3.5 Properties and applications of logarithmic functions
Activity 3.5.2
Solve each of the following equations exactly and then find an estimate that is accurate to 5 decimal places.
a. [latex]3^t = 5[/latex]
b. [latex]4 \cdot 2^t - 2 = 3[/latex]
c. [latex]3.7 \cdot (0.9)^{0.3t} + 1.5 = 2.1[/latex]
d. [latex]72 - 30(0.7)^{0.05t} = 60[/latex]
e. [latex]\ln(t) = -2[/latex]
f. [latex]3 + 2\log_{10}(t) = 3.5[/latex]
Show Solution
a. \begin{align*}
3^t &= 5 \\
\ln(3^t) & = \ln(5) \\
t\ln(3) & = \ln(5) \\
t & = \frac{\ln(5)}{\ln(3)} \\
& \approx 1.46497
\end{align*}
b. We first start by isolating[latex]2^t[/latex]
\begin{align*}
4 \cdot 2^t -2 & = 3 \\
4 \cdot 2^t & = 5 \\
2^t & = \frac{5}{4} \\
\ln(2^t) & = \ln(\frac{5}{4}) \\
t \ln(2) & = \ln(\frac{5}{4}) \\
t & = \frac{\ln(\frac{5}{4})}{\ln(2)} \\
& \approx 0.31293
\end{align*}
c. \begin{align*}
3.7(0.9)^{0.3t} + 1.5 & = 2.1 \\
3.7(0.9)^{0.3t} & = 0.6 \\
(0.9)^{0.3t} & = \frac{0.6}{3.7} \\
\ln(0.9)^{0.3t} & = \ln(\frac{0.6}{3.7}) \\
0.3t\ln(0.9) & = \ln(\frac{0.6}{3.7}) \\
t & = \frac{\ln(\frac{0.6}{3.7})}{0.3\ln(0.9}) \\
& \approx 57.55345 \\
\end{align*}
d. \begin{align*}
-30(0.7)^{0.05t} & = -12 \\
(0.7)^{0.05t} & = 12/30 \\
(0.7)^{0.05t} & = 2/5 \\
\ln(0.7^{0.05t}) & = \ln(2/5) \\
0.05t\ln(0.7) & = \ln(2/5) \\
t & = \frac{\ln(\frac{2}{5})}{0.05\ln(0.7)} \\
& \approx 51.37960 \\
\end{align*}
e. Start by rewriting the logarithmic equation in its equivalent exponential form [latex]t = e^{-2} \approx 0.13534.[/latex]
f. \begin{align*}
3+2log_{10}(t) & = 3.5 \\
2log_{10}(t) & = 0.5 \\
log_{10}(t) & = 0.25 \\
\end{align*}
Rewrite the logarithmic equation in its equivalent exponential form
\begin{align*}
t & = 10^{0.25} \\
& = \approx 1.77830 \\
\end{align*}
Activity 3.5.4
Solve each of the following equations for the exact value of k.
a. [latex]41 = 50e^{-k \cdot 7}[/latex]
b. [latex]65 = 34 + 47e^{-k \dot 45}[/latex]
c. [latex]7e^{2k-1} + 4 = 32[/latex]
d. [latex]\frac{5}{1+2e^{-10k}} = 4[/latex]
Show Solution
a. \begin{align*}
\frac{41}{50} & = e^{-k \cdot 7} \\
\ln(\frac{41}{50}) & = \ln(e^{-k \cdot 7} \\
\ln(\frac{41}{50}) & = -k \cdot 7 \ln(e) \\
\ln(\frac{41}{50}) & = -k \cdot 7 \\
k & = \frac{\ln(\frac{41}{50})}{-7} \\
\end{align*}
b. \begin{align*}
65 & = 34 + 47e^{-k \cdot 45} \\
31 & = 47e^{-k \cdot 45} \\
\frac{31}{47} & = e^{- k \cdot 45} \\
\ln(\frac{31}{47}) & = \ln(e^{-k \cdot 45}) \\
\ln(\frac{31}{47}) & = -k \cdot 45 \ln(e) \\
\ln(\frac{31}{47}) & = -k \cdot \\
k & = -\ln(\frac{31}{47})
\end{align*}
c. \begin{align*}
7e^{2k-1} +4 & = 32 \\
7e^{2k-1} & = 28 \\
e^{2k-1} & = 4 \\
\lne^{2k-1} & = \ln(4) \\
2k-1\ln(e) & = \ln(4) \\
2k-1 & = \ln(4) \\
2k & = \ln(4) + 1 \\
k & = \frac{ln(4)+1}{2}
\end{align*}
d. \begin{align*}
\frac{5}{1+2e^{-10k}} & = 4 \\
\end{align*}
Multiply the denominator on both sides
\begin{align*}
5 & = 4(1+2e^{-10k}) \\
5 & = 4+8e^{-10k} \\
1 & = 8e^{-10k} \\
\frac{1}{8} & = e^{-10k} \\
\ln(\frac{1}{8}) & = \ln(e^{-10k}) \\
\ln(\frac{1}{8}) & = -10k\ln(e) \\
\ln(\frac{1}{8}) & = -10k \\
k & = \frac{\ln(\frac{1}{8})}{10}
\end{align*}
3.5 Exercise 6
For a population that is growing exponentially according to a model of the form[latex]P(t) = Ae^{kt}[/latex], the doubling time is the amount of time that it takes the population to double. For each population described below, assume the function is growing exponentially according to a model [latex]P(t) = Ae^{kt}[/latex], where [latex]t[/latex]is measured in years.
a. Suppose that a certain population initially has 100 members and doubles after 3 years. What are the values of [latex]A[/latex] and[latex]k[/latex] in the model?
b. A different population is observed to satisfy [latex]P(4) = 250[/latex] and [latex]P(11) = 500[/latex]. What is the population’s doubling time? When will 2000 members of the population be present?
c. Another population is observed to have doubling time [latex]t = 21[/latex]. What is the value of [latex]k[/latex]in the model?
d. How is [latex]k[/latex] is related to a population’s doubling time, regardless of how long the doubling time is?
Show Solution
a. [latex]A[/latex] is 100 because that is the initial amount in the population. We know that the population doubles after 3 years which means the population would be 200 so we will plug in 3,200 into [latex]P(t)[/latex] to solve for [latex]k[/latex].
\begin{align*}
200 & = 100e^{3k} \\
2 & = e^{3k} \\
\ln(2) & = \ln(e^{3k}) \\
\ln(2) & = 3k\ln(e) \\
k & = \frac{ln(2)}{3} \\
\end{align*}
So [latex]A = 100[/latex] and[latex]k = \frac{\ln(2)}{3}.[/latex]
b. We take note that if the population is 250 in 4 years and then 500 in 11 years that is exactly double the amount. So, it took 11-4 = 7 years for the population to double.
Now we want to determine when will 2000 members be present. Well, we know that every 7 years the population doubles so in 11 years it will be 500 so in 18 years it will be 1000 and then in 25 years the population will be 2000.
c. We are given that the doubling time is [latex]t = 21[/latex]for a given population. So, if we start out with [latex]A[/latex] amount of people in 21 years the population will be [latex]2A[/latex]. We will plug this into [latex]P(t)[/latex] to get
\begin{align*}
2A & = Ae^{21k} \\
2 & = e^{21k} \\
\ln(2) & = \ln(e^{21k}) \\
\ln(2) & = 21t\ln(e) \\
\ln(2) & = 21k \\
k & = \frac{\ln(2)}{21}
\end{align*}
d. If [latex]t[/latex] is the doubling time of a population, then k = \frac{\ln(2)}{t}.