c cos k 1 · k 2 + 1 k 1 2 + 1 · k 2 2 + 1, where k 1 and k 2 are angle coefficients given straight lines. To obtain this record, formulas for determining the angle through the coordinates of normal vectors were used.
Example 4
There are two straight lines intersecting in a plane, given by equations y = - 3 5 x + 6 and y = - 1 4 x + 17 4 . Calculate the value of the intersection angle.
Solution
The angular coefficients of our lines are equal to k 1 = - 3 5 and k 2 = - 1 4. Let's add them to the formula α = a r c cos k 1 k 2 + 1 k 1 2 + 1 k 2 2 + 1 and calculate:
α = a r c cos - 3 5 · - 1 4 + 1 - 3 5 2 + 1 · - 1 4 2 + 1 = a r c cos 23 20 34 24 · 17 16 = a r c cos 23 2 34
Answer:α = a r c cos 23 2 34
In the conclusions of this paragraph, it should be noted that the formulas for finding the angle given here do not have to be learned by heart. To do this, it is enough to know the coordinates of the guides and/or normal vectors of given lines and be able to determine them by different types equations. But it’s better to remember or write down the formulas for calculating the cosine of an angle.
How to calculate the angle between intersecting lines in space
The calculation of such an angle can be reduced to calculating the coordinates of the direction vectors and determining the magnitude of the angle formed by these vectors. For such examples, the same reasoning that we gave before is used.
Let's say we have rectangular system coordinates located at three-dimensional space. It contains two straight lines a and b with an intersection point M. To calculate the coordinates of the direction vectors, we need to know the equations of these lines. Let us denote the direction vectors a → = (a x , a y , a z) and b → = (b x , b y , b z) . To calculate the cosine of the angle between them, we use the formula:
cos α = cos a → , b → ^ = a → , b → a → b → = a x b x + a y b y + a z b z a x 2 + a y 2 + a z 2 b x 2 + b y 2 + b z 2
To find the angle itself, we need this formula:
α = a r c cos a x b x + a y b y + a z b z a x 2 + a y 2 + a z 2 b x 2 + b y 2 + b z 2
Example 5
We have a line defined in three-dimensional space using the equation x 1 = y - 3 = z + 3 - 2. It is known that it intersects with the O z axis. Calculate the intercept angle and the cosine of that angle.
Solution
Let us denote the angle that needs to be calculated by the letter α. Let's write down the coordinates of the direction vector for the first straight line – a → = (1, - 3, - 2) . For the axis applicate we can take coordinate vector k → = (0, 0, 1) as a guide. We have received the necessary data and can add it to the desired formula:
cos α = cos a → , k → ^ = a → , k → a → k → = 1 0 - 3 0 - 2 1 1 2 + (- 3) 2 + (- 2) 2 0 2 + 0 2 + 1 2 = 2 8 = 1 2
As a result, we found that the angle we need will be equal to a r c cos 1 2 = 45 °.
Answer: cos α = 1 2 , α = 45 ° .
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With this video I begin a long series of lessons about logarithmic equations. Now you have three examples in front of you, on the basis of which we will learn to solve the most simple tasks, which are called so - protozoa.
log 0.5 (3x − 1) = −3
log (x + 3) = 3 + 2 log 5
Let me remind you that the simplest logarithmic equation is the following:
log a f (x) = b
In this case, it is important that the variable x is present only inside the argument, that is, only in the function f (x). And the numbers a and b are just numbers, and in no case are functions containing the variable x.
Basic solution methods
There are many ways to solve such structures. For example, most teachers at school offer this method: Immediately express the function f (x) using the formula f ( x ) = a b . That is, when you come across the simplest construction, you can immediately move on to the solution without additional actions and constructions.
Yes, of course, the decision will be correct. However, the problem with this formula is that most students do not understand, where it comes from and why we raise the letter a to the letter b.
As a result, I often see very annoying mistakes when, for example, these letters are swapped. This formula you need to either understand or cram, and the second method leads to mistakes at the most inopportune and most crucial moments: in exams, tests, etc.
That is why I suggest that all my students abandon the standard school formula and use logarithmic equations the second approach, which, as you probably guessed from the name, is called canonical form.
