Solve 2^n-1=1/16

$$2 ^ { n - 1 } = \frac { 1 } { 16 }$$

Solve for n

$$n=-3$$

Steps Using Definition of Logarithm

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

$$2^{n-1}=\frac{1}{16}$$

Take the logarithm of both sides of the equation.

$$\log(2^{n-1})=\log(\frac{1}{16})$$

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

$$\left(n-1\right)\log(2)=\log(\frac{1}{16})$$

Divide both sides by \log(2).

$$n-1=\frac{\log(\frac{1}{16})}{\log(2)}$$

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

$$n-1=\log_{2}\left(\frac{1}{16}\right)$$

Add 1 to both sides of the equation.

$$n=-4-\left(-1\right)$$