Solve 8^2x+1=32

$$8 ^ { 2 x + 1 } = 32$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Definition of Logarithm

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

$$8^{2x+1}=32$$

Take the logarithm of both sides of the equation.

$$\log(8^{2x+1})=\log(32)$$

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

$$\left(2x+1\right)\log(8)=\log(32)$$

Divide both sides by \log(8).

$$2x+1=\frac{\log(32)}{\log(8)}$$

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

$$2x+1=\log_{8}\left(32\right)$$

Subtract 1 from both sides of the equation.

$$2x=\frac{5}{3}-1$$

Divide both sides by 2.

$$x=\frac{\frac{2}{3}}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{3\ln(2)}+\frac{1}{3}$$

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