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.
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.
Solve the inequality $7^x>7^8$7x>78.
Solve the inequality $\left(\frac{1}{6}\right)^x\ge\left(\frac{1}{6}\right)^9$(16)x≥(16)9.
Solve the inequality $6^{-x}\ge6^2$6−x≥62.
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) |
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
Solve the inequality $6^{x+3}<6^7$6x+3<67.
Solve the inequality $3^x\le\sqrt[4]{3}$3x≤4√3.
Solve the inequality $25^x\le\frac{1}{5}$25x≤15.
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
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
Solve the inequality $6^{3x-2}<6^4$63x−2<64.
Solve the inequality $3^{3x-4}\le9$33x−4≤9.
Consider the inequality$\left(\frac{1}{27}\right)^{x+1}\ge\left(\frac{1}{81}\right)^{2x-3}$(127)x+1≥(181)2x−3.
What common (prime) base can $\frac{1}{27}$127 and $\frac{1}{81}$181 be written in terms of?
Hence solve the inequality.