Solve 8^2x-1=16

$$8 ^ { 2 x - 1 } = 16$$

Solve for x

$$x = \frac{7}{6} = 1\frac{1}{6} \approx 1.166666667$$

Steps Using Definition of Logarithm

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

$$8^{2x-1}=16$$

Take the logarithm of both sides of the equation.

$$\log(8^{2x-1})=\log(16)$$

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

$$\left(2x-1\right)\log(8)=\log(16)$$

Divide both sides by \log(8).

$$2x-1=\frac{\log(16)}{\log(8)}$$

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

$$2x-1=\log_{8}\left(16\right)$$

Add 1 to both sides of the equation.

$$2x=\frac{4}{3}-\left(-1\right)$$

Divide both sides by 2.

$$x=\frac{\frac{7}{3}}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{3\ln(2)}+\frac{7}{6}$$

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