Solve (1/16)^2x=64

$$( \frac { 1 } { 16 } ) ^ { 2 x } = 64$$

Solve for x

$$x=-\frac{3}{4}=-0.75$$

Steps Using Definition of Logarithm

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

$$\left(\frac{1}{16}\right)^{2x}=64$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{16}\right)^{2x})=\log(64)$$

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

$$2x\log(\frac{1}{16})=\log(64)$$

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

$$2x=\frac{\log(64)}{\log(\frac{1}{16})}$$

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

$$2x=\log_{\frac{1}{16}}\left(64\right)$$

Divide both sides by 2.

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

Solve for x (complex solution)

$$x=-\frac{i\pi n_{1}}{4\ln(2)}-\frac{3}{4}$$

$$n_{1}\in \mathrm{Z}$$