Negative bases, nested powers and mixed problems

The last les­son of the course gath­ers the remain­ing kinds of ques­tion: neg­a­tive frac­tional expo­nents on frac­tions, a neg­a­tive base under a cube root, pow­ers nested inside pow­ers, and quick equa­tions whose answer is used to find some­thing else. Each exam­ple is short; together they make a good revi­sion of all four­teen laws.

Neg­a­tive and frac­tional expo­nents together#

When an expo­nent is both neg­a­tive and a frac­tion, deal with it in three moves: the minus sign turns the base upside down (Law 8), the denom­i­na­tor of the expo­nent takes a root, and the numer­a­tor takes a power (Law 13).

Exam­ple 1#

Find $\left\{8^{-4/3} \div 2^{-2}\right\}^{1/2}$.

$$\displaystyle \begin{aligned} \left\{\frac{(2^{3})^{-4/3}}{2^{-2}}\right\}^{1/2} &= \left(\frac{2^{-4}}{2^{-2}}\right)^{1/2} \\ &= (2^{-4 + 2})^{1/2} = (2^{-2})^{1/2} \\ &= 2^{-1} = \frac{1}{2} \end{aligned}$$

Exam­ple 2#

Find $\displaystyle \left(\frac{100}{9}\right)^{-3/2}$.

$$\displaystyle \begin{aligned} \left(\frac{100}{9}\right)^{-3/2} &= \left(\frac{9}{100}\right)^{3/2} \\ &= \left[\left(\frac{3}{10}\right)^{2}\right]^{3/2} \\ &= \left(\frac{3}{10}\right)^{3} = \frac{27}{1000} \end{aligned}$$

Exam­ple 3#

Find $\left(\sqrt[3]{8}\right)^{-1/2}$.

First, $\sqrt[3]{8} = \sqrt[3]{2 \times 2 \times 2} = 2$. Then

$$\displaystyle 2^{-1/2} = \frac{1}{2^{1/2}} = \frac{1}{\sqrt{2}}$$

A neg­a­tive base#

Frac­tional expo­nents are nor­mally used with pos­i­tive bases. A neg­a­tive base can still be han­dled when the root is an odd root, such as a cube root, because odd roots of neg­a­tive num­bers exist: $\sqrt[3]{-27} = -3$, since $(-3)^{3} = -27$. Even roots of neg­a­tive num­bers, such as $\sqrt{-4}$, are not real num­bers.

Exam­ple 4#

Find $\displaystyle \left(-\frac{1}{27}\right)^{-2/3}$.

The minus in the expo­nent turns the frac­tion upside down: $\displaystyle \left(-\frac{1}{27}\right)^{-2/3} = (-27)^{2/3}$. Now take the cube root first, then square:

$$\begin{aligned} (-27)^{2/3} &= \left[(-3)^{3}\right]^{2/3} \\ &= (-3)^{2} = 9 \end{aligned}$$

The answer is $9$, not $\displaystyle \frac{1}{9}$: the neg­a­tive expo­nent was used up when the frac­tion was turned over. And it is pos­i­tive, because squar­ing removes the minus sign.

Pow­ers inside pow­ers#

Exam­ple 5#

If $\displaystyle \left[\left\{\left(\frac{1}{7^{2}}\right)^{-2}\right\}^{-1/3}\right]^{1/4} = 7^{m}$, find $m$.

Write $\displaystyle \frac{1}{7^{2}} = 7^{-2}$ and mul­ti­ply all the expo­nents together (Law 3):

$$\displaystyle \begin{aligned} \left[\left\{(7^{-2})^{-2}\right\}^{-1/3}\right]^{1/4} &= 7^{(-2)(-2)\left(-\frac{1}{3}\right)\left(\frac{1}{4}\right)} \\ &= 7^{-1/3} \end{aligned}$$

So $\displaystyle m = -\frac{1}{3}$. Keep track of the signs: two neg­a­tives make a pos­i­tive, and the third makes it neg­a­tive again.

Exam­ple 6#

If $\displaystyle \frac{x}{x^{1.5}} = 8x^{-1}$ and $x > 0$, find $x$.

