Practice 1 Solutions

1. A normal conversation has a loudness of about 60 dB. What is the intensity level?

$\begin{array}{rcl}60& =& 10·\log \left(\frac{\mathrm{I}}{{10}^{–12}}\right)\\ 6& =& \log \left(\frac{1}{{10}^{–12}}\right)\\ 6& =& \log \mathrm{I}–\log {10}^{–12}\\ 6& =& \log \mathrm{I} +12\\ –6& =& \log \mathrm{I}\\ \mathrm{I}& =& {10}^{–6}\end{array}$

$\mathrm{I}={10}^{–6 }\raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$

2. A rock concert has a loudness of about 120 dB. How many times more intense is sound at a rock concert than in a normal conversation?

$\begin{array}{rcl}120& =& 10·\log \left(\frac{\mathrm{I}}{{10}^{–12}}\right)\\ 12& =& \log \left(\frac{\mathrm{I}}{{10}^{–12}}\right)\\ 12& =& \log \mathrm{I}–\log {10}^{–12}\\ 12& =& \log \mathrm{I} +12\\ 0& =& \log \mathrm{I}\\ \mathrm{I}& =& 1\end{array}$

$\left(\frac{{\mathrm{I}}_{concert}}{{\mathrm{I}}_{conversation}}\right)=\left(\frac{1}{{10}^{–6}}\right)={10}^6$

Alternate solve:

$$\begin{gathered} 120 dB – 60 dB = 60 dB \\ 6 \mathrm{sets} \mathrm{of} 10 dB \mathrm{increases}. \\ \mathrm{Each} 10 dB \mathrm{increase} \mathrm{is} \mathrm{a} {10}^{1 }\mathrm{increase} \mathrm{of} \mathrm{intensity} \\ {10}^6 \end{gathered}$$

The sound at a rock concert is 1 million times more intense than a normal conversation.

3. The sound of a vacuum cleaner has a registered intensity of ${10}^{–4.5} \raisebox{1ex}{$W$}\!\left/ \!\raisebox{-1ex}{$m^2$}\right.$. What is the decibel level?

$$\begin{gathered} \mathrm{L}=10·\log \left(\frac{{10}^{–4.5}}{{10}^{–12}}\right) \\ \mathrm{L}=75 \end{gathered}$$

The vacuum cleaner has a level of 75 dB.

4. The intensity of a certain earthquake’s seismic waves diminishes continuously at a rate of 9% $(r=–0.09)$ per kilometer as they travel through the Earth’s crust. If the initial intensity of the waves is measured at 7.5 on the Richter scale, at what distance will the earthquake wave intensity reach 4.0 on the Richter scale? Round to the nearest hundredth.

$\begin{array}{rcl}y& =& 4.0, P=7.5, r=–0.09, t=\mathrm{distance} \mathrm{in} \mathrm{km}\\ 4.0& =& 7.5 e^{\left(–0.09\right)t}\\ \frac{4.0}{7.5}& =& e^{–0.09t}\\ \ln \left(\frac{4.0}{7.5}\right)& =& \ln e^{–0.09t}\\ \ln \left(\frac{4.0}{7.5}\right)& =& –0.09t\\ t& =& \frac{\ln 4.0–\ln 7.5}{–0.09}=6.9845\\ t& =& 6.98 km\end{array}$

The earthquake intensity is equal to 4.0 on the Richter scale at a distance of 6.98 kilometers.

5. Kristin invested $6000 in a savings account with a 6% annual interest rate, compounded continuously. How much money will she have in the account after 10 years?

$$\begin{gathered} y=6000 e^{0.06\left(10\right)} \\ y=10 932.7128 \end{gathered}$$

In 10 years, Kristin will have $10,932.71 in her account.

6. Kristin invested $6000 in a savings account that offers a 6% annual interest rate, compounded annually. How much money will she have in the account after 10 years?

$$\begin{gathered} A=6000{\left(1+\frac{0.06}{1}\right)}^{1\left(10\right)} \\ A=10 745.0861 \end{gathered}$$

When compounded yearly, in 10 years, Kristin will have $10,745.09 in her account.

7. Natalee invests $5000 in a savings account with a 6.5% interest rate that is compounded twice a year. How many years will it take for her to have $15,000 in her account?

$\begin{array}{rcl}15000& =& 5000{\left(1+\frac{0.065}{2}\right)}^{2t}\\ 3& =& {\left(1.0325\right)}^{2t}\\ \ln 3& =& 2t \ln \left(1.0325\right)\\ \frac{\ln 3}{\ln \left(1.0325\right)}& =& \frac{2t \ln \left(1.0325\right)}{\ln \left(1.0325\right)}\\ t& =& \frac{\ln 3}{2 \ln \left(1.035\right)}\\ t& =& 17.1749\end{array}$

In 17.17 years, Natalee will have $15,000 in her account.

8. A scientist discovers a new radioactive element, Element X. After observing a 100 gram sample for 10 days, they find that only 25 grams remain. What is the half-life of Element X?

$\begin{array}{rcl}25& =& 100{\left(\frac{1}{2}\right)}^{\frac{10}{h}}\\ 0.25& =& {\left(\frac{1}{2}\right)}^{\frac{10}{h}}\\ \ln 0.25& =& \frac{10}{h}\ln \left(\frac{1}{2}\right)\\ h \ln 0.25& =& 10 \ln \left(\frac{1}{2}\right)\\ h& =& \frac{10 \ln \left({\displaystyle \frac{1}{2}}\right)}{\ln 0.25}\\ h& =& 5 \end{array}$

The half-life of Element X is 5 days.

9. A biologist is studying a bacterial culture with a very short half-life of 30 minutes. If they start with a sample of 1 million bacteria, how many bacteria will be left after two hours?

$$\begin{gathered} A\left(2\right)=1,000,000{\left(\frac{1}{2}\right)}^{\frac{2}{0.5}} \\ A\left(2\right)=62 500 \end{gathered}$$

In two hours, only 62,500 bacteria will be left.

10. Determine the cooling constant, k. Round to the nearest hundredth.

$\begin{array}{rcl}250& =& 72+\left(350–72\right)e^{–k\left(15\right)}\\ 250& =& 72+278 e^{–15k}\\ 178& =& 278 e^{–15k}\\ \frac{178}{278}& =& e^{–15k}\\ \ln \left(\frac{178}{278}\right)& =& –15k\\ k& =& \frac{\ln \left(\frac{178}{278}\right)}{–15}=\frac{\left(\ln 178–\ln 278\right)}{–15}\\ k& =& 0.0297\end{array}$

$\begin{array}{rcl}k& =& 0.03\end{array}$

11. If the baker took the pie out of the oven at 6:00 p.m., when will it reach a temperature of $100°\mathrm{F}$? Round to the nearest whole minute.

$\begin{array}{rcl}100& =& 72+\left(350–72\right)e^{–0.03t}\\ 28& =& 278 e^{–0.03t}\\ \frac{28}{278}& =& e^{–0.03t}\\ \ln \left(\frac{28}{278}\right)& =& –0.03t\\ t& =& \frac{\ln \left(\frac{28}{278}\right)}{–0.03}=\frac{\left(\ln 28–\ln 278\right)}{–0.03}\\ t& =& 76.5138\\ t& =& 77 \mathrm{minutes}\\ & =& 1 \mathrm{hour} 17 \mathrm{minutes}\\ 6:00+1:17& =& 7:17\end{array}$

The pie will reach $100°\mathrm{F}$ at 7:17 pm.

12. Sketch the graph. Include labels.