Solve 4^2x-1=64

$$4 ^ { 2 x - 1 } = 64$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

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

$$4^{2x-1}=64$$

Take the logarithm of both sides of the equation.

$$\log(4^{2x-1})=\log(64)$$

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

$$\left(2x-1\right)\log(4)=\log(64)$$

Divide both sides by \log(4).

$$2x-1=\frac{\log(64)}{\log(4)}$$

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

$$2x-1=\log_{4}\left(64\right)$$

Add 1 to both sides of the equation.

$$2x=3-\left(-1\right)$$

Divide both sides by 2.

$$x=\frac{4}{2}$$

Solve for x (complex solution)

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

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