Solve 2^n-1=1/32

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

Solve for n

$$n=-4$$

Steps Using Definition of Logarithm

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

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

Take the logarithm of both sides of the equation.

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

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

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

Divide both sides by \log(2).

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

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

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

Add 1 to both sides of the equation.

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

Solve for n (complex solution)

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

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