The idea of the canonical form is simple. Let's look at our problem again: on the left we have log a, and by the letter a we mean a number, and in no case a function containing the variable x. Consequently, this letter is subject to all the restrictions that are imposed on the base of the logarithm. namely:
1 ≠ a > 0
On the other hand, from the same equation we see that the logarithm must be equal to the number b , and no restrictions are imposed on this letter, because it can take any values - both positive and negative. It all depends on what values the function f(x) takes.
And here we remember our wonderful rule that any number b can be represented as a logarithm to the base a of a to the power of b:
b = log a a b
How to remember this formula? Yes, very simple. Let's write the following construction:
b = b 1 = b log a a
Of course, in this case all the restrictions that we wrote down at the beginning arise. Now let's use the basic property of the logarithm and introduce the multiplier b as the power of a. We get:
b = b 1 = b log a a = log a a b
As a result, the original equation will be rewritten as follows:
log a f (x) = log a a b → f (x) = a b
That's all. New feature no longer contains a logarithm and can be solved using standard algebraic techniques.
Of course, someone will now object: why was it necessary to come up with some kind of canonical formula at all, why perform two additional unnecessary steps if it was possible to immediately move from the original design to the final formula? Yes, if only because most students do not understand where this formula comes from and, as a result, regularly make mistakes when applying it.
But this sequence of actions, consisting of three steps, allows you to solve the original logarithmic equation, even if you do not understand where the final formula comes from. By the way, this entry is called the canonical formula:
log a f (x) = log a a b
The convenience of the canonical form also lies in the fact that it can be used to solve a very wide class of logarithmic equations, and not just the simplest ones that we are considering today.
Examples of solutions
Now let's take a look real examples. So, let's decide:
log 0.5 (3x − 1) = −3
Let's rewrite it like this:
log 0.5 (3x − 1) = log 0.5 0.5 −3
Many students are in a hurry and try to immediately raise the number 0.5 to the power that came to us from the original problem. Indeed, when you are already well trained in solving such problems, you can immediately perform this step.
However, if you are now just starting to study this topic, it is better not to rush anywhere in order to avoid making offensive mistakes. So, we have the canonical form. We have:
3x − 1 = 0.5 −3
This is no longer a logarithmic equation, but linear with respect to the variable x. To solve it, let's first look at the number 0.5 to the power of −3. Note that 0.5 is 1/2.
(1/2) −3 = (2/1) 3 = 8
All decimals convert to ordinary ones when you solve a logarithmic equation.
We rewrite and get:
3x − 1 = 8
3x = 9
x = 3
That's it, we got the answer. The first problem has been solved.
Second task
Let's move on to the second task:
As we see, this equation is no longer the simplest. If only because there is a difference on the left, and not a single logarithm to one base.
Therefore, we need to somehow get rid of this difference. IN in this case everything is very simple. Let's take a closer look at the bases: on the left is the number under the root:
General recommendation: in all logarithmic equations, try to get rid of radicals, i.e., from entries with roots and move on to power functions, simply because the exponents of these powers are easily taken out of the sign of the logarithm and, ultimately, such a notation significantly simplifies and speeds up calculations. Let's write it down like this:
Now we remember wonderful property logarithm: powers can be derived from the argument, as well as from the base. In the case of grounds, the following happens:
log a k b = 1/k loga b
In other words, the number that was in the base power is brought forward and at the same time inverted, i.e. it becomes reciprocal number. In our case, the base degree was 1/2. Therefore, we can take it out as 2/1. We get:
5 2 log 5 x − log 5 x = 18
10 log 5 x − log 5 x = 18
Please note: under no circumstances should you get rid of logarithms at this step. Remember 4th-5th grade math and the order of operations: multiplication is performed first, and only then addition and subtraction. In this case, we subtract one of the same elements from 10 elements:
9 log 5 x = 18
log 5 x = 2
Now our equation looks as it should. This is the simplest construction, and we solve it using the canonical form:
log 5 x = log 5 5 2
x = 5 2
x = 25
That's all. The second problem has been solved.
Third example
Let's move on to the third task:
log (x + 3) = 3 + 2 log 5
Let me remind you of the following formula:
log b = log 10 b
If for some reason you are confused by the notation log b , then when performing all the calculations you can simply write log 10 b . You can work with decimal logarithms in the same way as with others: take powers, add and represent any numbers in the form lg 10.
