Solve 3.5=7e^-5k

$$3.5 = 7 e ^ { - 5 k }$$

Solve for k

$$k=\frac{\ln(2)}{5}\approx 0.138629436$$

Steps Using Definition of Logarithm

Divide both sides by 7.

$$\frac{3.5}{7}=e^{-5k}$$

Expand \frac{3.5}{7} by multiplying both numerator and the denominator by 10.

$$\frac{35}{70}=e^{-5k}$$

Reduce the fraction \frac{35}{70} to lowest terms by extracting and canceling out 35.

$$\frac{1}{2}=e^{-5k}$$

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

$$e^{-5k}=\frac{1}{2}$$

Take the logarithm of both sides of the equation.

$$\log(e^{-5k})=\log(\frac{1}{2})$$

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

$$-5k\log(e)=\log(\frac{1}{2})$$

Divide both sides by \log(e).

$$-5k=\frac{\log(\frac{1}{2})}{\log(e)}$$

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

$$-5k=\log_{e}\left(\frac{1}{2}\right)$$

Divide both sides by -5.

$$k=-\frac{\ln(2)}{-5}$$

Solve for k (complex solution)

$$k=-\frac{i\times 2\pi n_{1}}{5}+\frac{\ln(2)}{5}$$

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