Solve 32=(2)^n-1

$$32 = ( 2 ) ^ { n - 1 }$$

Solve for n

$$n=6$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

$$2^{n-1}=32$$

Take the logarithm of both sides of the equation.

$$\log(2^{n-1})=\log(32)$$

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

$$\left(n-1\right)\log(2)=\log(32)$$

Divide both sides by \log(2).

$$n-1=\frac{\log(32)}{\log(2)}$$

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

$$n-1=\log_{2}\left(32\right)$$

Add 1 to both sides of the equation.

$$n=5-\left(-1\right)$$