Solve 8^3x+1=1/8

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

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

Steps Using Definition of Logarithm

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

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

Take the logarithm of both sides of the equation.

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

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

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

Divide both sides by \log(8).

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

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

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

Subtract 1 from both sides of the equation.

$$3x=-1-1$$

Divide both sides by 3.

$$x=-\frac{2}{3}$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{9\ln(2)}-\frac{2}{3}$$

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