The log cheatsheet is there to encourage you to use logs for quick computations. The squares and powers of two are just for fun (OK it was written by geeks). We’re going to take a quick break from concepts to get the hang of computing with logs.
What is the probability of getting all sixes when rolling 10 six-sided dice? It’s 1 in \(6^{10} \) but how big is that? If we use base-10 logs, we see that
So the number we’re looking for is approximately \(10^{7.8} \) which is \(10^7 \times 10^{0.8} \) or \(10,000,000 \) times a shade over \(10^{.78} \text{,}\) this latter quantity being very close to 6 according to the one-digit logs you computed. So we’re looking at a little over sixty million to one odds against.
These are not just random examples, it is always the best way to get a quick idea of the size of a large power. When the base is 10 we already know how many digits is has, but when the base is something else, we quickly compute \(\log_{10} (b^a) =
a \cdot \log_{10} (b) \text{.}\)
Why is the value \(\ln (10) \approx 2.3 \) on your log cheatsheet so important? It converts back and forth between natural and base-10 logs. Remember, \(\log_{10} x = \ln x / \ln 10 \text{.}\) Thus the constant \(\ln 10 \) is an important conversion constant that just happens to be closer than it looks (the actual value is \(2.302\ldots\)). So for example,
A certain astronomical computation yields the number \(e^{24} \text{.}\) How many digits will this be? (Meaning, how many digits before the decimal point.)
Suppose, your company grows in value by 6% each year for 20 years. By what factor \(C \) does the value increase over this time? The answer is \(1.06^{20} \text{,}\) but about how big is that? For a quick answer, take logs. Using the fact that \(\ln 1.06 \approx 0.06 \) (that’s the logarithmic version of our estimate, with \(x=.06\)), we see that \(\ln C = \ln (1.06^{20}) \approx
20 \times 0.06 = 1.2 \text{.}\) We’d rather have this in base ten, so we compute
\begin{equation*}
\log_{10} C = \ln C / \ln 10 \approx \ln C / 2.3 \approx
1.2 / 2.3 \approx 0.5\ ,
\end{equation*}
maybe a little bigger like \(0.52 \) or so. Looking at the log cheatsheet shows this means \(C \) should be between 3 and 4, somewhat closer to 3. In fact to two significant figures, the growth factor is 3.2.
Historical economists look at real (inflation-adjusted) growth rates over periods of a century or more. If the real annual growth rate averages 2%, what should be the growth factor over the century and a half from 1870 to 2020? Use the method given in Example 1.13 and answer with a whole number.
If you ask someone to state a relationship between the numbers 20 and 30, the most common answer is that 30 is ten more than 20. A more fundamental answer is that 30 is 50% more, or equivalently that 30 is three halves of 20.
Aside
The section on proportionality is designed to emphasize multipicative thinking over additive thinking. Additive thinking is more common only because we find it computationally easier to add than to multiply. Saying that multiplicative thinking is more fundamental is not a precise mathematical statement, so there’s no way to prove it. One reason to believe it is that multiplicative statements remain the same no matter what units you use (as long as the 30 and the 20 are in the same units).
A small city has 40,000 households. To organize an emergency response system, the city wants to organize groups of households on a scale “halfway between the individual household and entire city scale.” What size of groups of households best fulfills this?
Exponentials and logarithms are built to express multiplicative facts. In fact the additive laws of exponentiation and logarithms basically convert multiplicative facts to additive facts, thereby converting the more fundamental fact to the type you can compute more easily.
A tip about setting up equations representing functional relationships: when a quantity has different values at different times such as "before" and "after", using one variable to represent both quantities can lead to mess and confusion. Better to use different names such as \(x_1 \) and \(x_2 \text{,}\) or \(x_{\rm init} \) and \(x_{\rm final} \text{,}\) or possibly \(x \) and \(x' \text{,}\) etc. Using this idea on the question above sets up an equation like this: \(\ln x_2 \approx \ln x_1 + 0.7 \text{.}\) From here, exponentiating leads to
So, if you observe \(\ln x \) increasing by about 0.7, you will know that \(x \) had approximately doubled. This is what it means that logarithms transfer multiplicative scales to additive ones. A multiplicative relation such as doubling transfers to an additive relation, namely addition of about 0.7.
One more thing to keep in mind about logarithms and exponentials is that they do not scale with units. If we change the units of \(x \) from inches to centimeters, and if \(y = e^x \text{,}\) then in the new units \(y' = e^{2.54 x'} = y^{2.54} \text{.}\) The new exponential appears to be the old one to the 2.54 power. What does that even mean? It is a tipoff that \(x \) should not be exponentiated: anything other than a unitless constant is likely to be meaningless when exponentiated. The same is true for logarithms and trig functions.
