Solve 2^x+8=1/32

$$2 ^ { x + 8 } = \frac { 1 } { 32 }$$

Solve for x

$$x=-13$$

Steps Using Definition of Logarithm

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

$$2^{x+8}=\frac{1}{32}$$

Take the logarithm of both sides of the equation.

$$\log(2^{x+8})=\log(\frac{1}{32})$$

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

$$\left(x+8\right)\log(2)=\log(\frac{1}{32})$$

Divide both sides by \log(2).

$$x+8=\frac{\log(\frac{1}{32})}{\log(2)}$$

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

$$x+8=\log_{2}\left(\frac{1}{32}\right)$$

Subtract 8 from both sides of the equation.

$$x=-5-8$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(2)}-13$$

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