Ways to solve systems of equations
To begin with, let us briefly recall what methods of solving systems of equations generally exist.
Exist four main ways solutions of systems of equations:
Substitution method: take any of these equations and express $y$ in terms of $x$, then $y$ is substituted into the equation of the system, from where the variable $x.$ is found. After that, we can easily calculate the variable $y.$
Addition method: in this method, it is necessary to multiply one or both equations by such numbers that when both are added together, one of the variables “disappears”.
Graphical method: both equations of the system are depicted on coordinate plane and find their point of intersection.
The method of introducing new variables: in this method, we do the replacement of some expressions to simplify the system, and then apply one of the above methods.
Systems of exponential equations
Definition 1
Systems of equations consisting of exponential equations are called a system of exponential equations.
We will consider the solution of systems of exponential equations using examples.
Example 1
Solve a system of equations
Picture 1.
Decision.
We will use the first method to solve this system. First, let's express $y$ in the first equation in terms of $x$.
Figure 2.
Substitute $y$ into the second equation:
\ \ \[-2-x=2\] \ \
Answer: $(-4,6)$.
Example 2
Solve a system of equations
Figure 3
Decision.
This system is equivalent to the system
Figure 4
We apply the fourth method for solving equations. Let $2^x=u\ (u >0)$ and $3^y=v\ (v >0)$, we get:
Figure 5
We solve the resulting system by the addition method. Let's add the equations:
\ \
Then from the second equation, we get that
Returning to the replacement, I received a new system of exponential equations:
Figure 6
We get:
Figure 7
Answer: $(0,1)$.
Systems of exponential inequalities
Definition 2
Systems of inequalities consisting of exponential equations are called a system exponential inequalities.
We will consider the solution of systems of exponential inequalities using examples.
Example 3
Solve the system of inequalities
Figure 8
Decision:
This system of inequalities is equivalent to the system
Figure 9
To solve the first inequality, recall the following equivalence theorem for exponential inequalities:
Theorem 1. The inequality $a^(f(x)) >a^(\varphi (x)) $, where $a >0,a\ne 1$ is equivalent to the set of two systems
\
Where the role of $b$ can be an ordinary number, or maybe something tougher. Examples? Yes please:
\[\begin(align) & ((2)^(x)) \gt 4;\quad ((2)^(x-1))\le \frac(1)(\sqrt(2));\ quad ((2)^(((x)^(2))-7x+14)) \lt 16; \\ & ((0,1)^(1-x)) \lt 0.01;\quad ((2)^(\frac(x)(2))) \lt ((4)^(\frac (4)(x))). \\\end(align)\]
I think the meaning is clear: there is an exponential function $((a)^(x))$, it is compared with something, and then asked to find $x$. In especially clinical cases, instead of the variable $x$, they can put some function $f\left(x \right)$ and thereby complicate the inequality a little. :)
Of course, in some cases, the inequality may look more severe. For example:
\[((9)^(x))+8 \gt ((3)^(x+2))\]
Or even this:
In general, the complexity of such inequalities can be very different, but in the end they still come down to a simple construction $((a)^(x)) \gt b$. And we will somehow deal with such a design (in especially clinical cases, when nothing comes to mind, logarithms will help us). Therefore, now we will learn how to solve such simple constructions.
Solution of the simplest exponential inequalities
Let's look at something very simple. For example, here it is:
\[((2)^(x)) \gt 4\]
Obviously, the number on the right can be rewritten as a power of two: $4=((2)^(2))$. Thus, the original inequality is rewritten in a very convenient form:
\[((2)^(x)) \gt ((2)^(2))\]
And now the hands are itching to "cross out" the deuces, standing in the bases of the degrees, in order to get the answer $x \gt 2$. But before we cross out anything, let's remember the powers of two:
\[((2)^(1))=2;\quad ((2)^(2))=4;\quad ((2)^(3))=8;\quad ((2)^( 4))=16;...\]
As we see what more stands in the exponent, the larger the output number. "Thanks, Cap!" one of the students will exclaim. Does it happen differently? Unfortunately, it happens. For example:
\[((\left(\frac(1)(2) \right))^(1))=\frac(1)(2);\quad ((\left(\frac(1)(2) \ right))^(2))=\frac(1)(4);\quad ((\left(\frac(1)(2) \right))^(3))=\frac(1)(8 );...\]
Here, too, everything is logical: the greater the degree, the more times the number 0.5 is multiplied by itself (that is, it is divided in half). Thus, the resulting sequence of numbers is decreasing, and the difference between the first and second sequences is only in the base:
- If the base of degree $a \gt 1$, then as the exponent $n$ grows, the number $((a)^(n))$ will also grow;
- Conversely, if $0 \lt a \lt 1$, then as the exponent $n$ grows, the number $((a)^(n))$ will decrease.
Summing up these facts, we get the most important statement, on which the entire solution of exponential inequalities is based:
If $a \gt 1$, then the inequality $((a)^(x)) \gt ((a)^(n))$ is equivalent to the inequality $x \gt n$. If $0 \lt a \lt 1$, then the inequality $((a)^(x)) \gt ((a)^(n))$ is equivalent to the inequality $x \lt n$.
In other words, if the base more than one, you can simply remove it - the inequality sign will not change. And if the base is less than one, then it can also be removed, but the sign of inequality will also have to be changed.
Note that we have not considered the $a=1$ and $a\le 0$ options. Because in these cases there is uncertainty. Suppose how to solve an inequality of the form $((1)^(x)) \gt 3$? A one to any power will again give a one - we will never get a three or more. Those. there are no solutions.
