Solve 4^2x-7=1/64

$$4 ^ { 2 x - 7 } = \frac { 1 } { 64 }$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

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

$$4^{2x-7}=\frac{1}{64}$$

Take the logarithm of both sides of the equation.

$$\log(4^{2x-7})=\log(\frac{1}{64})$$

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

$$\left(2x-7\right)\log(4)=\log(\frac{1}{64})$$

Divide both sides by \log(4).

$$2x-7=\frac{\log(\frac{1}{64})}{\log(4)}$$

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

$$2x-7=\log_{4}\left(\frac{1}{64}\right)$$

Add 7 to both sides of the equation.

$$2x=-3-\left(-7\right)$$

Divide both sides by 2.

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

Solve for x (complex solution)

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

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