Solve (1/16)^-3p=4^3

$$( \frac { 1 } { 16 } ) ^ { - 3 p } = 4 ^ { 3 }$$

Solve for p

$$p=\frac{1}{2}=0.5$$

Steps Using Definition of Logarithm

Use the rules of exponents and logarithms to solve the equation.

$$\left(\frac{1}{16}\right)^{-3p}=64$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{16}\right)^{-3p})=\log(64)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$-3p\log(\frac{1}{16})=\log(64)$$

Divide both sides by \log(\frac{1}{16}).

$$-3p=\frac{\log(64)}{\log(\frac{1}{16})}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$-3p=\log_{\frac{1}{16}}\left(64\right)$$

Divide both sides by -3.

$$p=-\frac{\frac{3}{2}}{-3}$$