Solve 5^-x+1=1

$$5 ^ { - x + 1 } = 1$$

Solve for x

$$x=1$$

Steps Using Definition of Logarithm

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

$$5^{-x+1}=1$$

Take the logarithm of both sides of the equation.

$$\log(5^{-x+1})=\log(1)$$

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

$$\left(-x+1\right)\log(5)=\log(1)$$

Divide both sides by \log(5).

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

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

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

Subtract 1 from both sides of the equation.

$$-x=-1$$

Divide both sides by -1.

$$x=-\frac{1}{-1}$$

Solve for x (complex solution)

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

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