Solve (2^7)^x-1=1

$$( 2 ^ { 7 } ) ^ { x - 1 } = 1$$

Solve for x

$$x=1$$

Steps Using Definition of Logarithm

Calculate 2 to the power of 7 and get 128.

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

Take the logarithm of both sides of the equation.

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

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

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

Divide both sides by \log(128).

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

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

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

Add 1 to both sides of the equation.

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

Solve for x (complex solution)

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

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