Evaluating expressions with exponents

With the four­teen laws in hand, expres­sions that look impos­si­ble with­out a cal­cu­la­tor, such as $\displaystyle \frac{25^{3/2} \times 243^{3/5}}{16^{5/4} \times 8^{4/3}}$, become a few lines of work. The secret is always the same: write every base as a power of a prime. This les­son shows the method on a range of numer­i­cal expres­sions.

The method#

1. Write each base as a power of a prime (or of a sim­ple frac­tion): $81 = 3^{4}$, $243 = 3^{5}$, $16 = 2^{4}$, $\displaystyle \frac{64}{125} = \left(\frac{4}{5}\right)^{3}$.
2. Use Law 3, $(a^{m})^{n} = a^{mn}$, to mul­ti­ply out the expo­nents. Frac­tional expo­nents usu­ally can­cel neatly at this step.
3. Col­lect pow­ers of the same prime with Laws 1 and 2.
4. Work out the value.

It helps to know the small pow­ers by heart:

Exam­ple 1#

Find $81^{2} \times 3^{-7}$.

$$\begin{aligned} 81^{2} \times 3^{-7} &= (3^{4})^{2} \times 3^{-7} \\ &= 3^{8} \times 3^{-7} \\ &= 3^{8 - 7} \\ &= 3^{1} = 3 \end{aligned}$$

Exam­ple 2#

Find $\displaystyle \frac{25^{3/2} \times 243^{3/5}}{16^{5/4} \times 8^{4/3}}$.

$$\displaystyle \begin{aligned} \frac{25^{3/2} \times 243^{3/5}}{16^{5/4} \times 8^{4/3}} &= \frac{(5^{2})^{3/2} \times (3^{5})^{3/5}}{(2^{4})^{5/4} \times (2^{3})^{4/3}} \\ &= \frac{5^{3} \times 3^{3}}{2^{5} \times 2^{4}} \\ &= \frac{125 \times 27}{512} = \frac{3375}{512} \end{aligned}$$

Each frac­tional expo­nent can­celled with the prime's own power: $\displaystyle 2 \times \frac{3}{2} = 3$, $\displaystyle 5 \times \frac{3}{5} = 3$, $\displaystyle 4 \times \frac{5}{4} = 5$ and $\displaystyle 3 \times \frac{4}{3} = 4$. The answer can also be writ­ten $\displaystyle \frac{15^{3}}{2^{9}}$, since $5^{3} \times 3^{3} = 15^{3}$ and $2^{5} \times 2^{4} = 2^{9}$.

Exam­ple 3#

Find $\displaystyle \left(\frac{64}{125}\right)^{-2/3} + \frac{1}{\left(\frac{256}{625}\right)^{1/4}} + \frac{\sqrt{25}}{\sqrt[3]{64}}$.

Write each frac­tion as a power of $\displaystyle \frac{4}{5}$: $\displaystyle \frac{64}{125} = \left(\frac{4}{5}\right)^{3}$ and $\displaystyle \frac{256}{625} = \left(\frac{4}{5}\right)^{4}$.

$$\displaystyle \begin{aligned} &= \left[\left(\frac{4}{5}\right)^{3}\right]^{-2/3} + \frac{1}{\left[\left(\frac{4}{5}\right)^{4}\right]^{1/4}} + \frac{\sqrt{5^{2}}}{\sqrt[3]{4^{3}}} \\ &= \left(\frac{4}{5}\right)^{-2} + \frac{1}{\frac{4}{5}} + \frac{5}{4} \\ &= \left(\frac{5}{4}\right)^{2} + \frac{5}{4} + \frac{5}{4} \\ &= \frac{25}{16} + \frac{10}{4} \\ &= \frac{25 + 40}{16} = \frac{65}{16} \end{aligned}$$

Exam­ple 4: zero and neg­a­tive expo­nents together#

Find $\displaystyle \left(\frac{1}{4}\right)^{-2} - 3 \times 8^{2/3} \times 4^{0} + \left(\frac{9}{16}\right)^{-1/2}$.

$$\displaystyle \begin{aligned} &= 4^{2} - 3 \times (2^{3})^{2/3} \times 1 + \left(\frac{16}{9}\right)^{1/2} \\ &= 16 - 3 \times 4 + \left[\left(\frac{4}{3}\right)^{2}\right]^{1/2} \\ &= 16 - 12 + \frac{4}{3} \\ &= 4 + \frac{4}{3} = \frac{16}{3} \end{aligned}$$

Laws 8 and 6 did the first step: $\displaystyle \left(\frac{1}{4}\right)^{-2} = 4^{2}$ and $4^{0} = 1$.

Exam­ple 5: sev­eral primes at once#

Find $\displaystyle \frac{2^{1/2} \times 3^{1/3} \times 4^{1/4}}{10^{-1/5} \times 5^{3/5}} \div \frac{3^{4/3} \times 5^{-7/5}}{4^{-3/5} \times 6}$.

Break every base into primes: $4 = 2^{2}$, $10 = 2 \times 5$ and $6 = 3 \times 2$. Then divid­ing by the sec­ond frac­tion means mul­ti­ply­ing by it turned upside down.

$$\displaystyle \begin{aligned} &= \frac{2^{1/2} \times 3^{1/3} \times (2^{2})^{1/4}}{(2 \times 5)^{-1/5} \times 5^{3/5}} \times \frac{(2^{2})^{-3/5} \times 3 \times 2}{3^{4/3} \times 5^{-7/5}} \\ &= \frac{2^{1/2} \times 3^{1/3} \times 2^{1/2}}{2^{-1/5} \times 5^{-1/5} \times 5^{3/5}} \times \frac{2^{-6/5} \times 3 \times 2}{3^{4/3} \times 5^{-7/5}} \end{aligned}$$

Now col­lect each prime. A power on the bot­tom moves to the top with its sign changed.

$$\displaystyle \begin{aligned} &= 2^{\frac{1}{2} + \frac{1}{2} + \frac{1}{5} - \frac{6}{5} + 1} \times 3^{\frac{1}{3} + 1 - \frac{4}{3}} \times 5^{\frac{1}{5} - \frac{3}{5} + \frac{7}{5}} \\ &= 2^{1} \times 3^{0} \times 5^{1} \\ &= 2 \times 5 = 10 \end{aligned}$$

Keep a col­umn for each prime when you do one of these; it stops terms going miss­ing.

Prac­tice#

1. Find $32^{2} \times 2^{-7}$.
2. Find $9^{3/2}$.
3. Find $16^{-3/4}$.
4. Find $\displaystyle \left(\frac{27}{8}\right)^{2/3}$.
5. Find $\displaystyle \left(\frac{49}{25}\right)^{-1/2}$.
6. Find $\displaystyle \frac{125^{2/3} \times 16^{3/4}}{32^{2/5}}$.
7. Find $\displaystyle \left(\frac{64}{729}\right)^{-1/6}$.
8. Find $\displaystyle \left(\frac{81}{16}\right)^{-3/4} + \left(\frac{8}{27}\right)^{-1/3}$.
9. Find $\displaystyle \left(\frac{1}{3}\right)^{-2} - 2 \times 27^{2/3} \times 5^{0} + \left(\frac{4}{9}\right)^{-1/2}$.
10. Find $\displaystyle \frac{2^{1/3} \times 3^{1/2} \times 4^{1/3}}{6^{1/2} \times 2^{-1/2}}$.

Answers#