Solve 4^-a=64

$$4 ^ { - a } = 64$$

Solve for a

$$a=-3$$

Steps Using Definition of Logarithm

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

$$4^{-a}=64$$

Take the logarithm of both sides of the equation.

$$\log(4^{-a})=\log(64)$$

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

$$-a\log(4)=\log(64)$$

Divide both sides by \log(4).

$$-a=\frac{\log(64)}{\log(4)}$$

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

$$-a=\log_{4}\left(64\right)$$

Divide both sides by -1.

$$a=\frac{3}{-1}$$