Solve 2^n+1=992

$$2 ^ { n + 1 } = 992$$

Solve for n

$$n=\log_{2}\left(31\right)+4\approx 8.95419631$$

Steps Using Definition of Logarithm

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

$$2^{n+1}=992$$

Take the logarithm of both sides of the equation.

$$\log(2^{n+1})=\log(992)$$

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

$$\left(n+1\right)\log(2)=\log(992)$$

Divide both sides by \log(2).

$$n+1=\frac{\log(992)}{\log(2)}$$

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

$$n+1=\log_{2}\left(992\right)$$

Subtract 1 from both sides of the equation.

$$n=\log_{2}\left(992\right)-1$$