Solve 2^2x-1:4=8^2

$$2 ^ { 2 x - 1 } : 4 = 8 ^ { 2 }$$

Solve for x

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Definition of Logarithm

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

$$\frac{1}{4}\times 2^{2x-1}=64$$

Multiply both sides by 4.

$$2^{2x-1}=256$$

Take the logarithm of both sides of the equation.

$$\log(2^{2x-1})=\log(256)$$

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

$$\left(2x-1\right)\log(2)=\log(256)$$

Divide both sides by \log(2).

$$2x-1=\frac{\log(256)}{\log(2)}$$

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

$$2x-1=\log_{2}\left(256\right)$$

Add 1 to both sides of the equation.

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

Divide both sides by 2.

$$x=\frac{9}{2}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(2)}+\frac{9}{2}$$

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