Practice 2 Solutions

1. In 1906, San Francisco was hit by a major earthquake with a magnitude of 7.8 on the Richter scale. What was the intensity of the earthquake?

$\begin{array}{rcl}7.8& =& \log \left(\frac{I}{1}\right)\\ 7.8& =& \log I – \log 1\\ 7.8 & =& \log I–0\\ 7.8& =& \log I\\ {10}^{7.8}& =& {10}^{\log I}\\ I& =& 63,095,734.45\end{array}$

The intensity of the earthquake was 63,095,734.45.

2. In 1989, another earthquake hit San Francisco, this time with a magnitude of 6.9. How many times more intense was the 1906 earthquake than the 1989 earthquake?

The difference in magnitudes: $7.8–6.9=0.9$

$$\begin{gathered} 0.9=\log \left(\frac{I}{1}\right) \\ 0.9=\log I \\ {10}^{0.9}=I \\ I=7.94 \end{gathered}$$

The 1906 earthquake was 7.94 times more intense than the 1989 earthquake.

1. Edmond invested $2500 in a compound interest account at his local bank. The bank offered an interest rate of 5% compounded monthly. When will Edmond’s account reach $7500? Round to the nearest whole year.

$\begin{array}{rcl}7500& =& 2500{\left(1+\frac{0.05}{12}\right)}^{12t}\\ 3& =& {\left(1+\frac{0.05}{12}\right)}^{12t}\\ \ln 3 & =& 12t \ln \left(1+\frac{0.05}{12}\right)\\ t& =& \frac{\ln 3}{12 \ln \left(1+\frac{0.05}{12}\right)}\\ t& =& 22.0179\end{array}$

Edmond will have $7500 in his account in approximately 22 years.

4. How much more would Edmond have in his account in the same amount of time if the bank offered 8% interest? Round to the nearest cent.

$$\begin{gathered} A=2500{\left(1+\frac{0.08}{12}\right)}^{12\left(22\right)} \\ A=14 446.4687 \end{gathered}$$

If the bank offers 8% interest, Edmond will have $6,946.47 more in his account over the same amount of time.

5. A solution has a pH of 4. What is the concentration of hydrogen ions in the solution?

$\begin{array}{rcl}4& =& –{\log }_{10}\left[H^{+}\right]\\ –4& =& {\log }_{10}\left[H^{+}\right]\\ {10}^{–4}& =& \left[H^{+}\right]\end{array}$

The concentration of hydrogen ions in the solution is 0.0001 moles per liter.

6. A substance had a concentration of hydrogen ions of $2.67\times {10}^{–8}$. Determine the pH to the nearest hundredth.

$$\begin{gathered} \mathrm{pH}=–{\log }_{10}\left[2.67\times {10}^{–8}\right] \\ \mathrm{pH}=7.5734 \end{gathered}$$

The substance has a pH of 7.57.

7. A biologist is studying a new strain of bacteria that exhibits continuous exponential growth. They begin with a small culture of 500 bacteria. After just 4 hours, the number of bacteria has skyrocketed to 27,000. What is the continuous hourly growth rate of this bacteria strain?

$\begin{array}{rcl}y& =& P{e}^{rt}\\ 27000& =& 500 e^{4r}\\ 54& =& e^{4r}\\ \ln 54& =& 4r\\ r& =& \frac{\ln 54}{4}=0.9972\end{array}$

The continuous growth rate of the bacteria is 0.997 or 99.7%.

8. Another bacteria grows continuously at a rate of 52.3%. The initial culture started with 1200 bacteria. What is the bacteria population after 8 hours?

$$\begin{gathered} y=1200 e^{0.523\left(8\right)} \\ y=78 753.4083 \end{gathered}$$

The population of the bacteria after 8 hours is 78,753.

9. A sample of a newly discovered radioactive substance has a mass of 300 grams. Scientists observe that after 5 years, only 50 grams of the new substance remain. What is the half-life of this mysterious substance?

$\begin{array}{rcl}50& =& 300{\left(\frac{1}{2}\right)}^{\frac{5}{h}}\\ \frac{1}{6}& =& {\left(\frac{1}{2}\right)}^{\frac{5}{h}}\\ \ln \left(\frac{1}{6}\right)& =& \frac{5}{h}\ln \left(\frac{1}{2}\right)\\ \frac{\ln \left(\frac{1}{6}\right)}{\ln \left(\frac{1}{2}\right)}& =& \frac{5}{h}\\ h& =& \frac{5 \ln \left(\frac{1}{2}\right)}{\ln \left(\frac{1}{6}\right)}\approx 1.934\end{array}$

The half-life of the new substance is 1.934 years.

10. A specific type of medication has a half-life of 4 hours in the human body. If a patient is given an initial dose of 400 milligrams, how long will it take for the amount of medication in their system to decrease to 25 milligrams?

$\begin{array}{rcl}25& =& 400{\left(\frac{1}{2}\right)}^{\frac{t}{4}}\\ \frac{1}{16}& =& {\left(\frac{1}{2}\right)}^{\frac{t}{4}}\\ \ln \left(\frac{1}{16}\right)& =& \frac{t}{4} \ln \left(\frac{1}{2}\right)\\ \frac{4 \ln \left({\displaystyle \frac{1}{16}}\right)}{\ln \left({\displaystyle \frac{1}{2}}\right)}& =& t\\ t& =& 16\end{array}$

It will take 16 hours for the medication to decrease to 25 mg.

11. Write the equation in the form $y=a{b}^x$.

$\begin{array}{rcl}& & \left(0, 4000\right), \left(10, 7000\right)\\ y& =& a{b}^x\\ 4000& =& a{b}^0\\ a& =& 4000\\ 7000& =& 4000b^{10}\\ \frac{7}{4}& =& b^{10}\\ \sqrt[10]{\frac{7}{4}}& =& b\\ b& =& 1.058\end{array}$

$y=4000{\left(1.058\right)}^x$

12. If this trend continues, what will the population be in the year 2030?

$$\begin{gathered} y=4000{\left(1.058\right)}^{30} \\ y=21708.5 \end{gathered}$$

In the year 2030, there will be approximately 21,709 people in the town.