TAOCP 1.2.2 Exercise 4

Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 4. ▶ [05] What is $(0.125)^{-2/3}$?

Verified: yes
Solve time: 49s

We have $(0.125)^{-2/3}$. First, write $0.125$ as a fraction:

$0.125 = \frac{1}{8}.$

Next, apply the definition of a rational exponent, $b^{p/q} = \sqrt[q]{b^p}$:

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

Raising $1/8$ to the $-2$ power gives

$(1/8)^{-2} = 8^2 = 64.$

Taking the cube root yields

$\sqrt[3]{64} = 4.$

Therefore,

$(0.125)^{-2/3} = \boxed{4}.$