Solve 29.3=33.6e^r(39)

$$29.3 = 33.6 e ^ { r ( 39 ) }$$

Solve for r

$$r=\frac{\ln(\frac{293}{336})}{39}\approx -0.003511245$$

Steps Using Definition of Logarithm

Divide both sides by 33.6.

$$\frac{29.3}{33.6}=e^{r\times 39}$$

Expand \frac{29.3}{33.6} by multiplying both numerator and the denominator by 10.

$$\frac{293}{336}=e^{r\times 39}$$

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

$$e^{r\times 39}=\frac{293}{336}$$

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

$$e^{39r}=\frac{293}{336}$$

Take the logarithm of both sides of the equation.

$$\log(e^{39r})=\log(\frac{293}{336})$$

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

$$39r\log(e)=\log(\frac{293}{336})$$

Divide both sides by \log(e).

$$39r=\frac{\log(\frac{293}{336})}{\log(e)}$$

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

$$39r=\log_{e}\left(\frac{293}{336}\right)$$

Divide both sides by 39.

$$r=\frac{\ln(\frac{293}{336})}{39}$$

Solve for r (complex solution)

$$r=\frac{i\times 2\pi n_{1}}{39}+\frac{\ln(\frac{293}{336})}{39}$$

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