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Stage 4 - Stage 5

# Solve Inequalities Involving Exponential Functions

Lesson

Recall that an exponential graph of the form $y=A\times b^{mx+c}+k$y=A×bmx+c+k (where $A\ne0$A0, $m\ne0$m0 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}$2x 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.

The function $y=2^x$y=2x is shown in green and $y=2^3=8$y=23=8 in blue.

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.

Case: $A>1$A>1

If two exponentials have the same base we can compare the indices separately from their base as follows:

If $A^mAm<An and$A>1$A>1 then$mm<n.

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.

The function $y=\left(\frac{1}{2}\right)^x$y=(12)x is shown in green and $y=\left(\frac{1}{2}\right)^3=0.125$y=(12)3=0.125 in blue.

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.

Case: $00<A<1 If two exponentials have the same base we can compare the indices separately from their base as follows: If$A^mAm<An and $00<A<1 then$m>n$m>n. Remember that the inequality sign flips when the base is between$0$0 and$1$1. #### Practice Questions ##### Question 1 Solve the inequality$7^x>7^8$7x>78. ##### Question 2 Solve the inequality$\left(\frac{1}{6}\right)^x\ge\left(\frac{1}{6}\right)^9$(16)x(16)9. ##### Question 3 Solve the inequality$6^{-x}\ge6^2$6x62. ### 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}$23x12. Think: Notice that both sides of the inequality are not using the same base. We can change the right-hand side so that it shares the same base as the left-hand side. Do: Rewrite$\frac{1}{2}$12 as$2^{-1}$21.$2^{3x}\ge2^{-1}$23x21 Both sides have the same base which is greater than$1$1 so we can compare the indices as below:$3x\ge-1$3x1 This inequality can be solved like any other inequality, in this case, by dividing both sides by$3$3.$x\ge-\frac{1}{3}$x13 #### Practice questions ##### Question 4 Solve the inequality$6^{x+3}<6^7$6x+3<67. ##### Question 5 Solve the inequality$3^x\le\sqrt[4]{3}$3x43. ##### question 6 Solve the inequality$25^x\le\frac{1}{5}$25x15. ### Three Step #### Worked Examples ##### Example 2 Solve the inequality$5^{6x-2}\le5^{2x+2}$56x252x+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$6x22x+6 Now we can solve this inequality as we normally would, be moving the constant terms to one side and the$x$x terms to the other side.$4x\le8$4x8 And finally placing$x$x by itself by dividing both sides by$4$4.$x\le2$x2 ##### Example 3 Solve the inequality$\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>32 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}$3x2>32 Now both sides of the inequality use the same base we can continue as a we would a two step inequality$\frac{x}{2}>-2$x2>2$x>-4$x>4 #### Practice questions ##### Question 7 Solve the inequality$6^{3x-2}<6^4$63x2<64. ##### Question 8 Solve the inequality$3^{3x-4}\le9$33x49. ##### Question 9 Consider the inequality$\left(\frac{1}{27}\right)^{x+1}\ge\left(\frac{1}{81}\right)^{2x-3}$(127)x+1(181)2x3. 1. What common (prime) base can$\frac{1}{27}$127 and$\frac{1}{81}\$181 be written in terms of?

2. Hence solve the inequality.