Solve 37=8^x+3

$$37 = 8 ^ { x + 3 }$$

Solve for x

$$x=\frac{\log_{2}\left(37\right)-9}{3}\approx -1.263515545$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

$$8^{x+3}=37$$

Take the logarithm of both sides of the equation.

$$\log(8^{x+3})=\log(37)$$

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

$$\left(x+3\right)\log(8)=\log(37)$$

Divide both sides by \log(8).

$$x+3=\frac{\log(37)}{\log(8)}$$

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

$$x+3=\log_{8}\left(37\right)$$

Subtract 3 from both sides of the equation.

$$x=\frac{\log_{2}\left(37\right)}{3}-3$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{3\ln(2)}+\frac{\log_{2}\left(37\right)}{3}-3$$

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