Drill 8.9.1.
Evaluate each of the following:
- \(\displaystyle \ln(e^2)\)
- \(\displaystyle \ln(e^{\sin(t)})\)
- \(\displaystyle e^{\ln(2)}\)
- \(\displaystyle e^{\ln(x+y)}\)
Section 8.9 Logarithms, Natural and Unnatural
DIGRESSION: Exponential and Logarithmic Notation.
Before reading further consider reviewing Inverse Function Notation. It will probably help.
As we’ve observed the natural logarithm function is, by definition, the inverse of the natural exponential in precisely the same way that \(\inverse\tan(x)\) is, by definition, the inverse of \(\tan(x)\) . Unfortunately the notation that is used for the natural exponential and the natural logarithm is not helpful.
It has become standard practice to notate a function using the first three letters of its name. That is why the trigonometric functions are denoted
\begin{align*}
\sin,\amp{}\amp{} \cos,\amp{}\amp{}
\tan,\amp{}\amp{} \cot,\amp{}\amp{} \sec,\amp{}\amp{}
\text{and} \amp{}\amp{}\csc.
\end{align*}
By that standard we should properly denote the natural exponential with “\(\exp(x)\)” and the natural logarithm with “\(\log(x)
\text{.}\)” But instead we use \(e^x\) and \(\ln(x) \text{.}\)
We have used \(\exp(x) \) for the natural exponential in the past and will continue to use it when we want to emphasize the functional nature of the natural exponential, but most of the time it is more helpful to think of it as the number “\(e\) raised to a power” and to denote it with \(e^x\) which, unfortunately, doesn’t look like the name of a function at all.
Even worse, when we talk among ourselves mathematicians tend to use the natural logarithm because for us it is the most important, but we often denote it as “\(\log(x)\text{.}\)” When we are talking to scientists and engineers we use whichever notation is appropriate to their discipline.
Sometimes there are reasons to use logarithms with bases other than \(10\) or \(e\text{.}\) For example, computer scientists frequently use logarithms with a base of \(2\) which they denote \(\text{lb}(x)\) (binary logarithm).
None of this affects what the natural logarithm is, how we use it, or how we think about it. Only how we write it. We only bring this up to warn you that because of this wide variation in usage logarithmic notation can be hard to get comfortable with. Be careful. The generic notation is \(\log_b(x)\) where \(b\) indicates the base. In this scheme the natural logarithm is \(\log_{e}(x)\text{,}\) the common logarithm is \(\log_{10}(x)\text{,}\) and the binary logarithm is \(\log_{2}(x)\text{.}\) If you find yourself getting confused it can help to revert to this generic notation.
It is, after all, just notation. Do not confuse the name of a concept with the concept itself. Conceptually the natural logarithm is the inverse of the natural exponential and you should think of it as such.
It is often helpful to think of \(\ln(x)\) as the function that peels the exponent off of \(e^x\) as in Logarithm Property #1 \(\ln(e^x)=x\text{.}\)
END OF DIGRESSION
Take the time to become very comfortable using all of the properties of logarithms, for exactly the same reason you took time to become very comfortable with the differentiation rules: Exponential and logarithmic functions are basic tools in science.
Problem 8.9.2.
Like Logarithm Property #3, Logarithm Property #5 is really just a convenience. Show that it is really an immediate consequence of Logarithm Property #4 and Property #6.
Hint.
Find a way to express \(\frac{a}{b}\) as a product.
The simplest use to which we can put a natural logarithm is to aid in solving an equation like:
\begin{equation*}
3^x=17.
\end{equation*}
The difficulty here is that the variable \(x\) is in the exponent so to isolate it we have to “undo” an exponential. This is exactly what logarithms do! By Logarithm Property #6 , \(\ln(3^x)=x\ln(3)\text{,}\) so we can solve our equation like this:
\begin{align*}
3^x\amp=17\\
\ln(3^x)\amp = \ln(17)\\
x\ln(3)\amp= \ln(17)\\\\
x\amp=\frac{\ln(17)}{\ln(3)}.
\end{align*}
At this point we can use technology to find that \(\ln(17) \approx 2.8332\) and \(\ln(3)\approx1.0986\) so that \(x=\frac{\ln(17)}{\ln(3)}\approx2.5789\text{.}\)
Drill 8.9.3.
