Solve 28=36e^-10k

$$28 = 36 e ^ { - 10 k }$$

Solve for k

$$k=\frac{\ln(3)}{5}-\frac{\ln(7)}{10}\approx 0.025131443$$

Steps Using Definition of Logarithm

Divide both sides by 36.

$$\frac{28}{36}=e^{-10k}$$

Reduce the fraction \frac{28}{36} to lowest terms by extracting and canceling out 4.

$$\frac{7}{9}=e^{-10k}$$

Swap sides so that all variable terms are on the left hand side.

$$e^{-10k}=\frac{7}{9}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-10k})=\log(\frac{7}{9})$$

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

$$-10k\log(e)=\log(\frac{7}{9})$$

Divide both sides by \log(e).

$$-10k=\frac{\log(\frac{7}{9})}{\log(e)}$$

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

$$-10k=\log_{e}\left(\frac{7}{9}\right)$$

Divide both sides by -10.

$$k=\frac{\ln(\frac{7}{9})}{-10}$$

Solve for k (complex solution)

$$k=-\frac{\pi n_{1}i}{5}+\frac{\ln(3)}{5}-\frac{\ln(7)}{10}$$

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