Solve 8^2+x=8

$${ 8 }^{ 2+x } = 8$$

Solve for x

$$x=-1$$

Steps Using Definition of Logarithm

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

$$8^{x+2}=8$$

Take the logarithm of both sides of the equation.

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

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

$$\left(x+2\right)\log(8)=\log(8)$$

Divide both sides by \log(8).

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

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

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

Subtract 2 from both sides of the equation.

$$x=1-2$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{3\ln(2)}-1$$

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