Solve frac{1}5^x+1=125

$$\frac { 1 } { 5 } ^ { x + 1 } = 125$$

Solve for x

$$x=-4$$

Steps Using Definition of Logarithm

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

$$\left(\frac{1}{5}\right)^{x+1}=125$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{5}\right)^{x+1})=\log(125)$$

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

$$\left(x+1\right)\log(\frac{1}{5})=\log(125)$$

Divide both sides by \log(\frac{1}{5}).

$$x+1=\frac{\log(125)}{\log(\frac{1}{5})}$$

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

$$x+1=\log_{\frac{1}{5}}\left(125\right)$$

Subtract 1 from both sides of the equation.

$$x=-3-1$$

Solve for x (complex solution)

$$x=-\frac{i\times 2\pi n_{1}}{\ln(5)}-4$$

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