Solve 16^2x-1=1/64

$$16 ^ { 2 x - 1 } = \frac { 1 } { 64 }$$

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.

$$16^{2x-1}=\frac{1}{64}$$

Take the logarithm of both sides of the equation.

$$\log(16^{2x-1})=\log(\frac{1}{64})$$

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

$$\left(2x-1\right)\log(16)=\log(\frac{1}{64})$$

Divide both sides by \log(16).

$$2x-1=\frac{\log(\frac{1}{64})}{\log(16)}$$

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

$$2x-1=\log_{16}\left(\frac{1}{64}\right)$$

Add 1 to both sides of the equation.

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

Divide both sides by 2.

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

Solve for x (complex solution)

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

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