1.
Find the domain and vertical asymptote of \(\displaystyle{ f(x)={\log\mathopen{}\left(x-8\right)} }\text{.}\) Enter the domain in interval notation.
Domain:
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Worksheet Logarithmic Functions
2.
Find the domain and range of \(\displaystyle{ f(x)={\ln\mathopen{}\left(2-x\right)} }\text{.}\) Enter your solutions in interval notation.
Domain:
Range:
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3.
Find the domain and vertical asymptote of \(\displaystyle{ f(x)={-\left(\log\mathopen{}\left(x-4\right)+4\right)} }\text{.}\) Enter your solutions in interval notation.
Domain:
4.
The decibel rating \(D\) is related to the sound intensity \(I\) by the formula \(\displaystyle D = 10 \log_{10} \left(
\frac{ I }{ 10^{-16} } \right)\) for the noise level in decibels.
(a) Let \(D\) and \(d\) represent the decibel ratings of sounds of intensity \(I\) and \(i\text{,}\) respectively. Using properties of logarithms, find a simplified formula for the difference between the two ratings, \(D - d\text{,}\) in terms of the two intensities \(I\) and \(i\text{.}\)
\(D - d =\)
(Enter log10 or logten for the base 10 logarithm.)
(b) If a sound’s intensity triples, how many decibels louder does the sound become?
decibels
Solution.
SOLUTION\(D_1=10\log\left(\frac{I_1}{I_0}\right)\)\(D_2=10\log\left(\frac{I_2}{I_0}\right)\text{.}\)
\begin{equation*}
\begin{aligned}
D_2-D_1 \amp =
10\log\left(\frac{I_2}{I_0}\right)-10\log\left(\frac{I_1}{I_0}\right)\\
\amp = 10\left(\log\left(\frac{I_2}{I_0}\right)-\log\left(\frac{I_1}{I_0}\right)\right)\quad\hbox{(by factoring)}\\
\amp = 10\log\left(\frac{I_2/I_0}{I_1/I_0}\right)\quad\hbox{
(by using a log property)}
\end{aligned}
\end{equation*}
\begin{equation*}
D_2-D_1=10\log\left(\frac{I_2}{I_1}\right).
\end{equation*}
\(I_1\)\(I_2\text{.}\)\(I_2=3 I_1\text{.}\)\(D_1\)\(D_2\)
\begin{equation*}
\begin{aligned}
\hbox{Increase in decibels } \amp = D_2-D_1\\
\amp =10\log\left(\frac{I_2}{I_1}\right)\quad\hbox{ (by using
formula from part (a))}\\
\amp =10\log\left(\frac{3 I_1}{I_1}\right)\\
\amp =10\log 3.
\end{aligned}
\end{equation*}
\(10\log 3\approx 4.77121\)
5.
A light, flashing regularly, consists of cycles, each cycle having a dark phase and a light phase. The frequency of this light is measured in cycles per second. As the frequency is increased, the eye initially perceives a series of flashes of light, then a coarse flicker, a fine flicker, and ultimately a steady light. The frequency at which the flickering disappears is called the fusion frequency. The table below shows the results of an experiment in which the fusion frequency \(F\) was measured as a function of the light intensity \(I\text{.}\) It is modeled by \(F=a \ \ln{I} +b\text{.}\)
| \(I\) | 0.8 | 1.9 | 4.4 | 10 | 21.4 | 48.4 | 92.5 | 218.7 | 437.3 | 980 |
| \(F\) | 8 | 12.1 | 15.2 | 18.5 | 21.7 | 25.3 | 28.3 | 31.9 | 35.2 | 38.2 |
Find \(\ln{I}\) for each value of \(I\) in the table above, and then use linear regression on a calculator to estimate \(a\) and \(b\) in the linear fit \(F= a \ \ln{I} +b\text{.}\) In the blanks below, enter the corresponding values for \(a\) and \(b\text{:}\)
\(a =\)
\(b =\)
(round values to 2 decimal places)
6.
Students in a fifth-grade class were given an exam. During the next 2 years, the same students were retested several times. The average score was given by the model
\begin{equation*}
f(t) = 89 - 7 \log_{10}(t+1), \qquad 0\leq t\leq 24
\end{equation*}
where \(t\) is the time in months.
(a) What is the average score on the original exam?
(b) What was the average score after 6 months?
(c) What was the average score after 18 months?
7.
Solve for \(\small{x}\) without using a calculating utility. Leave your answer in radical form, and write all fractions in lowest terms.
\(\ln\left(\large{\frac{6}{x}}\right) + \ln(8 x^3) = \ln 7\)
\(\small{x =}\)
Solution.
\(\small{\ln}\left(\large{\frac{6}{x}} \small{8 x^3}\right) \small{= \ln\left(48 x^2\right)}\text{.}\)\(\small{\ln\left(48 x^2\right) = \ln 7}\text{.}\)\(\small{48 x^2 = 7\Rightarrow x = \pm\sqrt{{\frac{7}{48}}} }\text{.}\)\(\small{x = \sqrt{{\frac{7}{48}}}}\text{.}\)
