Math 90: Week 2

Yes, although it’s the second week of class, this is after the first recitation.

In addition, I can now announce my office hours. They are from 7-8PM on Monday in Kassar House room 105 and 12:30-1:30PM on Wednesday in my office (number 18, my name is on the door) in the basement of the Kassar House. To get in the building on Monday, you should look at the MRC page, and in particular at this (shaky youtube) video.

After my evening recitation, I’ll post up the problems I gave out in class and their solutions. Please feel free to ask any questions you want here. The details are included after the fold:

In class today, I asked the following questions:

  1. In 1960, the Australian jackalope population was $latex 200000$, but has reduced by a third every $latex 4$ years.
    1. How many jackalopes are there $latex t$ years after the year 1960?
    2. How many jackalopes are there now?
    3. When will there only be a single, very lonely jackalope?
  2. Simplify the following:
    1. $latex \log_2 \left( e^{\ln (\ln e^4)}\right)$
    2. $latex \ln \left( e^{\log 100 – \log 1000} \right)$
    3. $latex 3^{\frac{\ln 1138}{\ln 3}}$ (it may be hard to tell, but that’s an exponent)
  3. Calculate the inverse functions of:
    1. $latex y = \dfrac{\sqrt x}{2 \sqrt x – 4}$
    2. $latex f(x) = \ln (x – 2) – \ln (x + 5)$

And you all did great! Let’s go over the solutions, one by one.

Question I

This is a classic form of exponential growth and decay. We expect the answer to be of the form $latex p(t) = ab^{rt}$ for some $latex a,b,r$. There are a few different ways of going about this, but we’ll just focus on one. We start with a population of $latex 200000$ jackalopes, but after $latex 4$ years, only two-thirds of the original population remains. If our time-measurement unit were groups of four years, then we might say something like $latex p(t) = 200000 \left( \frac{2}{3} \right)^{t}$. But we want to be more precise – our time-measurement unit will be years. So instead we use $latex p(t) = 200000 \left( \frac{2}{3} \right)^{t/4}$.

Does this make sense? Yes – after $latex 4$ years, we have only $latex 2/3$ of the population remaining. After another four years, our population decreases by another third. So it is of the form we want and passes our intuition check.

The two remaining parts are relatively simple from the formula $latex p(t) = 200000 \left( \frac{2}{3} \right)^{t/4}$. Here, $latex t$ represents years since 1960, so when we ask how many jackalopes there are now, we want to know $latex p(52) = 200000 \left( \frac{2}{3} \right) ^{52/4} \approx 1028$ jackalopes. When we ask when there will be only one jackalope, we want to solve for $latex t$ in $latex 1 = 200000 \left( \frac{2}{3} \right) ^ {t/4}$.

To do this, since we have our $latex t$ in the exponent, we divide by $latex 200000$, take the natural log of both sides, and simplify. $latex \ln (1/200000) = (t/4) \ln (2/3) \implies t = 4 \ln (1/200000) / \ln (2/3) \approx 120$ years. Poor jackalope.

Question II

This is the test of our understanding of logarithmic and exponential arithmetic rules (this may feel useless, but it’s something that will come up a lot in this course, so get these rules down). In particular, we’re going to use that the exponential and log functions are inverses of each other (so $latex e^{\ln x}$ and $\ln e^x = x$ for $latex x$ in appropriate domains) a log.

So let’s look at the first one: $latex \log_2 \left( e^{\ln (\ln e^4)}\right)$. First, note that $latex \ln e^4 = 4$, and $latex e^{\ln 4} = 4$. So we are left with $latex \log_2 4$, which is $latex 2$.

The second one is a both trickier and easier, because it is tempting to use many more rules then are necessary. One might proceed naively by writing $latex \ln \left( e^{\log 100 – \log 1000} \right)$ $latex = \ln e^{\log (100/1000)}$ $latex = \log (100/1000) = \log (1/10) = \log (10^{-1}) = -1$. And that isn’t wrong – so that’s great. Or you might also realize that $latex \log 100 = 2$ and that $latex \log 1000 = 3$, so we are asking $latex \ln e^{2 – 3} = -1$.

The last is a bit different. We’re going to do it in two different ways. We might try to cancel out the $latex \ln 3$ in the denominator of the exponent. One way to do this is to write $latex 3$ as $latex e^{\ln 3}$ (this is still using that the exponential and logs are inverses, but this time we are going from “simple” to “complicated” in a sense). Then $latex (3)^{\ln 1138/\ln 3} = e^{(\ln 3)(\ln 1138 / \ln 3)} = e^{\ln 1138} = 1138$.

Another way to do this is to remember the change-of-base formula, which states that $latex \ln 1138/\ln 3 = \log_3 1138$, so that we can use that $latex 3^x$ and $latex log_3 x$ are inverses to conclude that $latex 3^{\log_3 1138} = 1138$.

Question III

This was similar to the problem on the homework that seemed to give the most people the most trouble. Let’s proceed naively: $latex y = \dfrac{\sqrt x}{2 \sqrt x – 4}$, so $latex (2 \sqrt x – 4) y = \sqrt x$. Distributing, we get $latex 2y \sqrt x – 4y – \sqrt x = 0$. Factor out $latex \sqrt x$, we get $latex \sqrt x (2y – 1) = 4y$. Divide to get $latex \sqrt{x} = \dfrac{4y}{2y-1}$, and we can square to find $latex x = \dfrac{16y^2}{(2y-1)^2}$. And this “works,” sort of. Why don’t we specify the domain and range, as in the homework problem? What is the domain of the original function? $latex x > 0$ is necessary as real square roots can’t take in negative numbers. And $latex x \neq 4$, as the denominator can’t be zero. But everything else is fine.

What about its range?
To do this, we should graph the function. The graph looks like the graph below: It’s very similar to a rational function. But there is a key difference: the domain doesn’t extend to any negative value of $latex x$. There is an a horizontal asymptote at $latex y = 2$, and every value $latex y >2$ is in the range. But the left side stops at height $latex 0$, and so the range is $latex (-\infty,0]\cup(2,\infty)$.
The domain and range for the inverse are just the domain and range of the original function, but flipped. So there we are.

For the second, we won’t find the domain and range here. It’s a big pain. So let’s consider $latex y = \ln(x-2) – \ln(x+5)= \ln \left( \dfrac{x-2}{x+5} \right)$. How do we get rid of the $latex \ln$? We exponentiate!

$latex e^y = \dfrac{x-2}{x+5}$, so $latex (x+5)e^y = x-2$. Distributing and bringing it all to one side: $latex xe^y + 5e^y – x + 2 = 0$, so $latex x(e^y – 1) = -2 – 5e^y$. Dividing, we conclude: $latex x = \dfrac{-2-5e^y}{e^y – 1}$

And there we have it.

As (will become, but which I will pretend) is usual, if you have any questions on this week’s homework, please feel free to comment below. And if you haven’t commented on the last week’s post, please do so.

Until next week –

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