Chapter 3.4, Problem 107E

a.

To determine

To find:

An interval containing the solution of the equation by completing the table.

Answer to Problem 107E

Our required interval would be $(0.8,0.9)$ .

Explanation of Solution

Given:

A logarithmic equation and a table: $e^{3x}=12$ .

x	0.6	0.7	0.8	0.9	1.0
$e^{3x}$

Calculation:

To complete the table, we will substitute the values of x in our given expression as shown below:

  $\begin{array}{l}{e}^{3x}\\ \Rightarrow e^{3(0.6)}\\ \Rightarrow e^{1.8}\\ \Rightarrow 6.04964746\approx 6.050\end{array}$

  $\begin{array}{l}{e}^{3x}\\ \Rightarrow e^{3(0.7)}\\ \Rightarrow e^{2.1}\\ \Rightarrow 8.16616991\approx 8.166\end{array}$

  $\begin{array}{l}{e}^{3x}\\ \Rightarrow e^{3(0.8)}\\ \Rightarrow e^{2.4}\\ \Rightarrow 11.02317638\approx 11.023\end{array}$

  $\begin{array}{l}{e}^{3x}\\ \Rightarrow e^{3(0.9)}\\ \Rightarrow e^{2.7}\\ \Rightarrow 14.87973172\approx 14.880\end{array}$

  $\begin{array}{l}{e}^{3x}\\ \Rightarrow e^{3(1.0)}\\ \Rightarrow e^{3.0}\\ \Rightarrow 20.0855369\approx 20.086\end{array}$

The complete table would be:

x	0.6	0.7	0.8	0.9	1.0
$e^{3x}$	6.050	8.166	11.023	14.880	20.086

Since 12 lies between 11.023and 14.880, therefore our required interval would be $(0.8,0.9)$ .

b.

To determine

To graph:

The given function using a graphing utility and approximate the solution.

Answer to Problem 107E

The solution of our given equation would be $x=0.828$ .

Explanation of Solution

Given:

A logarithmic equation: $e^{3x}=12$ .

Calculation:

Upon graphing our given function, we will get our required graph as shown below:

  

We can see that both lines intersect at $x=0.828$ , therefore, $x=0.828$ is a solution for our given function.

c.

To determine

To solve:

The given equation algebraically.

Answer to Problem 107E

The solution of our given equation would be $x\approx 0.828$ .

Explanation of Solution

Given:

A logarithmic equation: $e^{3x}=12$ .

Calculation:

We will use logarithmic properties to solve our given equation.

  $\begin{array}{l}{e}^{3x}=12\left(\text{Given equation}\right)\\ \ln \left(e^{3x}\right)=\ln (12)\left(\text{Taking natural log on both sides}\right)\\ 3x\cdot \ln (e)=\ln (12)\left(\text{Using property ln}\left(x^m\right)=m\ln (x)\right)\\ 3x=\ln (12)\left(\text{Using property ln(}e\text{)}=\text{1}\right)\\ \frac{3x}{3}=\frac{2.484906649}{3}\left(\text{Dividing both sides by 3}\right)\\ x=0.828302216\\ x\approx 0.828\left(\text{Rounding to three decimal places}\right)\end{array}$

Therefore, the solution of our given equation would be $x\approx 0.828$ .