Change the dec­i­mal expo­nent to a frac­tion: $\displaystyle 1.5 = \frac{15}{10} = \frac{3}{2}$. Mul­ti­ply both sides by $x$:

$$\displaystyle \begin{aligned} \frac{x}{x^{3/2}} &= \frac{8}{x} \\ \frac{x^{2}}{x^{3/2}} &= 8 \\ x^{2 - 3/2} &= 8 \\ x^{1/2} &= 8 \\ x &= 8^{2} = 64 \end{aligned}$$

Solve, then use the answer#

Exam­ple 7#

If $8^{x + 1} = 64$, find $3^{2x + 1}$.

$$8^{x + 1} = 8^{2}, \quad \text{so } x + 1 = 2 \text{ and } x = 1$$

Then $3^{2x + 1} = 3^{2(1) + 1} = 3^{3} = 27$.

Exam­ple 8#

If $16^{2x + 3} = 64^{x + 3}$, find $4^{2x - 2}$.

Both $16$ and $64$ are pow­ers of $4$.

$$\begin{aligned} (4^{2})^{2x + 3} &= (4^{3})^{x + 3} \\ 4^{4x + 6} &= 4^{3x + 9} \\ 4x + 6 &= 3x + 9 \\ x &= 3 \end{aligned}$$

Then $4^{2x - 2} = 4^{6 - 2} = 4^{4} = 256$.

Exam­ple 9#

If $\displaystyle \frac{3^{5x} \times 81^{2} \times 6561}{3^{2x}} = 3^{7}$, find $x$.

Write every num­ber as a power of $3$: $81 = 3^{4}$, and divid­ing $6561$ by $3$ repeat­edly shows $6561 = 3^{8}$.

$$\begin{aligned} 3^{5x - 2x} \times (3^{4})^{2} \times 3^{8} &= 3^{7} \\ 3^{3x} \times 3^{8} \times 3^{8} &= 3^{7} \\ 3^{3x + 16} &= 3^{7} \\ 3x + 16 &= 7 \\ 3x &= -9 \\ x &= -3 \end{aligned}$$

Exam­ple 10#

If $\sqrt{2^{n}} = 1024$, find $\displaystyle 3^{2\left(\frac{n}{4} - 4\right)}$.

$\sqrt{2^{n}} = 2^{n/2}$ and $1024 = 2^{10}$, so $\displaystyle \frac{n}{2} = 10$ and $n = 20$. Then

$$\displaystyle 3^{2\left(\frac{20}{4} - 4\right)} = 3^{2(5 - 4)} = 3^{2} = 9$$

Exam­ple 11: tak­ing out the small­est power#

Sim­plify $\displaystyle \frac{5^{n + 2} - 6 \times 5^{n + 1}}{13 \times 5^{n} - 2 \times 5^{n + 1}}$.

Every term con­tains $5^{n}$. Take it out of the top and the bot­tom, using $5^{n + 2} = 5^{n} \times 25$ and $5^{n + 1} = 5^{n} \times 5$:

$$\displaystyle \frac{5^{n}(25 - 6 \times 5)}{5^{n}(13 - 2 \times 5)} = \frac{25 - 30}{13 - 10} = -\frac{5}{3}$$

The answer does not depend on $n$ at all.

Prac­tice#

1. Find $\left\{27^{-2/3} \div 3^{-4}\right\}^{1/2}$.
2. Find $\displaystyle \left(\frac{49}{16}\right)^{-3/2}$.
3. Find $\left(\sqrt[3]{27}\right)^{-1/2}$.
4. Find $\displaystyle \left(-\frac{1}{8}\right)^{-2/3}$.
5. If $\displaystyle \left[\left\{\left(\frac{1}{5^{3}}\right)^{-2}\right\}^{1/3}\right]^{-1/2} = 5^{m}$, find $m$.
6. If $\displaystyle \frac{x^{2.5}}{x} = 27$ and $x > 0$, find $x$.
7. If $9^{x + 1} = 243$, find $2^{2x}$.
8. If $9^{x + 2} = 27^{x + 1}$, find $2^{x - 1}$.
9. If $\displaystyle \frac{3^{4x} \times 9^{3}}{3^{x}} = 3^{9}$, find $x$.
10. Sim­plify $\displaystyle \frac{2^{n + 3} - 2 \times 2^{n + 1}}{2^{n + 2} + 2^{n}}$.

Answers#