Use your favorite computational tool to confirm that \(3^{2.5789}\approx 17\text{.}\)
Drill 8.9.4.
Find approximate solutions for each of the following equations:
- \(\displaystyle e^x=5\)
- \(\displaystyle e^x=\frac{1}{\sqrt{e}}\)
- \(\displaystyle e^{3x}=7\)
- \(\displaystyle e^{x/3}=12\)
- \(\displaystyle 2^{x}=3^x\)
- \(\displaystyle e^{x-5}=e^5\)
- \(\displaystyle 7^{x^2-x}=e\)
- \(\displaystyle 7e^{x+3}=2\)
- \(\displaystyle e^{2x}-3e^x+2=0\)
- \(\displaystyle e^{2x}- e^x-2=0\)
- \(\displaystyle \sin(e^x)=\frac{1}{\sqrt{2}}\)
Similarly if we needed to solve the equation \(\ln(2x-4)=0\) we would take advantage of Logarithm Property #2. Taking the exponential of both side we see that:
\begin{align*}
e^{\ln(2x-4)}\amp =e^{0}\\\\
2x-4\amp =1\\
x\amp =5/2.
\end{align*}
Drill 8.9.5.
Solve each of the following equations exactly, without the use of technology.
- \(\displaystyle \ln(2x)-\ln(x+1)=4\)
- \(\displaystyle 2\ln(x)=\ln(3x-2)\)
- \(\displaystyle -\ln(2x)=8\)
- \(\displaystyle -\ln\left(\frac{1}{2x}\right)=8\)
- \(\displaystyle \ln(2x)+\ln(2)=8\)
- \(\displaystyle \ln(5x)=\ln(x)+x\)
- \(\displaystyle [\ln(x)]^2+3\ln(x)+2=0\)
- \(\displaystyle [\ln(2x)]^2- \ln(x)-2=0\)
- \(\displaystyle \ln(x)\cdot\ln(2x)=1\)
- \(\displaystyle \ln(2x)\cdot\ln(x)=\ln(3x)\)
Problem 8.9.6.
(a)
A particle is moving on the \(x\)–axis so that its position at time \(t\ge 0\) is given by
\begin{equation*}
x(t) = \sin(e^t).
\end{equation*}
- When is the first time the particle stops moving forward and starts moving backward?
- When does the particle start moving forward again?
(b)
A particle is moving on the \(x\)–axis so that its position at time \(t\ge 0\) is given by \(y(t) =
e^t-\cos(e^t)\text{.}\)
- Show that this particle never moves backward.
- At which times does it stop moving forward?
Let’s return to the question that motivated our look into the logarithm function in the first place: “How long will it take for my money to double?”
Problem 8.9.7. Find the Pattern.
(a)
Suppose that \(P_0\) dollars are invested in two separate accounts, \(A\) and \(B\text{,}\) where the interest in compounded continuously at the nominal rates of \(5\%\) and \(10\%\text{,}\) respectively. Let \(t_A\) be the time it takes for the initial investment in account \(A\) to double, and let \(t_B\) be the time it takes for the initial investment in account \(B\) to double. Is \(t_B=\frac12 t_A\text{?}\)
(b)
At the end of the last section we asked you to guess the answers to part (a). Did you guess correctly? What was the intuition that lead you to your correct, or incorrect, guess?
(c)
Suppose that \(P_0\) dollars are invested in two separate accounts, \(A\) and \(B\text{,}\) where the interest in compounded continuously at the rates \(r_A\) and \(r_B\text{,}\) respectively. Let \(t_A\) be the time it takes for the initial investment in account \(A\) to reach \(nP_0\text{,}\) and let \(t_B\) be the time it takes for the initial investment in account \(B\) to reach \(nP_0\text{,}\) were \(n\) is some positive number. How are \(t_A\) and \(t_B\) related?
When we were modeling exponential population growth, we always stated the relative growth rate for you, but in real applications this rate will rarely be known a priori. Usually we will have to compute the relative growth rate based on other information from the problem. For this, logarithms are needed. An example will help to clarify what we mean.
Example 8.9.8.
Suppose we have a bacteria culture which grows at a rate proportional to the amount of bacteria present. Suppose further that it has been observed that we have \(2\) grams of the bacteria initially, then \(24\) hours later we have \(3\) grams. Can we predict how much would we have in \(48\) hours?
