Solve 32=2^2x-7

$$32 = 2 ^ { 2 x - 7 }$$

Solve for x

$$x=6$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

$$2^{2x-7}=32$$

Take the logarithm of both sides of the equation.

$$\log(2^{2x-7})=\log(32)$$

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

$$\left(2x-7\right)\log(2)=\log(32)$$

Divide both sides by \log(2).

$$2x-7=\frac{\log(32)}{\log(2)}$$

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

$$2x-7=\log_{2}\left(32\right)$$

Add 7 to both sides of the equation.

$$2x=5-\left(-7\right)$$

Divide both sides by 2.

$$x=\frac{12}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(2)}+6$$

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