Solve Inequalities Involving Exponential Functions

Recall that an exponential graph of the form $y=A\times b^{mx+c}+k$y=A×bmx+c+k (where $A\ne0$A≠0, $m\ne0$m≠0 and $b>1$b>1) will have one of four general shapes.

$A>0$A>0, $m>0$m>0	$A>0$A>0, $m<0$m<0
$A<0$A<0, $m>0$m>0	$A<0$A<0, $m<0$m<0

 

$k$k will horizontally translate our exponential, and $c$c will vertically translate it. Remember also that if $0$0$<$<$b<1$b<1 we can rearrange our exponential to an equivalent exponential where the base $b>1$b>1. For instance, $\left(\frac{1}{2}\right)^x$(12​)x is the same as $2^{-x}$2−x and $\left(\frac{4}{5}\right)^{3x}$(45​)3x is the same as $\left(\frac{5}{4}\right)^{-3x}$(54​)−3x.

We are now ready to solve some inequalities involving exponentials.

 

Inequalities involving exponential functions can often be reduced to a simpler inequality. First, this involves identifying whether the base of the exponential is greater than or less than $1$1. Then we can reduce the inequality so that it only contains exponent terms.

Exploration

Let’s look at the inequality $2^x<2^3$2x<23. To examine this inequality we will look at the graphs of $y=2^x$y=2x and $y=2^3=8$y=23=8, the two sides of the inequality.

We can see from the two graphs that for any value of $x$x less than $3$3, the graph of $2^x$2x will be below the line and therefore $2^x<2^3$2x<23 will hold. For values of $x$x greater than $3$3, $2^x>2^3$2x>23 would be true as the graph only ever increases. So the solution to $2^x<2^3$2x<23 is $x<3$x<3.

Notice than the solution is simply the indices of the original inequality with the same sign in between. The sign could be $>$>, $<$<, $\le$≤ or $\ge$≥ and it will still be consistent in the final inequality. This is true of any exponential inequality if the bases are equal and greater than $1$1.

 

We have to be careful when our base is less than $1$1 though, for example, $\left(\frac{1}{2}\right)^x<\left(\frac{1}{2}\right)^3$(12​)x<(12​)3. Let’s look at the graphs $y=\left(\frac{1}{2}\right)^x$y=(12​)x and $y=\left(\frac{1}{2}\right)^3=\frac{1}{8}=0.125$y=(12​)3=18​=0.125.

If we compare this graph to the previous one, we can see that graph of $y=\left(\frac{1}{2}\right)^x$y=(12​)x is sloping in the opposite direction, it decreases as $x$x increases. For all $x$x values greater than $3$3, the graph of $y=\left(\frac{1}{2}\right)^x$y=(12​)x will be below the line $y=\left(\frac{1}{2}\right)^3$y=(12​)3 . So the solution to $\left(\frac{1}{2}\right)^x<\left(\frac{1}{2}\right)^3$(12​)x<(12​)3 is $x>3$x>3. Notice that the solution is simply the indices of the original inequality with the sign between having been flipped.

 

 

Practice Questions

Question 1

Question 2

Question 3

Two Step

An inequality such as $3^x<9$3x<9 features an exponential on one side and on the other side an exponential in a different base. However the right-hand side can be rewritten as a power of $3$3, so both sides will share the same base . That is, $9$9 can be rewritten as $3^2$32 so we can rewrite the inequality as a one step inequality and solve as shown below.

$3^x$3x	$<$<	$9$9	(Writing the inequality)
$3^x$3x	$<$<	$3^2$32	(Making both bases the same)
$x$x	$<$<	$2$2	(Simplifying the inequality)

There are many ways to represent an exponential term. We can rewrite these terms to have a desired base.

$\sqrt[3]{2}$3√2	can be written as	$2^{\frac{1}{3}}$213​
$\frac{1}{6}$16​	can be written as	$6^{-1}$6−1

From one step exponentials we also saw that an inequality with a base between $0$0 and $1$1 will require the sign to be flipped when the indices are compared as seen below.

$\left(\frac{1}{3}\right)^x$(13​)x	$\le$≤	$\left(\frac{1}{3}\right)^2$(13​)2
$x$x	$\ge$≥	$2$2

Some exponentials may require more steps as the indices contain more complicated expressions such as:

$\left(\frac{1}{6}\right)^{x+3}<\left(\frac{1}{6}\right)^6$(16​)x+3<(16​)6

Because both sides have the same base the indices can be brought down and then solved as a normal inequality. Making sure to flip the inequality sign because the bases are between $0$0 and $1$1.

$\left(\frac{1}{6}\right)^{x+3}$(16​)x+3	$<$<	$\left(\frac{1}{6}\right)^6$(16​)6	(Rewriting the inequality)
$x+3$x+3	$>$>	$6$6	(Compare indices and flip sign)
$x$x	$>$>	$3$3	(Move constants to one side)

 

Worked example

Solve the inequality $2^{3x}\ge\frac{1}{2}$23x≥12​.

Think: Notice that both sides of the inequality are not using the same base. We can change the right-hand side so that it shares the same base as the left-hand side.

Do: Rewrite $\frac{1}{2}$12​ as $2^{-1}$2−1.

$2^{3x}\ge2^{-1}$23x≥2−1

Both sides have the same base which is greater than $1$1 so we can compare the indices as below:

$3x\ge-1$3x≥−1

This inequality can be solved like any other inequality, in this case, by dividing both sides by $3$3.

$x\ge-\frac{1}{3}$x≥−13​

Practice questions

Question 4

Question 5

question 6

Three Step

Worked Examples 

Example 2

Solve the inequality $5^{6x-2}\le5^{2x+2}$56x−2≤52x+2.

Think: The base on both sides is equal and greater than $1$1.

Do: Create a simpler inequality using the indices and the original inequality sign

$6x-2\le2x+6$6x−2≤2x+6

Now we can solve this inequality as we normally would, be moving the constant terms to one side and the $x$x terms to the other side.

$4x\le8$4x≤8

And finally placing $x$x by itself by dividing both sides by $4$4.

$x\le2$x≤2

Example 3

Solve the inequality $\sqrt{3^x}>\frac{1}{9}$√3x>19​.

Think: Both sides of the inequality need to be in a common base to simplify the inequality.

Do: Rewrite $\sqrt{3^x}$√3x and $\frac{1}{9}$19​ in the form $3^{\editable{}}$3.

$\left(3^{\frac{1}{2}}\right)^x>3^{-2}$(312​)x>3−2

Now by using the index law dealing with raising a power to another power we can simplify the inequality to 

$3^{\frac{x}{2}}>3^{-2}$3x2​>3−2

Now both sides of the inequality use the same base we can continue as a we would a two step inequality

$\frac{x}{2}>-2$x2​>−2

$x>-4$x>−4

Practice questions

Question 7

Question 8

Question 9