Let \(B(t)\) represent the amount of bacteria present. This time our model is
\begin{align*}
\dfdx{B}{t}\amp =rB, \text{ (The growth rate is proportional to the amount of bacteria present.)} \\
B(0)\amp =2, \text{ (Initially we have }2 \text{ grams of bacteria.)}\\\\
B(24)\amp =3.\text{ (There are } 3 \text{ grams of bacteria
present after }24 \text{ hours.)}
\end{align*}
Problem 8.9.9.
Before we proceed, try to guess how much bacteria we will have after \(48\) hours. Will it be \(4\) grams? \(6\) grams? More? Less? Write down your best guess. We will revisit this in Problem 8.9.10, as soon as we have a formula for \(B(t)\text{.}\)
Note that our model is an IVP plus the extra datum: \(B(24)=3\text{.}\) The first two pieces of our model, \(\dfdx{B}{t}=rB\text{,}\) and \(B(0)=2\text{,}\) define an IVP very similar to IVP (8.18). In fact the only difference is that where we had \(0.3\) before, we now have the unknown parameter, \(r\text{.}\) In Problem 8.6.4 we saw that \(y(t)=10e^{0.3t}, \) so it appears that the solution of the problem in this example must be
\begin{equation*}
y(t)=2e^{rt}
\end{equation*}
but we don’t know the value of \(r\) (yet).
To determine \(r\text{,}\) we use the second datum. Since \(B(24)=3\) we have \(\frac32=e^{24r}\text{.}\) Had we not taken the trouble to invent the natural logarithm it would be very difficult to proceed from this point. Since we did take the trouble, the next step is straightforward. Taking the natural logarithm of both sides, we see that \(\ln\left(\frac{3}{2}\right)=24r\) or
\begin{equation*}
\frac{1}{24}\ln\left(\frac{3}{2}\right)=r.
\end{equation*}
Thus our growth model for this bacterial growth culture is
\begin{equation*}
B(t)=2e^{\left(\frac{1}{24} \ln \frac{3}{2}\right)t}.
\end{equation*}
Expressing it in this form is cumbersome. It is not wrong, just awkward. We can use the properties of exponents to clean it up at bit. This is not strictly necessary, but in general the simpler we keep our notation the better. Using the properties of exponents and the mutually inverse nature of the natural exponential and the natural logarithm, we have
\begin{equation*}
B(t)=2e^{\left(\frac{1}{24} \ln\left(3/2\right)t\right)} =
2\left(e^{\ln\left(\frac32\right)}\right)^\frac{t}{24}
=2\left(\frac{3}{2}\right)^{\frac{t}{24}}.
\end{equation*}
So after \(48\) hours we have
\begin{equation*}
B(48)=2\left(\frac{3}{2}\right)^{48/24}=2\left(\frac{3}{2}\right)^2=2\left(\frac{9}{4}\right)=\frac{9}{2}=4.5
\text{ grams}.
\end{equation*}
Problem 8.9.10.
How good was your guess in Problem 8.9.9? Whether it was good or bad isn’t really the point. Either way take a moment to hone your guessing skills by considering why you guessed the way you did. What was your intuition telling you about the problem?
A natural question to ask is, “How long will it take for the culture to reach \(4\) grams?” Translating this into a mathematical question, we want to find \(t\) when \(B(t)=4\text{,}\) That is, we need to solve
\begin{equation*}
4=2\left(\frac{3}{2}\right)^{\frac{t}{24}}
\end{equation*}
for \(t\text{.}\)
Problem 8.9.11.
Show that \(t=\frac{24 \ln(2)}{\ln \left(\frac{3}{2}\right)}
\approx 41.03 \text{ hours.} \)
Problem 8.9.12.
At a certain time a Petri dish contains \(3\) grams of bacteria. Three hours earlier it contained \(0.5\) grams of bacteria.
(a)
Find a formula for the function, \(A(t)\text{,}\) which gives the amount of bacteria at any time, \(t\gt0\text{.}\)
(b)
How long does it take for the dish to contain \(8\) grams of bacteria?
Problem 8.9.13.
Starting with an unknown number of bacteria a culture is growing at a nominal (relative) rate of \(0.25 \frac{\text{grams}}{\text{hour}}\text{.}\)
(a)
How long does it take for the culture to double in size?
(b)
How long does it take for the culture to triple in size?
