Solve 25.3=20+7e^-k2

$$25.3 = 20 + 7 e ^ { - k 2 }$$

Solve for k_2

$$k_{2}=\ln(\frac{70}{53})\approx 0.278203328$$

Steps Using Definition of Logarithm

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

$$20+7e^{-k_{2}}=25.3$$

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

$$7e^{-k_{2}}+20=25.3$$

Subtract 20 from both sides of the equation.

$$7e^{-k_{2}}=5.3$$

Divide both sides by 7.

$$e^{-k_{2}}=\frac{53}{70}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-k_{2}})=\log(\frac{53}{70})$$

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

$$-k_{2}\log(e)=\log(\frac{53}{70})$$

Divide both sides by \log(e).

$$-k_{2}=\frac{\log(\frac{53}{70})}{\log(e)}$$

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

$$-k_{2}=\log_{e}\left(\frac{53}{70}\right)$$

Divide both sides by -1.

$$k_{2}=\frac{\ln(\frac{53}{70})}{-1}$$

Solve for k_2 (complex solution)

$$k_{2}=-i\times 2\pi n_{1}+\ln(\frac{70}{53})$$

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