Chapter 0.5, Problem 34E

To determine

To solve: An equation algebrically and verify solution graphically.

Answer to Problem 34E

The required algebraic and graphical solutions are $t=21.972$ .

Explanation of Solution

Given data: The given equation is $e^{0.05t}=3$ .

Formula used:

The log of a number $a^m$ to the base ‘ e’ is $\ln \left(a^m\right)=m\ln \left(a\right)$ .

Calculation:

The given equation is $e^{0.05t}=3$ .Taking log of both sides,

  $\left.\begin{array}{c}\ln \left(e^{0.05t}\right)=\ln \left(3\right)\\ 0.05t\cdot \ln e=\ln 3\end{array}\right\}$ … (1)

Since, $\ln \left(e\right)=1$ and $\ln \left(3\right)=1.0986$ , so substitute $1$ for $\ln \left(e\right)$ and $1.0986$ for $\ln \left(3\right)$ in result (1),

  $\begin{array}{c}0.05t\left(1\right)=1.0986\\ t=\frac{1.0986}{0.05}\\ t=21.972\end{array}$

Now, write given equation in functional form as,

  $y\left(t\right)=e^{0.05t}-3$

The solution $y\left(t\right)=0$ is the solution of the equation $e^{0.05t}-3$ or $e^{0.05t}=3$ .

The graph of the function $y\left(t\right)=e^{0.05t}-3$ is shown in Fig. 1.

The curve $y\left(t\right)=e^{0.05t}-3$ intersect t -axis at point $t=21.972$ , $y\left(t\right)=0$ .

  

Hence, $t=21.972$ is the solution of the equation $e^{0.05t}=3$ .

Thus, the algebraic and graphical both the solutions of the equation $e^{0.05t}=3$ are $t=21.972$ .