Exponentials

Hong Kong

Stage 4 - Stage 5

Lesson

Recall that an exponential graph of the form $y=A\times b^{mx+c}+k$`y`=`A`×`b``m``x`+`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)3`x` is the same as $\left(\frac{5}{4}\right)^{-3x}$(54)−3`x`.

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$2`x`<23. To examine this inequality we will look at the graphs of $y=2^x$`y`=2`x` 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$2`x` will be below the line and therefore $2^x<2^3$2`x`<23 will hold. For values of $x$`x` greater than $3$3, $2^x>2^3$2`x`>23 would be true as the graph only ever increases. So the solution to $2^x<2^3$2`x`<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

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$7`x`>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$3`x`<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}$23`x`≥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}$23`x`≥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$3`x`≥−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$6`x`+3<67.

Solve the inequality $3^x\le\sqrt[4]{3}$3`x`≤^{4}√3.

Solve the inequality $25^x\le\frac{1}{5}$25`x`≤15.

Solve the inequality $5^{6x-2}\le5^{2x+2}$56`x`−2≤52`x`+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$6`x`−2≤2`x`+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$4`x`≤8

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

$x\le2$`x`≤2

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

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

**Do:** Rewrite $\sqrt{3^x}$√3`x` 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}$3`x`2>3−2

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

$\frac{x}{2}>-2$`x`2>−2

$x>-4$`x`>−4

Solve the inequality $6^{3x-2}<6^4$63`x`−2<64.

Solve the inequality $3^{3x-4}\le9$33`x`−4≤9.

Consider the inequality$\left(\frac{1}{27}\right)^{x+1}\ge\left(\frac{1}{81}\right)^{2x-3}$(127)`x`+1≥(181)2`x`−3.

What common (prime) base can $\frac{1}{27}$127 and $\frac{1}{81}$181 be written in terms of?

Hence solve the inequality.