With negative bases, it's even more interesting. Consider, for example, the following inequality:
\[((\left(-2 \right))^(x)) \gt 4\]
At first glance, everything is simple:
Correctly? But no! It is enough to substitute a couple of even and a couple of odd numbers instead of $x$ to make sure that the solution is wrong. Take a look:
\[\begin(align) & x=4\Rightarrow ((\left(-2 \right))^(4))=16 \gt 4; \\ & x=5\Rightarrow ((\left(-2 \right))^(5))=-32 \lt 4; \\ & x=6\Rightarrow ((\left(-2 \right))^(6))=64 \gt 4; \\ & x=7\Rightarrow ((\left(-2 \right))^(7))=-128 \lt 4. \\\end(align)\]
As you can see, the signs alternate. But there is more fractional powers and other tin. How, for example, would you order to count $((\left(-2 \right))^(\sqrt(7)))$ (minus two raised to the root of seven)? No way!
Therefore, for definiteness, we assume that in all exponential inequalities (and equations, by the way, too) $1\ne a \gt 0$. And then everything is solved very simply:
\[((a)^(x)) \gt ((a)^(n))\Rightarrow \left[ \begin(align) & x \gt n\quad \left(a \gt 1 \right), \\ & x \lt n\quad \left(0 \lt a \lt 1 \right). \\\end(align) \right.\]
In general, once again remember the main rule: if the base in the exponential equation is greater than one, you can simply remove it; and if the base is less than one, it can also be removed, but this will change the inequality sign.
Solution examples
So, consider a few simple exponential inequalities:
\[\begin(align) & ((2)^(x-1))\le \frac(1)(\sqrt(2)); \\ & ((0,1)^(1-x)) \lt 0.01; \\ & ((2)^(((x)^(2))-7x+14)) \lt 16; \\ & ((0,2)^(1+((x)^(2))))\ge \frac(1)(25). \\\end(align)\]
The primary task is the same in all cases: to reduce the inequalities to the simplest form $((a)^(x)) \gt ((a)^(n))$. This is what we will now do with each inequality, and at the same time repeat the properties of the powers and exponential function. So let's go!
\[((2)^(x-1))\le \frac(1)(\sqrt(2))\]
What can be done here? Well, on the left we already have a demonstrative expression - nothing needs to be changed. But on the right there is some kind of crap: a fraction, and even a root in the denominator!
However, remember the rules for working with fractions and powers:
\[\begin(align) & \frac(1)(((a)^(n)))=((a)^(-n)); \\ & \sqrt[k](a)=((a)^(\frac(1)(k))). \\\end(align)\]
What does it mean? First, we can easily get rid of the fraction by turning it into a negative exponent. And secondly, since the denominator is the root, it would be nice to turn it into a degree - this time with a fractional exponent.
Let's apply these actions sequentially to the right side of the inequality and see what happens:
\[\frac(1)(\sqrt(2))=((\left(\sqrt(2) \right))^(-1))=((\left(((2)^(\frac( 1)(3))) \right))^(-1))=((2)^(\frac(1)(3)\cdot \left(-1 \right)))=((2)^ (-\frac(1)(3)))\]
Do not forget that when raising a degree to a power, the exponents of these degrees are added. And in general, when working with exponential equations and inequalities, it is absolutely necessary to know at least the simplest rules for working with powers:
\[\begin(align) & ((a)^(x))\cdot ((a)^(y))=((a)^(x+y)); \\ & \frac(((a)^(x)))(((a)^(y)))=((a)^(x-y)); \\ & ((\left(((a)^(x)) \right))^(y))=((a)^(x\cdot y)). \\\end(align)\]
Actually, we just applied the last rule. Therefore, our original inequality will be rewritten as follows:
\[((2)^(x-1))\le \frac(1)(\sqrt(2))\Rightarrow ((2)^(x-1))\le ((2)^(-\ frac(1)(3)))\]
Now we get rid of the deuce at the base. Since 2 > 1, the inequality sign remains the same:
\[\begin(align) & x-1\le -\frac(1)(3)\Rightarrow x\le 1-\frac(1)(3)=\frac(2)(3); \\ & x\in \left(-\infty ;\frac(2)(3) \right]. \\\end(align)\]
That's the whole solution! The main difficulty is not at all in the exponential function, but in the competent transformation of the original expression: you need to carefully and as quickly as possible bring it to its simplest form.
Consider the second inequality:
\[((0,1)^(1-x)) \lt 0,01\]
So-so. Here we are waiting for decimal fractions. As I have said many times, in any expressions with powers, you should get rid of decimal fractions - often this is the only way to see a quick and easy solution. Here's what we'll get rid of:
\[\begin(align) & 0,1=\frac(1)(10);\quad 0,01=\frac(1)(100)=((\left(\frac(1)(10) \ right))^(2)); \\ & ((0,1)^(1-x)) \lt 0,01\Rightarrow ((\left(\frac(1)(10) \right))^(1-x)) \lt ( (\left(\frac(1)(10) \right))^(2)). \\\end(align)\]
Before us is again the simplest inequality, and even with the base 1/10, i.e. less than one. Well, we remove the bases, simultaneously changing the sign from "less" to "greater", and we get:
\[\begin(align) & 1-x \gt 2; \\ & -x \gt 2-1; \\ & -x \gt 1; \\& x \lt -1. \\\end(align)\]
We got the final answer: $x\in \left(-\infty ;-1 \right)$. Please note that the answer is exactly the set, and in no case is the construction of the form $x \lt -1$. Because formally such a construction is not a set at all, but an inequality with respect to the variable $x$. Yes, it's very simple, but it's not the answer!