It is these properties that we will now use to solve the problem, since it is not the simplest one that we wrote down at the very beginning of our lesson.
First, note that the factor 2 in front of lg 5 can be added and becomes a power of base 5. In addition, the free term 3 can also be represented as a logarithm - this is very easy to observe from our notation.
Judge for yourself: any number can be represented as log to base 10:
3 = log 10 10 3 = log 10 3
Let's rewrite the original problem taking into account the obtained changes:
log (x − 3) = log 1000 + log 25
log (x − 3) = log 1000 25
log (x − 3) = log 25,000
Before us again is the canonical form, and we obtained it without going through the transformation stage, i.e. the simplest logarithmic equation did not appear anywhere.
This is exactly what I talked about at the very beginning of the lesson. The canonical form allows you to solve a wider class of problems than the standard one school formula, which is given by most school teachers.
That's it, let's get rid of the sign decimal logarithm, and we get a simple linear construction:
x + 3 = 25,000
x = 24,997
All! The problem is solved.
A note on scope
Here I would like to make an important remark regarding the scope of definition. Surely now there will be students and teachers who will say: “When we solve expressions with logarithms, we must remember that the argument f (x) must be greater than zero!” In this regard, a logical question arises: why did we not require this inequality to be satisfied in any of the problems considered?
Do not worry. None extra roots will not occur in these cases. And this is another great trick that allows you to speed up the solution. Just know that if in the problem the variable x occurs only in one place (or rather, in one single argument of a single logarithm), and nowhere else in our case does the variable x appear, then write down the domain of definition no need, because it will be executed automatically.
Judge for yourself: in the first equation we got that 3x − 1, i.e. the argument should be equal to 8. This automatically means that 3x − 1 will be greater than zero.
With the same success we can write that in the second case x should be equal to 5 2, i.e. it is certainly greater than zero. And in the third case, where x + 3 = 25,000, i.e., again, obviously greater than zero. In other words, the scope is satisfied automatically, but only if x occurs only in the argument of only one logarithm.
That's all you need to know to solve the simplest problems. This rule alone, together with the transformation rules, will allow you to solve a very wide class of problems.
But let's be honest: in order to finally understand this technique, to learn how to apply the canonical form of the logarithmic equation, it is not enough to just watch one video lesson. So download the options right now for independent decision, which are attached to this video lesson and start solving at least one of these two independent works.
It will take you literally a few minutes. But the effect of such training will be much higher than if you simply watched this video lesson.
I hope this lesson will help you understand logarithmic equations. Use the canonical form, simplify expressions using the rules for working with logarithms - and you won’t be afraid of any problems. That's all I have for today.
Taking into account the domain of definition
Now let's talk about the domain of definition logarithmic function, as well as how this affects the solution of logarithmic equations. Consider a construction of the form
log a f (x) = b
Such an expression is called the simplest - it contains only one function, and the numbers a and b are just numbers, and in no case a function that depends on the variable x. It can be solved very simply. You just need to use the formula:
b = log a a b
This formula is one of the key properties of the logarithm, and when substituting into our original expression we get the following:
log a f (x) = log a a b
f (x) = a b
This is a familiar formula from school textbooks. Many students will probably have a question: since in the original expression the function f (x) is under the log sign, the following restrictions are imposed on it:
f(x) > 0
This limitation applies because the logarithm of negative numbers does not exist. So, perhaps, as a result of this limitation, a check on answers should be introduced? Perhaps they need to be inserted into the source?
No, in the simplest logarithmic equations additional checking is unnecessary. And that's why. Take a look at our final formula:
f (x) = a b
The fact is that the number a is in any case greater than 0 - this requirement is also imposed by the logarithm. The number a is the base. In this case, no restrictions are imposed on the number b. But this doesn’t matter, because no matter what degree we raise positive number, we will still get a positive number at the output. Thus, the requirement f (x) > 0 is satisfied automatically.
What's really worth checking is the domain of the function under the log sign. There may be quite complex structures, and you definitely need to keep an eye on them during the solution process. Let's get a look.
