Chapter 3.5, Problem 13E

Solution

To find: The solution of the given equation and then check it by substituting it into the original equation.

The exact solution of the given equation $50e^{0.035x}=200$ is $x\approx 39.6084$ .

Given information:

The given equation is $50e^{0.035x}=200$ .

Concept Used:

Step involved in solving the logarithm equation:

Isolate the exponential expression.

Take the natural logarithm on both sides.

Apply power rule of logarithm and then solve.

Calculation:

Divide by $50$ on both sides.

  $\begin{array}{c}\frac{50}{50}{e}^{0.035x}=\frac{200}{50}\\ e^{0.035x}=4\end{array}$

Take natural logarithms on both sides.

  $\begin{array}{c}\ln e^{0.035x}=\ln 4\\ 0.035x\ln e=\ln 4\\ 0.035x=2\ln 2\left[\ln e=1\right]\\ x=\frac{2\ln 2}{0.035}\end{array}$

Use the TI-83 calculator for calculating the value of $\frac{2\left(\ln 2\right)}{0.035}$ .

  $\frac{2\left(\ln 2\right)}{0.035}\approx 39.6084$

So,

  $x\approx 39.6084$

Now, check the result while substituting the value of $x$ in the given equation.

  $50e^{0.035\left(39.6084\right)}\stackrel{?}{=}200$

Use the TI-83 calculator for calculating the value of $50\left(e^{0.035\left(39.6084\right)}\right)$ .

  $\begin{array}{c}50\left(e^{0.035\left(39.6084\right)}\right)\approx 200\\ 200=200\left(\text{True}\right)\end{array}$

Hence, the value of $x$ is $\approx 39.6084$ .