Important note. This inequality could be solved in another way - by reducing both parts to a power with a base greater than one. Take a look:
\[\frac(1)(10)=((10)^(-1))\Rightarrow ((\left(((10)^(-1)) \right))^(1-x)) \ lt ((\left(((10)^(-1)) \right))^(2))\Rightarrow ((10)^(-1\cdot \left(1-x \right))) \lt ((10)^(-1\cdot 2))\]
After such a transformation, we again get an exponential inequality, but with a base of 10 > 1. And this means that you can simply cross out the ten - the inequality sign will not change. We get:
\[\begin(align) & -1\cdot \left(1-x \right) \lt -1\cdot 2; \\ & x-1 \lt-2; \\ & x \lt -2+1=-1; \\ & x \lt -1. \\\end(align)\]
As you can see, the answer is exactly the same. At the same time, we saved ourselves from the need to change the sign and generally remember some rules there. :)
\[((2)^(((x)^(2))-7x+14)) \lt 16\]
However, don't let that scare you. Whatever is in the indicators, the technology for solving the inequality itself remains the same. Therefore, we note first that 16 = 2 4 . Let's rewrite the original inequality taking this fact into account:
\[\begin(align) & ((2)^(((x)^(2))-7x+14)) \lt ((2)^(4)); \\ & ((x)^(2))-7x+14 \lt 4; \\ & ((x)^(2))-7x+10 \lt 0. \\\end(align)\]
Hooray! We got the usual square inequality! The sign has not changed anywhere, since the base is a deuce - a number greater than one.
Function zeros on the number line
We arrange the signs of the function $f\left(x \right)=((x)^(2))-7x+10$ - obviously, its graph will be a parabola with branches up, so there will be “pluses” on the sides. We are interested in the region where the function is less than zero, i.e. $x\in \left(2;5 \right)$ is the answer to the original problem.
Finally, consider another inequality:
\[((0,2)^(1+((x)^(2))))\ge \frac(1)(25)\]
Again we see an exponential function with a decimal fraction in the base. Let's convert this fraction to a common fraction:
\[\begin(align) & 0,2=\frac(2)(10)=\frac(1)(5)=((5)^(-1))\Rightarrow \\ & \Rightarrow ((0 ,2)^(1+((x)^(2))))=((\left(((5)^(-1)) \right))^(1+((x)^(2) )))=((5)^(-1\cdot \left(1+((x)^(2)) \right)))\end(align)\]
In this case, we took advantage of the remark made earlier - we reduced the base to the number 5\u003e 1 in order to simplify our further decision. Let's do the same with the right side:
\[\frac(1)(25)=((\left(\frac(1)(5) \right))^(2))=((\left(((5)^(-1)) \ right))^(2))=((5)^(-1\cdot 2))=((5)^(-2))\]
Let's rewrite the original inequality, taking into account both transformations:
\[((0,2)^(1+((x)^(2))))\ge \frac(1)(25)\Rightarrow ((5)^(-1\cdot \left(1+ ((x)^(2)) \right)))\ge ((5)^(-2))\]
The bases on both sides are the same and greater than one. There are no other terms on the right and left, so we just “cross out” the fives and we get a very simple expression:
\[\begin(align) & -1\cdot \left(1+((x)^(2)) \right)\ge -2; \\ & -1-((x)^(2))\ge -2; \\ & -((x)^(2))\ge -2+1; \\ & -((x)^(2))\ge -1;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))\le 1. \\\end(align)\]
This is where you have to be careful. Many students like to simply extract Square root both parts of the inequality and write something like $x\le 1\Rightarrow x\in \left(-\infty ;-1 \right]$. You should never do this, since the root of an exact square is module, and in no case the original variable:
\[\sqrt(((x)^(2)))=\left| x\right|\]
However, working with modules is not the most pleasant experience, right? So we won't work. Instead, we simply move all the terms to the left and solve the usual inequality using the interval method:
$\begin(align) & ((x)^(2))-1\le 0; \\ & \left(x-1 \right)\left(x+1 \right)\le 0 \\ & ((x)_(1))=1;\quad ((x)_(2)) =-1; \\\end(align)$
Again, we mark the obtained points on the number line and look at the signs:
Please note: dots are shaded. Since we were solving a non-strict inequality, all points on the graph are shaded. Therefore, the answer will be: $x\in \left[ -1;1 \right]$ is not an interval, but a segment.
In general, I would like to note that there is nothing complicated in exponential inequalities. The meaning of all the transformations that we performed today boils down to a simple algorithm:
- Find the base to which we will reduce all degrees;
- Carefully perform transformations to get an inequality of the form $((a)^(x)) \gt ((a)^(n))$. Of course, instead of the variables $x$ and $n$, there can be much more complex functions, but the meaning of this will not change;
- Cross out the bases of the degrees. In this case, the inequality sign may change if the base $a \lt 1$.
Essentially, this universal algorithm solutions to all such inequalities. And everything else that will be told to you on this topic is just specific tricks and tricks to simplify and speed up the transformation. Here's one of those tricks we'll talk about now. :)
rationalization method
Consider another batch of inequalities:
\[\begin(align) & ((\text( )\!\!\pi\!\!\text( ))^(x+7)) \gt ((\text( )\!\!\pi \!\!\text( ))^(((x)^(2))-3x+2)); \\ & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1; \\ & ((\left(\frac(1)(3) \right))^(((x)^(2))+2x)) \gt ((\left(\frac(1)(9) \right))^(16-x)); \\ & ((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt 1. \\\end(align)\]
Well, what is so special about them? They are also lightweight. Although, stop! Is pi raised to a power? What kind of nonsense?
And how to raise the number $2\sqrt(3)-3$ to a power? Or $3-2\sqrt(2)$? The compilers of the problems obviously drank too much "Hawthorn" before sitting down to work. :)
In fact, there is nothing wrong with these tasks. Let me remind you: an exponential function is an expression of the form $((a)^(x))$, where the base $a$ is any positive number, except for one. The number π is positive - we already know this. The numbers $2\sqrt(3)-3$ and $3-2\sqrt(2)$ are also positive - this is easy to see if we compare them with zero.
It turns out that all these “terrifying” inequalities are no different from the simple ones discussed above? And they do it the same way? Yes, absolutely right. However, using their example, I would like to consider one trick that saves a lot of time on independent work and exams. We will talk about the method of rationalization. So attention:
Any exponential inequality of the form $((a)^(x)) \gt ((a)^(n))$ is equivalent to the inequality $\left(x-n \right)\cdot \left(a-1 \right) \gt 0 $.
