Solve e^-2k=11/15

$$e ^ { - 2 k } = \frac { 11 } { 15 }$$

Solve for k

$$k=\frac{\ln(\frac{15}{11})}{2}\approx 0.155077464$$

Steps Using Definition of Logarithm

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

$$e^{-2k}=\frac{11}{15}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-2k})=\log(\frac{11}{15})$$

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

$$-2k\log(e)=\log(\frac{11}{15})$$

Divide both sides by \log(e).

$$-2k=\frac{\log(\frac{11}{15})}{\log(e)}$$

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

$$-2k=\log_{e}\left(\frac{11}{15}\right)$$

Divide both sides by -2.

$$k=\frac{\ln(\frac{11}{15})}{-2}$$

Solve for k (complex solution)

$$k=-i\pi n_{1}+\frac{\ln(\frac{15}{11})}{2}$$

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