Test 22 (Lessons 43–44): Logarithmic Functions and Their Applications Solutions

1. Paulo saved $5500 from a summer job. He finds a high-yield savings account with 4.75% interest that compounds continuously. If Paulo does not add any more money to the account, how long will it take to save $6000? $y=P{e}^{rt}$

$\begin{array}{rcl}& & \begin{array}{ccc}y=6000& & y=P{e}^{rt}\\ P=5500& & \frac{6000}{5500}=\frac{5500}{5500}\left(e^{0.0475t}\right)\\ r=0.0475& & \frac{12}{11}=e^{0.0475t}\\ & & \ln \left(\frac{12}{11}\right)=\ln e^{0.0475t}\\ & & \frac{\ln \left({\displaystyle \frac{12}{11}}\right)}{0.0475}=\frac{0.0475t}{0.0475}\\ & & t=1.83\end{array}\\ & & \end{array}$

It will take Paulo 1.83 years to have $6000 in the savings account.

2. Write the inverse of h(x).

   $h\left(x\right)=8^{x–5}$

$$\begin{gathered} y=8^{x–5} \\ x=8{ }^{y–5} \\ {\log }_{8}x={\log }_{8}8{ }^{y–5} \\ {\log }_{8}x=y–5 \\ y={\log }_{8}x+5 \end{gathered}$$

$h^{–1}\left(x\right)={\log }_8\left(x\right)+5$

3. Name the domain and range of the inverse of q(x).
   $q\left(x\right)=e^x+3$

$$\begin{gathered} y=e^x+3 \\ x=e^y+3 \\ x–3=e^y \\ \ln \left(x–3\right)=\ln e^y \\ y=\ln \left(x–3\right) \end{gathered}$$

or  

$$\begin{gathered} {\mathrm{domain}}_q\left\{x|x\in \mathrm{ℝ}\right\} \\ {\mathrm{range}}_q\left\{y|y\in \mathrm{ℝ}, \mathrm{y}>3\right\} \end{gathered}$$

$$\begin{gathered} q^{–1}\left(x\right) \\ \mathrm{domain}: \left\{x|x\in \mathrm{ℝ}, x>3\right\} \\ \mathrm{range}: \left\{y|y\in \mathrm{ℝ}\right\} \end{gathered}$$

4. Describe the transformation of the functions f(x) to g(x).

   $f\left(x\right)=\ln \left(x\right)+5$      $g\left(x\right)=\ln \left(x–1\right)$

$$\begin{gathered} f\left(x\right): a=1, h=0, k=5 \\ g\left(x\right): a=1, h=1, k=0 \end{gathered}$$

From f(x) to g(x), the graph shifts right one unit and down five units.

6. Name the end behavior of$y={\log }_2 x.$

$\mathrm{As} x\to +\infty , f\left(x\right)\to +\infty , \mathrm{and} \mathrm{as} x\to 0, f\left(x\right)\to –\infty$

7. Write the equation.

$C_0=182, h=5$

$C\left(t\right)=182{\left(0.5\right)}^{\frac{t}{5}}$

8. Sketch a graph. Include labels.

9. Determine the approximate amount of caffeine in Douglas’ system at 8:30 a.m. the next day.  

$h=24$

$C\left(24\right)=182{\left(0.5\right)}^{\left(\frac{24}{5}\right)}=6.5332$

Douglas will have about 6.5 mg of caffeine at 8:30 a.m. the next day.

10. After how many hours will Douglas have one-fourth of the amount of caffeine that he started with that day?

$\begin{array}{rcl}182\left(0.25\right)& =& 182{\left(0.5\right)}^{\frac{t}{5}}\\ 0.25& =& {\left(0.5\right)}^{\frac{t}{5}}\\ \log 0.25& =& \log {\left(0.5\right)}^{\frac{t}{5}}\\ \log 0.25& =& \frac{t}{5}\log \left(0.5\right)\\ \frac{\log \left(0.25\right)}{\log \left(0.5\right)}& =& \frac{t}{5}\\ t& =& \frac{5 \log \left(0.25\right)}{\log \left(0.5\right)}\end{array}$

After 10 hours, one-quarter of the initial amount of caffeine will be in Douglas’ system.