Solve 4^-x-16=0

$${ 4 }^{ -x } -16 = 0$$

Solve for x

$$x=-2$$

Steps Using Definition of Logarithm

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

$$4^{-x}-16=0$$

Add 16 to both sides of the equation.

$$4^{-x}=16$$

Take the logarithm of both sides of the equation.

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

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

$$-x\log(4)=\log(16)$$

Divide both sides by \log(4).

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

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

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

Divide both sides by -1.

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

Solve for x (complex solution)

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

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