First task:
First step: convert the fraction on the right. We get:
We get rid of the logarithm sign and get the usual irrational equation:
Of the obtained roots, only the first one suits us, since the second root is less than zero. The only answer will be the number 9. That's it, the problem is solved. No additional checks are required to ensure that the expression under the logarithm sign is greater than 0, because it is not just greater than 0, but according to the condition of the equation it is equal to 2. Therefore, the requirement “greater than zero” is satisfied automatically.
Let's move on to the second task:
Everything is the same here. We rewrite the construction, replacing the triple:
We get rid of the logarithm signs and get an irrational equation:
We square both sides taking into account the restrictions and get:
4 − 6x − x 2 = (x − 4) 2
4 − 6x − x 2 = x 2 + 8x + 16
x 2 + 8x + 16 −4 + 6x + x 2 = 0
2x 2 + 14x + 12 = 0 |:2
x 2 + 7x + 6 = 0
We solve the resulting equation through the discriminant:
D = 49 − 24 = 25
x 1 = −1
x 2 = −6
But x = −6 does not suit us, because if we substitute this number into our inequality, we get:
−6 + 4 = −2 < 0
In our case, it is required that it be greater than 0 or, in extreme cases, equal. But x = −1 suits us:
−1 + 4 = 3 > 0
The only answer in our case will be x = −1. That's the solution. Let's go back to the very beginning of our calculations.
The main takeaway from this lesson is that you don't need to check constraints on a function in simple logarithmic equations. Because during the solution process all constraints are satisfied automatically.
However, this in no way means that you can forget about checking altogether. In the process of working on a logarithmic equation, it may well turn into an irrational one, which will have its own restrictions and requirements for the right side, which we have seen today in two different examples.
Feel free to solve such problems and be especially careful if there is a root in the argument.
Logarithmic equations with different bases
We continue to study logarithmic equations and look at two more quite interesting techniques with which it is fashionable to solve more complex designs. But first, let’s remember how the simplest problems are solved:
log a f (x) = b
In this entry, a and b are numbers, and in the function f (x) the variable x must be present, and only there, that is, x must only be in the argument. We will transform such logarithmic equations using the canonical form. To do this, note that
b = log a a b
Moreover, a b is precisely an argument. Let's rewrite this expression as follows:
log a f (x) = log a a b
This is exactly what we are trying to achieve, so that there is a logarithm to base a on both the left and the right. In this case, we can, figuratively speaking, cross out the log signs, and from a mathematical point of view we can say that we are simply equating the arguments:
f (x) = a b
As a result, we will get a new expression that will be much easier to solve. Let's apply this rule to our problems today.
So, the first design:
First of all, I note that on the right is a fraction whose denominator is log. When you see an expression like this, it’s a good idea to remember a wonderful property of logarithms:
Translated into Russian, this means that any logarithm can be represented as the quotient of two logarithms with any base c. Of course 0< с ≠ 1.
So: this formula has one wonderful special case, when the variable c is equal to the variable b. In this case we get a construction like:
This is exactly the construction we see from the sign on the right in our equation. Let's replace this construction with log a b , we get:
In other words, in comparison with the original task, we swapped the argument and the base of the logarithm. Instead, we had to reverse the fraction.
We recall that any degree can be derived from the base according to the following rule:
In other words, the coefficient k, which is the power of the base, is expressed as an inverted fraction. Let's render it as an inverted fraction:
The fractional factor cannot be left in front, because in this case we will not be able to represent this entry as a canonical form (after all, in the canonical form there is no additional factor before the second logarithm). Therefore, let's add the fraction 1/4 to the argument as a power:
Now we equate arguments whose bases are the same (and our bases are really the same), and write:
x + 5 = 1
x = −4
That's all. We got the answer to the first logarithmic equation. Please note: in the original problem, the variable x appears in only one log, and it appears in its argument. Therefore, there is no need to check the domain, and our number x = −4 is indeed the answer.
Now let's move on to the second expression:
log 56 = log 2 log 2 7 − 3log (x + 4)
Here, in addition to the usual logarithms, we will have to work with log f (x). How to solve such an equation? To an unprepared student it may seem like this is some kind of tough task, but in fact everything can be solved in an elementary way.