That's the whole method. :) Did you think that there would be some kind of next game? Nothing like this! But this simple fact, written literally in one line, will greatly simplify our work. Take a look:
\[\begin(matrix) ((\text( )\!\!\pi\!\!\text( ))^(x+7)) \gt ((\text( )\!\!\pi\ !\!\text( ))^(((x)^(2))-3x+2)) \\ \Downarrow \\ \left(x+7-\left(((x)^(2)) -3x+2 \right) \right)\cdot \left(\text( )\!\!\pi\!\!\text( )-1 \right) \gt 0 \\\end(matrix)\]
Here are no more exponential functions! And you don't have to remember whether the sign changes or not. But a new problem arises: what to do with the fucking multiplier \[\left(\text( )\!\!\pi\!\!\text( )-1 \right)\]? We don't know what the exact value of pi is. However, the captain seems to hint at the obvious:
\[\text( )\!\!\pi\!\!\text( )\approx 3,14... \gt 3\Rightarrow \text( )\!\!\pi\!\!\text( )-1 \gt 3-1=2\]
In general, the exact value of π doesn’t bother us much - it’s only important for us to understand that in any case $\text( )\!\!\pi\!\!\text( )-1 \gt 2$, t .e. is a positive constant, and we can divide both sides of the inequality by it:
\[\begin(align) & \left(x+7-\left(((x)^(2))-3x+2 \right) \right)\cdot \left(\text( )\!\! \pi\!\!\text( )-1 \right) \gt 0 \\ & x+7-\left(((x)^(2))-3x+2 \right) \gt 0; \\ & x+7-((x)^(2))+3x-2 \gt 0; \\ & -((x)^(2))+4x+5 \gt 0;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))-4x-5 \lt 0; \\ & \left(x-5 \right)\left(x+1 \right) \lt 0. \\\end(align)\]
As you can see, at a certain point, we had to divide by minus one, and the inequality sign changed. At the end, I expanded the square trinomial according to the Vieta theorem - it is obvious that the roots are equal to $((x)_(1))=5$ and $((x)_(2))=-1$. Then everything is solved by the classical method of intervals:
We solve the inequality by the method of intervals All points are punctured because the original inequality is strict. We are interested in the area with negative values, so the answer is $x\in \left(-1;5 \right)$. That's the solution. :)
Let's move on to the next task:
\[((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1\]
Everything is simple here, because there is a unit on the right. And we remember that a unit is any number raised to the power of zero. Even if this number is irrational expression standing at the base on the left:
\[\begin(align) & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt 1=((\left(2 \sqrt(3)-3\right))^(0)); \\ & ((\left(2\sqrt(3)-3 \right))^(((x)^(2))-2x)) \lt ((\left(2\sqrt(3)-3 \right))^(0)); \\\end(align)\]
So let's rationalize:
\[\begin(align) & \left(((x)^(2))-2x-0 \right)\cdot \left(2\sqrt(3)-3-1 \right) \lt 0; \\ & \left(((x)^(2))-2x-0 \right)\cdot \left(2\sqrt(3)-4 \right) \lt 0; \\ & \left(((x)^(2))-2x-0 \right)\cdot 2\left(\sqrt(3)-2 \right) \lt 0. \\\end(align)\ ]
It remains only to deal with the signs. The multiplier $2\left(\sqrt(3)-2 \right)$ does not contain the variable $x$ - it's just a constant, and we need to figure out its sign. To do this, note the following:
\[\begin(matrix) \sqrt(3) \lt \sqrt(4)=2 \\ \Downarrow \\ 2\left(\sqrt(3)-2 \right) \lt 2\cdot \left(2 -2 \right)=0 \\\end(matrix)\]
It turns out that the second factor is not just a constant, but a negative constant! And when dividing by it, the sign of the original inequality will change to the opposite:
\[\begin(align) & \left(((x)^(2))-2x-0 \right)\cdot 2\left(\sqrt(3)-2 \right) \lt 0; \\ & ((x)^(2))-2x-0 \gt 0; \\ & x\left(x-2 \right) \gt 0. \\\end(align)\]
Now everything becomes quite obvious. Roots square trinomial on the right: $((x)_(1))=0$ and $((x)_(2))=2$. We mark them on the number line and look at the signs of the function $f\left(x \right)=x\left(x-2 \right)$:
The case when we are interested in lateral intervals We are interested in the intervals marked with a plus sign. It remains only to write down the answer:
Let's move on to the next example:
\[((\left(\frac(1)(3) \right))^(((x)^(2))+2x)) \gt ((\left(\frac(1)(9) \ right))^(16-x))\]
Well, everything is quite obvious here: the bases are powers of the same number. Therefore, I will write everything briefly:
\[\begin(matrix) \frac(1)(3)=((3)^(-1));\quad \frac(1)(9)=\frac(1)(((3)^( 2)))=((3)^(-2)) \\ \Downarrow \\ ((\left(((3)^(-1)) \right))^(((x)^(2) )+2x)) \gt ((\left(((3)^(-2)) \right))^(16-x)) \\\end(matrix)\]
\[\begin(align) & ((3)^(-1\cdot \left(((x)^(2))+2x \right))) \gt ((3)^(-2\cdot \ left(16-x\right))); \\ & ((3)^(-((x)^(2))-2x)) \gt ((3)^(-32+2x)); \\ & \left(-((x)^(2))-2x-\left(-32+2x \right) \right)\cdot \left(3-1 \right) \gt 0; \\ & -((x)^(2))-2x+32-2x \gt 0; \\ & -((x)^(2))-4x+32 \gt 0;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))+4x-32 \lt 0; \\ & \left(x+8 \right)\left(x-4 \right) \lt 0. \\\end(align)\]
