Solve 4^-x=1/16

$$4 ^ { - x } = \frac { 1 } { 16 }$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

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

$$4^{-x}=\frac{1}{16}$$

Take the logarithm of both sides of the equation.

$$\log(4^{-x})=\log(\frac{1}{16})$$

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

$$-x\log(4)=\log(\frac{1}{16})$$

Divide both sides by \log(4).

$$-x=\frac{\log(\frac{1}{16})}{\log(4)}$$

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

$$-x=\log_{4}\left(\frac{1}{16}\right)$$

Divide both sides by -1.

$$x=-\frac{2}{-1}$$

Solve for x (complex solution)

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

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