Solve 28=36e^-k10

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

Solve for k_10

$$k_{10}=\ln(\frac{9}{7})\approx 0.251314428$$

Steps Using Definition of Logarithm

Divide both sides by 36.

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

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

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

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

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

Take the logarithm of both sides of the equation.

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

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

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

Divide both sides by \log(e).

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

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

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

Divide both sides by -1.

$$k_{10}=\frac{\ln(\frac{7}{9})}{-1}$$

Solve for k_10 (complex solution)

$$k_{10}=-2\pi n_{1}i+\ln(\frac{9}{7})$$

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