3.5 Properties and applications of logarithmic functions (continued)

Putting logarithms to work

Summary

A population of bacteria cells is growing at a rate proportionate to the number of cells present at a given time $t$ (in hours). Suppose that the number of cells, $P$, in the population is measured in millions of cells and we know that $P(0) = 2.475$ and $P(10) = 4.298$. Find a model of the form $P(t) = Ae^{kt}$ that fits this data and use it to determine the value of $k$ and how long it will take for the population to reach $1$ billion cells.

$41 = 50e^{-k \cdot 7}$

$65 = 34 + 47e^{-k \cdot 45}$

$7e^{2k-1} + 4 = 32$

$\frac{5}{1+2e^{-10k}} = 4$

For a population that is growing exponentially according to a model of the form $P(t) = Ae^{kt}$, 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 $P(t) = Ae^{kt}$, where $t$ is measured in years.

• Suppose that a certain population initially has $100$ members and doubles after $3$ years. What are the values of $A$ and $k$ in the model? - A different population is observed to satisfy $P(4) = 250$ and $P(11) = 500$. What is the population's doubling time? When will $2000$ members of the population be present? - Another population is observed to have doubling time $t = 21$. What is the value of $k$ in the model? - How is $k$ related to a population's doubling time, regardless of how long the doubling time is?

A new car is purchased for ＄$28000$. Exactly $1$ year later, the value of the car is ＄$23200$. Assume that the car's value in dollars, $V$, $t$ years after purchase decays exponentially according to a model of form $V(t) = Ae^{-kt}$.

• Determine the exact values of $A$ and $k$ in the model. - How many years will it take until the car's value is ＄$10000$? - Suppose that rather than having the car's value decay all the way to ＄$0$, the lowest dollar amount its value ever approaches is ＄$500$. Explain why a model of the form $V(t) = Ae^{-kt} + c$ is more appropriate. - Under the original assumptions ($V(0) = 28000$ and $V(1) = 23200$) along with the condition in (c) that the car's value will approach ＄$500$ in the long-term, determine the exact values of $A$, $k$, and $c$ in the model $V(t) = Ae^{-kt} + c$. Are the values of $A$ and $k$ the same or different from the model explored in (a)? Why?

In Exercise 3.4.2, we explored graphically how the function $y = \log_b(x)$ can be thought of as a vertical stretch of the nautral logarithm, $y = \ln(x)$. In this exercise, we determine the exact value of the vertical stretch that is needed.

Recall that $\log_b(x)$ is the power to which we raise $b$ to get $x$.

• Write the equation $y = \log_b(x)$ as an equivalent equation involving exponents with no logarithms present. - Take the equation you found in (a) and take the natural logarithm of each side. - Use rules and properties of logarithms appropriately to solve the equation from (b) for $y$. Your result here should express $y$ in terms of $\ln(x)$ and $\ln(b)$. - Recall that $y = \log_b(x)$. Explain why the following equation (often called the Golden Rule for Logarithms) is true:

. - What is the value of $k$ that allows us to express the function $y = \log_b(x)$ as a vertical stretch of the function $y = \ln(x)$?