Solve 4^-3a=16

$$4 ^ { - 3 a } = 16$$

Solve for a

$$a=-\frac{2}{3}\approx -0.666666667$$

Steps Using Definition of Logarithm

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

$$4^{-3a}=16$$

Take the logarithm of both sides of the equation.

$$\log(4^{-3a})=\log(16)$$

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

$$-3a\log(4)=\log(16)$$

Divide both sides by \log(4).

$$-3a=\frac{\log(16)}{\log(4)}$$

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

$$-3a=\log_{4}\left(16\right)$$

Divide both sides by -3.

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

Solve for a (complex solution)

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

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