As you can see, in the process of transformations, we had to multiply by a negative number, so the inequality sign has changed. At the very end, I again applied Vieta's theorem to factorize a square trinomial. As a result, the answer will be the following: $x\in \left(-8;4 \right)$ - those who wish can verify this by drawing a number line, marking points and counting signs. In the meantime, we will move on to the last inequality from our “set”:
\[((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt 1\]
As you can see, the base is again an irrational number, and the unit is again on the right. Therefore, we rewrite our exponential inequality as follows:
\[((\left(3-2\sqrt(2) \right))^(3x-((x)^(2)))) \lt ((\left(3-2\sqrt(2) \ right))^(0))\]
Let's rationalize:
\[\begin(align) & \left(3x-((x)^(2))-0 \right)\cdot \left(3-2\sqrt(2)-1 \right) \lt 0; \\ & \left(3x-((x)^(2))-0 \right)\cdot \left(2-2\sqrt(2) \right) \lt 0; \\ & \left(3x-((x)^(2))-0 \right)\cdot 2\left(1-\sqrt(2) \right) \lt 0. \\\end(align)\ ]
However, it is quite obvious that $1-\sqrt(2) \lt 0$, since $\sqrt(2)\approx 1.4... \gt 1$. Therefore, the second factor is again a negative constant, by which both parts of the inequality can be divided:
\[\begin(matrix) \left(3x-((x)^(2))-0 \right)\cdot 2\left(1-\sqrt(2) \right) \lt 0 \\ \Downarrow \ \\end(matrix)\]
\[\begin(align) & 3x-((x)^(2))-0 \gt 0; \\ & 3x-((x)^(2)) \gt 0;\quad \left| \cdot \left(-1 \right) \right. \\ & ((x)^(2))-3x \lt 0; \\ & x\left(x-3 \right) \lt 0. \\\end(align)\]
Change to another base
A separate problem in solving exponential inequalities is the search for the “correct” basis. Unfortunately, at the first glance at the task it is far from always obvious what to take as a basis, and what to do as the degree of this basis.
But do not worry: there is no magic and "secret" technologies here. In mathematics, any skill that cannot be algorithmized can be easily developed through practice. But for this you will have to solve problems of different levels of complexity. For example, these are:
\[\begin(align) & ((2)^(\frac(x)(2))) \lt ((4)^(\frac(4)(x))); \\ & ((\left(\frac(1)(3) \right))^(\frac(3)(x)))\ge ((3)^(2+x)); \\ & ((\left(0,16 \right))^(1+2x))\cdot ((\left(6,25 \right))^(x))\ge 1; \\ & ((\left(\frac(27)(\sqrt(3)) \right))^(-x)) \lt ((9)^(4-2x))\cdot 81. \\\ end(align)\]
Complicated? Scary? Yes, it's easier than a chicken on the asphalt! Let's try. First inequality:
\[((2)^(\frac(x)(2))) \lt ((4)^(\frac(4)(x)))\]
Well, I think everything is clear here:
We rewrite the original inequality, reducing everything to the base "two":
\[((2)^(\frac(x)(2))) \lt ((2)^(\frac(8)(x)))\Rightarrow \left(\frac(x)(2)- \frac(8)(x) \right)\cdot \left(2-1 \right) \lt 0\]
Yes, yes, you understood correctly: I just applied the rationalization method described above. Now we need to work carefully: we got a fractional-rational inequality (this is one that has a variable in the denominator), so before equating something to zero, you need to bring everything to common denominator and get rid of the constant multiplier.
\[\begin(align) & \left(\frac(x)(2)-\frac(8)(x) \right)\cdot \left(2-1 \right) \lt 0; \\ & \left(\frac(((x)^(2))-16)(2x) \right)\cdot 1 \lt 0; \\ & \frac(((x)^(2))-16)(2x) \lt 0. \\\end(align)\]
Now we use the standard interval method. Numerator zeros: $x=\pm 4$. The denominator goes to zero only when $x=0$. In total, there are three points that should be marked on the number line (all points are punched out, because the inequality sign is strict). We get:
More complicated case: three roots As you might guess, hatching marks the intervals at which the expression on the left takes negative values. Therefore, two intervals will go into the final answer at once:
The ends of the intervals are not included in the answer because the original inequality was strict. No further validation of this answer is required. In this regard, exponential inequalities are much simpler than logarithmic ones: no DPV, no restrictions, etc.
Let's move on to the next task:
\[((\left(\frac(1)(3) \right))^(\frac(3)(x)))\ge ((3)^(2+x))\]
There are no problems here either, since we already know that $\frac(1)(3)=((3)^(-1))$, so the whole inequality can be rewritten like this:
\[\begin(align) & ((\left(((3)^(-1)) \right))^(\frac(3)(x)))\ge ((3)^(2+x ))\Rightarrow ((3)^(-\frac(3)(x)))\ge ((3)^(2+x)); \\ & \left(-\frac(3)(x)-\left(2+x \right) \right)\cdot \left(3-1 \right)\ge 0; \\ & \left(-\frac(3)(x)-2-x \right)\cdot 2\ge 0;\quad \left| :\left(-2\right)\right. \\ & \frac(3)(x)+2+x\le 0; \\ & \frac(((x)^(2))+2x+3)(x)\le 0. \\\end(align)\]
Please note: in the third line, I decided not to waste time on trifles and immediately divide everything by (−2). Minul went into the first bracket (now there are pluses everywhere), and the deuce was reduced with a constant multiplier. This is exactly what you should do when making real calculations on independent and control work- no need to paint directly every action and transformation.
