Question

Use algebra to solve the following equation. Round decimal to two places.20 = 95 • e^-0.03t

Answer 1

$20=95\cdot e^{-0.03t}$

Dividing by 95 at both sides of the equation:

$\begin{gathered} \frac{20}{95}=\frac{95\cdot e^{-0.03t}}{95} \\ \frac{4}{19}=e^{-0.03t} \end{gathered}$

Taking natural logarithm at both sides of the equation:

$\begin{gathered} \ln (\frac{4}{19})=\ln (e^{-0.03t}) \\ \ln (\frac{4}{19})=-0.03t\cdot\ln (e^{}) \\ \ln (\frac{4}{19})=-0.03t \end{gathered}$

Dividing by -0.03 at both sides of the equation:

$\begin{gathered} \frac{\ln (\frac{4}{19})}{-0.03}=\frac{-0.03t}{-0.03} \\ \frac{\ln(\frac{4}{19})}{-0.03}=t \\ 51.94\approx t \end{gathered}$