Solve 8^x+3=1/16

$$8 ^ { x + 3 } = \frac { 1 } { 16 }$$

Solve for x

$$x = -\frac{13}{3} = -4\frac{1}{3} \approx -4.333333333$$

Steps Using Definition of Logarithm

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

$$8^{x+3}=\frac{1}{16}$$

Take the logarithm of both sides of the equation.

$$\log(8^{x+3})=\log(\frac{1}{16})$$

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

$$\left(x+3\right)\log(8)=\log(\frac{1}{16})$$

Divide both sides by \log(8).

$$x+3=\frac{\log(\frac{1}{16})}{\log(8)}$$

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

$$x+3=\log_{8}\left(\frac{1}{16}\right)$$

Subtract 3 from both sides of the equation.

$$x=-\frac{4}{3}-3$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{3\ln(2)}-\frac{13}{3}$$

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