Solve 2^2x-5=32

$$2 ^ { 2 x - 5 } = 32$$

Solve for x

$$x=5$$

Steps Using Definition of Logarithm

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

$$2^{2x-5}=32$$

Take the logarithm of both sides of the equation.

$$\log(2^{2x-5})=\log(32)$$

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

$$\left(2x-5\right)\log(2)=\log(32)$$

Divide both sides by \log(2).

$$2x-5=\frac{\log(32)}{\log(2)}$$

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

$$2x-5=\log_{2}\left(32\right)$$

Add 5 to both sides of the equation.

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

Divide both sides by 2.

$$x=\frac{10}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(2)}+5$$

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