Solve 20^-x=1/7

$${ 20 }^{ -x } = \frac{ 1 }{ 7 }$$

Solve for x

$$x=\log_{20}\left(7\right)\approx 0.649560766$$

Steps Using Definition of Logarithm

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

$$20^{-x}=\frac{1}{7}$$

Take the logarithm of both sides of the equation.

$$\log(20^{-x})=\log(\frac{1}{7})$$

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

$$-x\log(20)=\log(\frac{1}{7})$$

Divide both sides by \log(20).

$$-x=\frac{\log(\frac{1}{7})}{\log(20)}$$

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

$$-x=\log_{20}\left(\frac{1}{7}\right)$$

Divide both sides by -1.

$$x=-\frac{\log_{20}\left(7\right)}{-1}$$

Solve for x (complex solution)

$$x=-\frac{2\pi n_{1}i}{\ln(20)}+\log_{20}\left(7\right)$$

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