Solve 4^2x+1=8

$$4 ^ { 2 x + 1 } = 8$$

Solve for x

$$x=\frac{1}{4}=0.25$$

Steps Using Definition of Logarithm

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

$$4^{2x+1}=8$$

Take the logarithm of both sides of the equation.

$$\log(4^{2x+1})=\log(8)$$

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

$$\left(2x+1\right)\log(4)=\log(8)$$

Divide both sides by \log(4).

$$2x+1=\frac{\log(8)}{\log(4)}$$

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

$$2x+1=\log_{4}\left(8\right)$$

Subtract 1 from both sides of the equation.

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

Divide both sides by 2.

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

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{2\ln(2)}+\frac{1}{4}$$

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