Solve 128^x-2=1

$$128 ^ { x - 2 } = 1$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

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

$$128^{x-2}=1$$

Take the logarithm of both sides of the equation.

$$\log(128^{x-2})=\log(1)$$

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

$$\left(x-2\right)\log(128)=\log(1)$$

Divide both sides by \log(128).

$$x-2=\frac{\log(1)}{\log(128)}$$

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

$$x-2=\log_{128}\left(1\right)$$

Add 2 to both sides of the equation.

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

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{7\ln(2)}+2$$

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