Chapter 3.4, Problem 118E

a.

To determine

To find:

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

Answer to Problem 118E

Our required interval would be $(160,165)$ .

Explanation of Solution

Given:

A logarithmic equation and a table: $5{\log }_{10}\left(x-2\right)=11$ .

x	150	155	160	165	170
$5{\log }_{10}\left(x-2\right)$

Calculation:

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

  $\begin{array}{l}5{\log }_{10}\left(x-2\right)\\ \Rightarrow 5{\log }_{10}\left(150-2\right)\\ \Rightarrow 5{\log }_{10}\left(148\right)\\ \Rightarrow 10.8513085\approx 10.851\end{array}$

  $\begin{array}{l}5{\log }_{10}\left(x-2\right)\\ \Rightarrow 5{\log }_{10}\left(155-2\right)\\ \Rightarrow 5{\log }_{10}\left(153\right)\\ \Rightarrow 10.9234571\approx 10.923\end{array}$

  $\begin{array}{l}5{\log }_{10}\left(x-2\right)\\ \Rightarrow 5{\log }_{10}\left(160-2\right)\\ \Rightarrow 5{\log }_{10}\left(158\right)\\ \Rightarrow 10.9932854\approx 10.993\end{array}$

  $\begin{array}{l}5{\log }_{10}\left(x-2\right)\\ \Rightarrow 5{\log }_{10}\left(165-2\right)\\ \Rightarrow 5{\log }_{10}\left(163\right)\\ \Rightarrow 11.060938\approx 11.061\end{array}$

  $\begin{array}{l}5{\log }_{10}\left(x-2\right)\\ \Rightarrow 5{\log }_{10}\left(170-2\right)\\ \Rightarrow 5{\log }_{10}\left(168\right)\\ \Rightarrow 11.1265464\approx 11.127\end{array}$

The complete table would be:

x	150	155	160	165	170
$5{\log }_{10}\left(x-2\right)$	10.851	10.923	10.993	11.061	11.127

Since 11 lies between 10.933and11.127, therefore our required interval would be $(160,165)$ .

b.

To determine

To graph:

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

Answer to Problem 118E

The solution of our given equation would be $x=160.489$ .

Explanation of Solution

Given:

A logarithmic equation: $5{\log }_{10}\left(x-2\right)=11$ .

Calculation:

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

  

We can see that both lines intersect at $x=160.489$ , therefore, $x=160.489$ is a solution for our given function.

c.

To determine

To solve:

The given equation algebraically.

Answer to Problem 118E

The solution of our given equation would be $x\approx 160.489$ .

Explanation of Solution

Given:

A logarithmic equation: $5{\log }_{10}\left(x-2\right)=11$ .

Calculation:

We will use logarithmic properties to solve our given equation.

  $\begin{array}{l}5{\log }_{10}\left(x-2\right)=11\left(\text{Given equation}\right)\\ \frac{5{\log }_{10}\left(x-2\right)}{5}=\frac{11}{5}\left(\text{Dividing both sides by 5}\right)\\ {\log }_{10}\left(x-2\right)=\frac{11}{5}\\ x-2={10}^{\frac{11}{5}}\left(\text{Converting log to exponential form}\right)\\ x-2=158.489319246\\ x-2+2=158.489319246+2\\ x=160.489319246\\ x\approx 160.489\left(\text{Rounding to three decimal places}\right)\end{array}$

Therefore, the solution of our given equation would be $x\approx 160.489$ .