Solve 16^x-1-4^3=0

$$16 ^ { x - 1 } - 4 ^ { 3 } = 0$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Definition of Logarithm

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

$$16^{x-1}-64=0$$

Add 64 to both sides of the equation.

$$16^{x-1}=64$$

Take the logarithm of both sides of the equation.

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

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

$$\left(x-1\right)\log(16)=\log(64)$$

Divide both sides by \log(16).

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

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

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

Add 1 to both sides of the equation.

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

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{2\ln(2)}+\frac{5}{2}$$

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