If a quantity \(Q \) increases at a constant additive rate, it means that if you wait one unit of time, \(Q \) always increases by the same additive amount. In fact, between any two times \(s \) and \(t \) the increase will be \(c (t-s) \text{.}\)
If a quantity \(Q \) increases at a constant multiplicative rate, it means waiting one unit of time always multiples \(Q \) by the same amount, and in general, between times \(s \) and \(t \text{,}\) the factor by which \(Q \) increases will be \(c^{t-s} \) where \(c \) is the factor by which \(Q \) increases in one unit of time.
To get back to the question of what it means about logs relating additive to multiplicative growth, if \(\log x = a + bt \) (constant additive growth over time) then \(x = e^{a + bt} = e^a e^{bt} = A B^t \) where \(A = e^a \) and \(B = e^b \text{.}\) This is constant multiplicative growth.
Constant multiplicative growth rates occur in a lot of applications. This is also called exponential growth because the formula for a quantity growing multiplicatively is \(A e^{bt} \) (also \(e^{a+bt} \) or \(A B^t\)). When \(b \lt 0 \text{,}\) it is called exponential decay or decrease.
If an item is hotter or colder than its environment then the temperature difference between the object and its environment, as a function of time, decreases exponentially.
If we get to assume a nice clean exponential model, and can observe at more than one time point, then exponential growth/decay models are nearly as easy to solve as linear growth models (a highlight of eighth grade math). You should learn this both conceptually and as a mindless skill. In Math 104, this is taught as an elementary example of differential equations. We’ll get to those next semester, but really this kind of growth/decay model is far more fundamental and should be discussed now. Here’s an example of how the computation goes.
A viral infection is spreading exponentially through the community. On the first day that the outbreak had a name, there were 25 infections. A week later there were 40 infections. How many infections will there be in another two weeks? When will the number of infections reach 200,000, which is the size of the entire local population?
Let \(N(t) \) denote the number of infections after \(t \) weeks. Our model is \(N(t) = A e^{bt} \text{.}\) The given information is that plugging in \(t=0 \) and \(t=1 \) give \(N = 25 \) and \(N=40 \) respectively. Because \(e^0 = 1 \text{,}\) we have \(25 = A \text{,}\) while \(40 = A e^b \text{.}\) This gives \(e^b = 40/25 = 8/5 \text{,}\) hence \(b = \ln (8/5) \text{.}\) In another two weeks we will have \(t=3 \text{,}\) so
If we use the growth factor \(B \) in the equation \(A B^t \) instead of the exponential constant \(b \) in \(A e^{bt} \) we may get away without logs. In a week the increase was from 25 to 40, a factor of \(8/5 \) so clearly \(B = 8/5 \text{.}\) Thus \(N = 25 (8/5)^t \text{.}\) In three weeks we have \(N(3) = 25 (8/5)^3 = 512/5 = 102.5 \text{.}\) Evidently the expression \(25 e^{3 \ln (8/5)} \) can be simplified! The time needed to get to 200,000, a growth factor of 8000, is \(t \) such that \((8/5)^t = 8000 \text{.}\) This is, by definition \(\log_{8/5} 8000 \text{,}\) which is equal to \(\log_b 8000 / \log_b (8/5) \) for any base \(b\text{.}\) If we pick \(b=10\text{,}\) we get
\begin{equation*}
t = \frac{\log 8000}{\log (8/5)}\approx 19.12 \, .
\end{equation*}
Notice that that the answer given by plugging in logs is a special case of the growth factor approach, using base \(e \text{.}\) But the ratio of two logs is the same in any base! In Solution 2, we can use what we know about \(\log_{10}\) to compute
So it should take between nineteen and twenty weeks to saturate the city. That’s close enough to the more precise answer that we obtained from a calculator.
The following exercises are a little more involved than the usual self-check exercise. It will be more readily solved if you realize that computations in the exponential model for equilibrating temperatures work better if you keep track of the temperature difference between substance and the environment rather than the temperature of the substance on its own.
Room temperature is \(70^\circ \) F. The temperature of a cup of coffee is \(205^\circ \) F (it’s from McDonald’s, where they make strikingly hot coffee). In 10 minutes it’s a barely sippable \(160^\circ \) F. About how many minutes after purchase is a cup of coffee gulpable (\(130^\circ \) F)? Body temperature (\(98.6^\circ\))? Approximate answers are fine, but please don’t forget to name your variables, give units, etc.