Next, the familiar method of intervals comes into play. Zeros of the numerator: but there are none. Because the discriminant will be negative. In turn, the denominator is set to zero only when $x=0$ — just like last time. Well, it is clear that the fraction will take positive values to the right of $x=0$, and negative ones to the left. Since we are only interested in negative values, the final answer is $x\in \left(-\infty ;0 \right)$.
\[((\left(0,16 \right))^(1+2x))\cdot ((\left(6,25 \right))^(x))\ge 1\]
And what should be done with decimal fractions in exponential inequalities? That's right: get rid of them by converting them into ordinary ones. Here we are translating:
\[\begin(align) & 0,16=\frac(16)(100)=\frac(4)(25)\Rightarrow ((\left(0,16 \right))^(1+2x)) =((\left(\frac(4)(25) \right))^(1+2x)); \\ & 6,25=\frac(625)(100)=\frac(25)(4)\Rightarrow ((\left(6,25 \right))^(x))=((\left(\ frac(25)(4) \right))^(x)). \\\end(align)\]
Well, what did we get in the bases of exponential functions? And we got two mutually reciprocal numbers:
\[\frac(25)(4)=((\left(\frac(4)(25) \right))^(-1))\Rightarrow ((\left(\frac(25)(4) \ right))^(x))=((\left(((\left(\frac(4)(25) \right))^(-1)) \right))^(x))=((\ left(\frac(4)(25) \right))^(-x))\]
Thus, the original inequality can be rewritten as follows:
\[\begin(align) & ((\left(\frac(4)(25) \right))^(1+2x))\cdot ((\left(\frac(4)(25) \right) )^(-x))\ge 1; \\ & ((\left(\frac(4)(25) \right))^(1+2x+\left(-x \right)))\ge ((\left(\frac(4)(25) \right))^(0)); \\ & ((\left(\frac(4)(25) \right))^(x+1))\ge ((\left(\frac(4)(25) \right))^(0) ). \\\end(align)\]
Of course, when multiplying powers with the same base, their indicators add up, which happened in the second line. In addition, we have represented the unit on the right, also as a power in base 4/25. It remains only to rationalize:
\[((\left(\frac(4)(25) \right))^(x+1))\ge ((\left(\frac(4)(25) \right))^(0)) \Rightarrow \left(x+1-0 \right)\cdot \left(\frac(4)(25)-1 \right)\ge 0\]
Note that $\frac(4)(25)-1=\frac(4-25)(25) \lt 0$, i.e. the second factor is a negative constant, and when divided by it, the inequality sign will change:
\[\begin(align) & x+1-0\le 0\Rightarrow x\le -1; \\ & x\in \left(-\infty ;-1 \right]. \\\end(align)\]
Finally, the last inequality from the current "set":
\[((\left(\frac(27)(\sqrt(3)) \right))^(-x)) \lt ((9)^(4-2x))\cdot 81\]
In principle, the idea of the solution here is also clear: all the exponential functions that make up the inequality must be reduced to the base "3". But for this you have to tinker a little with roots and degrees:
\[\begin(align) & \frac(27)(\sqrt(3))=\frac(((3)^(3)))(((3)^(\frac(1)(3)) ))=((3)^(3-\frac(1)(3)))=((3)^(\frac(8)(3))); \\ & 9=((3)^(2));\quad 81=((3)^(4)). \\\end(align)\]
Given these facts, the original inequality can be rewritten as follows:
\[\begin(align) & ((\left(((3)^(\frac(8)(3))) \right))^(-x)) \lt ((\left(((3) ^(2)) \right))^(4-2x))\cdot ((3)^(4)); \\ & ((3)^(-\frac(8x)(3))) \lt ((3)^(8-4x))\cdot ((3)^(4)); \\ & ((3)^(-\frac(8x)(3))) \lt ((3)^(8-4x+4)); \\ & ((3)^(-\frac(8x)(3))) \lt ((3)^(4-4x)). \\\end(align)\]
Pay attention to the 2nd and 3rd lines of calculations: before doing something with inequality, be sure to bring it to the form that we talked about from the very beginning of the lesson: $((a)^(x)) \lt ((a)^(n))$. As long as you have left or right left multipliers, extra constants, etc., no rationalization and "crossing out" of the grounds can be performed! Countless tasks have been done wrong due to a misunderstanding of this simple fact. I myself constantly observe this problem with my students when we are just starting to analyze exponential and logarithmic inequalities.
But back to our task. Let's try this time to do without rationalization. Remember: the base of the degree is greater than one, so the triples can simply be crossed out - the inequality sign will not change. We get:
\[\begin(align) & -\frac(8x)(3) \lt 4-4x; \\ & 4x-\frac(8x)(3) \lt 4; \\ & \frac(4x)(3) \lt 4; \\ & 4x \lt 12; \\ & x \lt 3. \\\end(align)\]
That's all. Final answer: $x\in \left(-\infty ;3 \right)$.
Highlighting a stable expression and replacing a variable
In conclusion, I propose to solve four more exponential inequalities, which are already quite difficult for unprepared students. To cope with them, you need to remember the rules for working with degrees. In particular, putting common factors out of brackets.
But the most important thing is to learn to understand: what exactly can be bracketed. Such an expression is called stable - it can be denoted by a new variable and thus get rid of the exponential function. So, let's look at the tasks:
\[\begin(align) & ((5)^(x+2))+((5)^(x+1))\ge 6; \\ & ((3)^(x))+((3)^(x+2))\ge 90; \\ & ((25)^(x+1,5))-((5)^(2x+2)) \gt 2500; \\ & ((\left(0,5 \right))^(-4x-8))-((16)^(x+1,5)) \gt 768. \\\end(align)\]
Let's start with the very first line. Let's write this inequality separately:
\[((5)^(x+2))+((5)^(x+1))\ge 6\]
Note that $((5)^(x+2))=((5)^(x+1+1))=((5)^(x+1))\cdot 5$, so the right side can be rewrite:
Note that there are no other exponential functions except for $((5)^(x+1))$ in the inequality. And in general, the variable $x$ does not occur anywhere else, so let's introduce a new variable: $((5)^(x+1))=t$. We get the following construction:
\[\begin(align) & 5t+t\ge 6; \\ & 6t\ge 6; \\ & t\ge 1. \\\end(align)\]
We return to the original variable ($t=((5)^(x+1))$), and at the same time remember that 1=5 0 . We have:
\[\begin(align) & ((5)^(x+1))\ge ((5)^(0)); \\ &x+1\ge 0; \\ & x\ge -1. \\\end(align)\]
That's the whole solution! Answer: $x\in \left[ -1;+\infty \right)$. Let's move on to the second inequality:
\[((3)^(x))+((3)^(x+2))\ge 90\]
Everything is the same here. Note that $((3)^(x+2))=((3)^(x))\cdot ((3)^(2))=9\cdot ((3)^(x))$ . Then the left side can be rewritten:
\[\begin(align) & ((3)^(x))+9\cdot ((3)^(x))\ge 90;\quad \left| ((3)^(x))=t \right. \\&t+9t\ge 90; \\ & 10t\ge 90; \\ & t\ge 9\Rightarrow ((3)^(x))\ge 9\Rightarrow ((3)^(x))\ge ((3)^(2)); \\ & x\ge 2\Rightarrow x\in \left[ 2;+\infty \right). \\\end(align)\]
This is approximately how you need to draw up a decision on real control and independent work.
