Solve e^-3r=21/20^2

$$e ^ { - 3 r } = \frac { 21 } { 20 ^ { 2 } }$$

Solve for r

$$r=\frac{2\ln(5)+4\ln(2)-\ln(21)}{3}\approx 0.982314036$$

Steps Using Definition of Logarithm

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

$$e^{-3r}=\frac{21}{400}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-3r})=\log(\frac{21}{400})$$

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

$$-3r\log(e)=\log(\frac{21}{400})$$

Divide both sides by \log(e).

$$-3r=\frac{\log(\frac{21}{400})}{\log(e)}$$

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

$$-3r=\log_{e}\left(\frac{21}{400}\right)$$

Divide both sides by -3.

$$r=\frac{\ln(\frac{21}{400})}{-3}$$

Solve for r (complex solution)

$$r=-\frac{i\times 2\pi n_{1}}{3}+\frac{2\ln(20)}{3}-\frac{\ln(21)}{3}$$

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