Take a close look at the term lg 2 log 2 7. What can we say about it? The bases and arguments of log and lg are the same, and this should give some ideas. Let's remember once again how powers are taken out from under the sign of the logarithm:
log a b n = nlog a b
In other words, what was a power of b in the argument becomes a factor in front of log itself. Let's apply this formula to the expression lg 2 log 2 7. Don't be scared by lg 2 - this is the most common expression. You can rewrite it as follows:
All the rules that apply to any other logarithm are valid for it. In particular, the factor in front can be added to the degree of the argument. Let's write it down:
Very often, students do not see this action directly, because it is not good to enter one log under the sign of another. In fact, there is nothing criminal about this. Moreover, we get a formula that is easy to calculate if you remember an important rule:
This formula can be considered both as a definition and as one of its properties. In any case, if you are converting a logarithmic equation, you should know this formula just like you would know the log representation of any number.
Let's return to our task. We rewrite it taking into account the fact that the first term to the right of the equal sign will be simply equal to lg 7. We have:
lg 56 = lg 7 − 3lg (x + 4)
Let's move lg 7 to the left, we get:
lg 56 − lg 7 = −3lg (x + 4)
We subtract the expressions on the left because they have the same base:
lg (56/7) = −3lg (x + 4)
Now let's take a closer look at the equation we got. It is practically the canonical form, but there is a factor −3 on the right. Let's add it to the right lg argument:
log 8 = log (x + 4) −3
Before us is the canonical form of the logarithmic equation, so we cross out the lg signs and equate the arguments:
(x + 4) −3 = 8
x + 4 = 0.5
That's all! We solved the second logarithmic equation. In this case, no additional checks are required, because in the original problem x was present in only one argument.
I'll list it again key points this lesson.
The main formula that is taught in all the lessons on this page dedicated to solving logarithmic equations is the canonical form. And don’t be scared by the fact that in most school textbooks you are taught to solve similar tasks differently. This tool works very effectively and allows you to solve a much wider class of problems than the simplest ones that we studied at the very beginning of our lesson.
In addition, to solve logarithmic equations it will be useful to know the basic properties. Namely:
- The formula for moving to one base and the special case when we reverse log (this was very useful to us in the first problem);
- Formula for adding and subtracting powers from the logarithm sign. Here, many students get stuck and do not see that the degree taken out and introduced can itself contain log f (x). Nothing wrong with that. We can introduce one log according to the sign of the other and at the same time significantly simplify the solution of the problem, which is what we observe in the second case.
In conclusion, I would like to add that it is not necessary to check the domain of definition in each of these cases, because everywhere the variable x is present in only one sign of log, and at the same time is in its argument. As a consequence, all requirements of the scope are fulfilled automatically.
Problems with variable base
Today we will look at logarithmic equations, which for many students seem non-standard, if not completely unsolvable. It's about about expressions based not on numbers, but on variables and even functions. We will solve such constructions using our standard technique, namely through the canonical form.
To begin with, let us remember how the simplest problems are solved, based on regular numbers. So, the simplest construction is called
log a f (x) = b
To solve such problems we can use the following formula:
b = log a a b
We rewrite our original expression and get:
log a f (x) = log a a b
Then we equate the arguments, i.e. we write:
f (x) = a b
Thus, we get rid of the log sign and solve the usual problem. In this case, the roots obtained from the solution will be the roots of the original logarithmic equation. In addition, a record when both the left and the right are in the same logarithm with the same base is precisely called the canonical form. It is to such a record that we will try to reduce today's designs. So, let's go.
First task:
log x − 2 (2x 2 − 13x + 18) = 1
Replace 1 with log x − 2 (x − 2) 1 . The degree that we observe in the argument is actually the number b that stood to the right of the equal sign. Thus, let's rewrite our expression. We get:
log x − 2 (2x 2 − 13x + 18) = log x − 2 (x − 2)
What do we see? Before us is the canonical form of the logarithmic equation, so we can safely equate the arguments. We get:
2x 2 − 13x + 18 = x − 2
But the solution doesn't end there, because given equation not equivalent to the original one. After all, the resulting construction consists of functions that are defined on the entire number line, and our original logarithms are not defined everywhere and not always.
Therefore, we must write down the domain of definition separately. Let's not split hairs and first write down all the requirements:
First, the argument of each of the logarithms must be greater than 0:
2x 2 − 13x + 18 > 0
x − 2 > 0
Secondly, the base must not only be greater than 0, but also different from 1:
x − 2 ≠ 1
As a result, we get the system:
But don’t be alarmed: when processing logarithmic equations, such a system can be significantly simplified.