Well, let's try something more difficult. For example, here is an inequality:
\[((25)^(x+1,5))-((5)^(2x+2)) \gt 2500\]
What is the problem here? First of all, the bases of the exponential functions on the left are different: 5 and 25. However, 25 \u003d 5 2, so the first term can be transformed:
\[\begin(align) & ((25)^(x+1,5))=((\left(((5)^(2)) \right))^(x+1,5))= ((5)^(2x+3)); \\ & ((5)^(2x+3))=((5)^(2x+2+1))=((5)^(2x+2))\cdot 5. \\\end(align )\]
As you can see, at first we brought everything to the same base, and then we noticed that the first term is easily reduced to the second - it is enough just to expand the exponent. Now we can safely introduce a new variable: $((5)^(2x+2))=t$, and the whole inequality will be rewritten like this:
\[\begin(align) & 5t-t\ge 2500; \\ & 4t\ge 2500; \\ & t\ge 625=((5)^(4)); \\ & ((5)^(2x+2))\ge ((5)^(4)); \\ & 2x+2\ge 4; \\ & 2x\ge 2; \\ & x\ge 1. \\\end(align)\]
Again, no problem! Final answer: $x\in \left[ 1;+\infty \right)$. Moving on to the final inequality in today's lesson:
\[((\left(0,5 \right))^(-4x-8))-((16)^(x+1,5)) \gt 768\]
The first thing to note is, of course, decimal at the base of the first degree. It is necessary to get rid of it, and at the same time bring all exponential functions to the same base - the number "2":
\[\begin(align) & 0,5=\frac(1)(2)=((2)^(-1))\Rightarrow ((\left(0,5 \right))^(-4x- 8))=((\left(((2)^(-1)) \right))^(-4x-8))=((2)^(4x+8)); \\ & 16=((2)^(4))\Rightarrow ((16)^(x+1,5))=((\left(((2)^(4)) \right))^( x+1.5))=((2)^(4x+6)); \\ & ((2)^(4x+8))-((2)^(4x+6)) \gt 768. \\\end(align)\]
Great, we have taken the first step - everything has led to the same foundation. Now we need to highlight the stable expression. Note that $((2)^(4x+8))=((2)^(4x+6+2))=((2)^(4x+6))\cdot 4$. If we introduce a new variable $((2)^(4x+6))=t$, then the original inequality can be rewritten as follows:
\[\begin(align) & 4t-t \gt 768; \\ & 3t \gt 768; \\ & t \gt 256=((2)^(8)); \\ & ((2)^(4x+6)) \gt ((2)^(8)); \\ & 4x+6 \gt 8; \\ & 4x \gt 2; \\ & x \gt \frac(1)(2)=0.5. \\\end(align)\]
Naturally, the question may arise: how did we find out that 256 = 2 8 ? Unfortunately, here you just need to know the powers of two (and at the same time the powers of three and five). Well, or divide 256 by 2 (you can divide, because 256 - even number) until we get the result. It will look something like this:
\[\begin(align) & 256=128\cdot 2= \\ & =64\cdot 2\cdot 2= \\ & =32\cdot 2\cdot 2\cdot 2= \\ & =16\cdot 2 \cdot 2\cdot 2\cdot 2= \\ & =8\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2= \\ & =4\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2= \\ & =2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2= \\ & =((2)^(8)).\end(align )\]
The same is with the three (numbers 9, 27, 81 and 243 are its powers), and with the seven (numbers 49 and 343 would also be nice to remember). Well, the five also have “beautiful” degrees that you need to know:
\[\begin(align) & ((5)^(2))=25; \\ & ((5)^(3))=125; \\ & ((5)^(4))=625; \\ & ((5)^(5))=3125. \\\end(align)\]
Of course, if you wish, all these numbers can be restored in your mind by simply multiplying them one by one. However, when you have to solve several exponential inequalities, and each next one is more difficult than the previous one, then the last thing you want to think about is the powers of some numbers there. And in this sense, these problems are more complex than the "classical" inequalities, which are solved by the interval method.
I hope this lesson helped you in mastering this topic. If something is not clear, ask in the comments. And see you in the next tutorials. :)
Lesson and presentation on the topic: "Exponential equations and exponential inequalities"
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Definition of exponential equations
Guys, we studied exponential functions, learned their properties and built graphs, analyzed examples of equations in which exponential functions were encountered. Today we will study exponential equations and inequalities.Definition. Equations of the form: $a^(f(x))=a^(g(x))$, where $a>0$, $a≠1$ are called exponential equations.
Remembering the theorems that we studied in the topic "Exponential function", we can introduce a new theorem:
Theorem. exponential equation$a^(f(x))=a^(g(x))$, where $a>0$, $a≠1$ is equivalent to the equation $f(x)=g(x)$.
Examples of exponential equations
Example.Solve Equations:
a) $3^(3x-3)=27$.
b) $((\frac(2)(3)))^(2x+0,2)=\sqrt(\frac(2)(3))$.
c) $5^(x^2-6x)=5^(-3x+18)$.
