Chapter 6.4, Problem 50E

(a)

To determine

To determine when will the events have about same attendance

Answer to Problem 50E

Events will have about same attendance in year 2007

Explanation of Solution

Given:

The graph shows the annual attendance at two events:

  

Concept Used:

The solution is given by intersection point of graph

From the given graph both the events will have about same attendance at the point of intersection that is $\left(3,8000\right)$

Conclusion:

As year $\left(0\leftrightarrow 2004\right)$

Therefore,

X-intercept 3 represents year 2007

Hence,

Events will have about same attendance in year 2007

(b)

To determine

To verify answer in part (a)

Explanation of Solution

Given:

The graph shows the annual attendance at two events:

  

Concept Used:

Put y=8000 in both equations to verify result for x

Calculation:

For $y=4000{\left(1.25\right)}^x$

  $\begin{array}{l}y=4000{\left(1.25\right)}^x\\ \text{put }y=8000\\ 4000{\left(1.25\right)}^x=8000\\ {\left(1.25\right)}^x=\frac{8000}{4000}\\ {\left(1.25\right)}^x=2\\ \text{take log both sides}\\ \log {\left(1.25\right)}^x=\log \left(2\right)\\ x\log \left(1.25\right)=\log \left(2\right)\left(\text{using log}{\left(a\right)}^b=b\log a\right)\\ x=\frac{\log \left(2\right)}{\log \left(1.25\right)}\\ x\approx 3\end{array}$

For $y=12000{\left(0.87\right)}^x$

  $\begin{array}{l}y=12000{\left(0.87\right)}^x\\ \text{put }y=8000\\ 12000{\left(0.87\right)}^x=8000\\ {\left(0.87\right)}^x=\frac{8000}{12000}\\ {\left(0.87\right)}^x=\frac{2}{3}\\ \text{take log both sides}\\ \log {\left(0.87\right)}^x=\log \left(\frac{2}{3}\right)\\ x\log \left(0.87\right)=\log \left(\frac{2}{3}\right)\left(\text{using log}{\left(a\right)}^b=b\log a\right)\\ x=\log \left(\frac{2}{3}\right)÷\log \left(0.87\right)\\ x\approx 3\end{array}$

Conclusion:

As answer in both cases is about same

Therefore,

Verified the answer given by part (a)