Solve 2^x/2^x-1=8^y

$$\frac { 2 ^ { x } } { 2 ^ { x - 1 } } = 8 ^ { y }$$

Solve for y

$$y = \frac{1}{3} = 0.3333333333333333$$

Steps Using Definition of Logarithm

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

$$8^{y}=\frac{2^{x}}{2^{x-1}}$$

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

$$8^{y}=2$$

Take the logarithm of both sides of the equation.

$$\log(8^{y})=\log(2)$$

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

$$y\log(8)=\log(2)$$

Divide both sides by \log(8).

$$y=\frac{\log(2)}{\log(8)}$$

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

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

Solve for x (complex solution)

$$x\in \mathrm{C}$$

$$\exists n_{1}\in \mathrm{Z}\text{ : }y=\frac{2\pi n_{1}i}{3\ln(2)}+\frac{1}{3}$$

Solve for y (complex solution)

$$y=\frac{2\pi n_{1}i}{3\ln(2)}+\frac{1}{3}$$

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

Solve for x

$$x\in \mathrm{R}$$

$$y = \frac{1}{3} = 0.3333333333333333$$