Decision.
a) We know well that $27=3^3$.
Let's rewrite our equation: $3^(3x-3)=3^3$.
Using the theorem above, we get that our equation reduces to the equation $3x-3=3$, solving this equation, we get $x=2$.
Answer: $x=2$.
B) $\sqrt(\frac(2)(3))=((\frac(2)(3)))^(\frac(1)(5))$.
Then our equation can be rewritten: $((\frac(2)(3)))^(2x+0,2)=((\frac(2)(3)))^(\frac(1)(5) )=((\frac(2)(3)))^(0,2)$.
$2x+0.2=$0.2.
$x=0$.
Answer: $x=0$.
C) The original equation is equivalent to the equation: $x^2-6x=-3x+18$.
$x^2-3x-18=0$.
$(x-6)(x+3)=0$.
$x_1=6$ and $x_2=-3$.
Answer: $x_1=6$ and $x_2=-3$.
Example.
Solve the equation: $\frac(((0.25))^(x-0.5))(\sqrt(4))=16*((0.0625))^(x+1)$.
Decision:
We will sequentially perform a series of actions and bring both parts of our equation to the same bases.
Let's perform a series of operations on the left side:
1) $((0.25))^(x-0.5)=((\frac(1)(4)))^(x-0.5)$.
2) $\sqrt(4)=4^(\frac(1)(2))$.
3) $\frac(((0.25))^(x-0.5))(\sqrt(4))=\frac(((\frac(1)(4)))^(x-0 ,5))(4^(\frac(1)(2)))= \frac(1)(4^(x-0.5+0.5))=\frac(1)(4^x) =((\frac(1)(4)))^x$.
Let's move on to the right side:
4) $16=4^2$.
5) $((0.0625))^(x+1)=\frac(1)((16)^(x+1))=\frac(1)(4^(2x+2))$.
6) $16*((0.0625))^(x+1)=\frac(4^2)(4^(2x+2))=4^(2-2x-2)=4^(-2x )=\frac(1)(4^(2x))=((\frac(1)(4)))^(2x)$.
The original equation is equivalent to the equation:
$((\frac(1)(4)))^x=((\frac(1)(4)))^(2x)$.
$x=2x$.
$x=0$.
Answer: $x=0$.
Example.
Solve the equation: $9^x+3^(x+2)-36=0$.
Decision:
Let's rewrite our equation: $((3^2))^x+9*3^x-36=0$.
$((3^x))^2+9*3^x-36=0$.
Let's make a change of variables, let $a=3^x$.
In new variable equation takes the form: $a^2+9a-36=0$.
$(a+12)(a-3)=0$.
$a_1=-12$ and $a_2=3$.
Let's perform the reverse change of variables: $3^x=-12$ and $3^x=3$.
In the last lesson, we learned that exponential expressions can only take positive values, remember the graph. This means that the first equation has no solutions, the second equation has one solution: $x=1$.
Answer: $x=1$.
Let's make a memo of ways to solve exponential equations:
1. Graphic method. We represent both parts of the equation as functions and build their graphs, find the intersection points of the graphs. (We used this method in the last lesson).
2. The principle of equality of indicators. The principle is based on the fact that two expressions with the same bases are equal if and only if the degrees (exponents) of these bases are equal. $a^(f(x))=a^(g(x))$ $f(x)=g(x)$.
3. Change of variables method. This method should be used if the equation, when changing variables, simplifies its form and is much easier to solve.
Example.
Solve the system of equations: $\begin (cases) (27)^y*3^x=1, \\ 4^(x+y)-2^(x+y)=12. \end(cases)$.
Decision.
Consider both equations of the system separately:
$27^y*3^x=1$.
$3^(3y)*3^x=3^0$.
$3^(3y+x)=3^0$.
$x+3y=0$.
Consider the second equation:
$4^(x+y)-2^(x+y)=12$.
$2^(2(x+y))-2^(x+y)=12$.
Let's use the change of variables method, let $y=2^(x+y)$.
Then the equation will take the form:
$y^2-y-12=0$.
$(y-4)(y+3)=0$.
$y_1=4$ and $y_2=-3$.
Let's move on to the initial variables, from the first equation we get $x+y=2$. The second equation has no solutions. Then our initial system of equations is equivalent to the system: $\begin (cases) x+3y=0, \\ x+y=2. \end(cases)$.
Subtract the second equation from the first equation, we get: $\begin (cases) 2y=-2, \\ x+y=2. \end(cases)$.
$\begin (cases) y=-1, \\ x=3. \end(cases)$.
Answer: $(3;-1)$.
exponential inequalities
Let's move on to inequalities. When solving inequalities, it is necessary to pay attention to the base of the degree. There are two possible scenarios for the development of events when solving inequalities.Theorem. If $a>1$, then the exponential inequality $a^(f(x))>a^(g(x))$ is equivalent to the inequality $f(x)>g(x)$.
If $0
Example.
Solve inequalities:
a) $3^(2x+3)>81$.
b) $((\frac(1)(4)))^(2x-4) c) $(0,3)^(x^2+6x)≤(0,3)^(4x+15)$ .
Decision.
a) $3^(2x+3)>81$.
$3^(2x+3)>3^4$.
Our inequality is equivalent to the inequality:
$2x+3>4$.
$2x>1$.
$x>0.5$.
B) $((\frac(1)(4)))^(2x-4) $((\frac(1)(4)))^(2x-4) In our equation, the base with a degree less than 1, then when replacing an inequality with an equivalent one, it is necessary to change the sign.
$2x-4>2$.
$x>3$.
C) Our inequality is equivalent to the inequality:
$x^2+6x≥4x+15$.
$x^2+2x-15≥0$.
$(x-3)(x+5)≥0$.
Let's use the interval solution method:
Answer: $(-∞;-5]U)