Judge for yourself: on the one hand, we are required that the quadratic function be greater than zero, and on the other hand, this quadratic function is equated to a certain linear expression, which is also required to be greater than zero.
In this case, if we require that x − 2 > 0, then the requirement 2x 2 − 13x + 18 > 0 will automatically be satisfied. Therefore, we can safely cross out the inequality containing quadratic function. Thus, the number of expressions contained in our system will be reduced to three.
Of course, we could just as well cross out linear inequality, that is, cross out x − 2 > 0 and demand that 2x 2 − 13x + 18 > 0. But you must agree that solving the simplest linear inequality is much faster and easier than quadratic, even if as a result of solving the entire this system we will get the same roots.
In general, try to optimize calculations whenever possible. And in the case of logarithmic equations, cross out the most difficult inequalities.
Let's rewrite our system:
Here is a system of three expressions, two of which we, in fact, have already dealt with. Let's write it down separately quadratic equation and let's solve it:
2x 2 − 14x + 20 = 0
x 2 − 7x + 10 = 0
Given before us quadratic trinomial and, therefore, we can use Vieta's formulas. We get:
(x − 5)(x − 2) = 0
x 1 = 5
x 2 = 2
Now we return to our system and find that x = 2 does not suit us, because we are required that x be strictly greater than 2.
But x = 5 suits us quite well: the number 5 is greater than 2, and at the same time 5 is not equal to 3. Therefore, the only solution of this system will be x = 5.
That's it, the problem is solved, including taking into account the ODZ. Let's move on to the second equation. More interesting and informative calculations await us here:
First step: as in last time, we bring this whole matter to canonical form. To do this, we can write the number 9 as follows:
You don’t have to touch the base with the root, but it’s better to transform the argument. Let's go from root to power c rational indicator. Let's write down:
Let me not rewrite our entire large logarithmic equation, but just immediately equate the arguments:
x 3 + 10x 2 + 31x + 30 = x 3 + 9x 2 + 27x + 27
x 2 + 4x + 3 = 0
Before us is a newly reduced quadratic trinomial, let’s use Vieta’s formulas and write:
(x + 3)(x + 1) = 0
x 1 = −3
x 2 = −1
So, we got the roots, but no one guaranteed us that they would fit the original logarithmic equation. After all, log signs impose additional restrictions(here we should have written down the system, but due to the cumbersome nature of the whole structure, I decided to calculate the domain of definition separately).
First of all, remember that the arguments must be greater than 0, namely:
These are the requirements imposed by the scope of definition.
Let us immediately note that since we equate the first two expressions of the system to each other, we can cross out any of them. Let's cross out the first one because it looks more threatening than the second one.
In addition, note that the solution to the second and third inequalities will be the same sets (the cube of some number is greater than zero, if this number itself is greater than zero; similarly, with a root of the third degree - these inequalities are completely analogous, so we can cross it out).
But with the third inequality this will not work. Let's get rid of the radical sign on the left by raising both parts to a cube. We get:
So we get the following requirements:
− 2 ≠ x > −3
Which of our roots: x 1 = −3 or x 2 = −1 meets these requirements? Obviously, only x = −1, because x = −3 does not satisfy the first inequality (since our inequality is strict). So, returning to our problem, we get one root: x = −1. That's it, problem solved.
Once again, the key points of this task:
- Feel free to apply and solve logarithmic equations using canonical form. Students who write this way, rather than go directly from the original problem to a construction like log a f (x) = b, allow much less mistakes than those who are in a hurry somewhere, skipping intermediate steps of calculations;
- As soon as the logarithm appears variable base, the task ceases to be the simplest. Therefore, when solving it, it is necessary to take into account the domain of definition: the arguments must be greater than zero, and the bases must not only be greater than 0, but they also must not be equal to 1.
The final requirements can be applied to the final answers in different ways. For example, you can solve an entire system containing all the requirements for the domain of definition. On the other hand, you can first solve the problem itself, and then remember the domain of definition, separately work it out in the form of a system and apply it to the obtained roots.
Which method to choose when solving a particular logarithmic equation is up to you. In any case, the answer will be the same.