Solve 4^-x+2=19

$$4 ^ { - x + 2 } = 19$$

Solve for x

$$x=-\frac{\log_{2}\left(19\right)}{2}+2\approx -0.123963757$$

Steps Using Definition of Logarithm

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

$$4^{-x+2}=19$$

Take the logarithm of both sides of the equation.

$$\log(4^{-x+2})=\log(19)$$

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

$$\left(-x+2\right)\log(4)=\log(19)$$

Divide both sides by \log(4).

$$-x+2=\frac{\log(19)}{\log(4)}$$

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

$$-x+2=\log_{4}\left(19\right)$$

Subtract 2 from both sides of the equation.

$$-x=\frac{\log_{2}\left(19\right)}{2}-2$$

Divide both sides by -1.

$$x=\frac{\frac{\log_{2}\left(19\right)}{2}-2}{-1}$$

Solve for x (complex solution)

$$x=-\frac{\pi n_{1}i}{\ln(2)}-\frac{\log_{2}\left(19\right)}{2}+2$$

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