Mathochism: Logs, damn logs!

One woman’s attempt to revisit the math that plagued her in school. But can determination make up for 25 years of math neglect?

The scheduling gods decided to give me a break, so I got to take that third pre-calc exam after all. I did my best to study for it, bringing all my books on the romantic mid-week getaway I had been planning with my spouse for months. (Poor man. At least he got to drink good wine while his math-obsessed wife solved problems in the tasting room gift shop.)

But this test was really dense. It only covered about a chapter and a half, but that chapter and a half included exponential equations, logarithms and trigonometry.

I’ve written before about logarithms, and how much they confused me. The first time I studied them in Algebra II, things were so much harder, since I was being taught by the Brofessor. Luckily, the Youthful Professor’s approach was much clearer, and I was gaining confidence.

But then I got sciatica, and had to miss a lecture, and it just happened to be the lecture on change of base formulas and solving log equations. I studied on my own as best I could. I watched web seminars. I read both pre-calc books.

I thought I understood how to do it. But then I took the test, and the befuddlement began. There were three problems I simply couldn’t solve. And a few more I sort of got, but sort of guessed at.

Happily, I seem to have much more of an aptitude for trig, and sailed through the rest of the test, moving through quadrants and secants with ease. SOHCAHTOA this, suckers!

Long story short, I continue in the B-Zone. I got an 83, which is the lowest score to date. But I’m hoping the next exam, which should be trig-heavy, will be my chance for redemption. I plan to go to YP’s lair before the next class, and ask him to give me a brief primer on logs. I know I will meet them again on the final.

Damn logs … why can’t I quit you?

All text copyrighted by A.K. Whitney, and cannot be used without permission.

6 comments

  • Do you know what it is that confuses you about logs?

  • Mostly change of base stuff. And some of the equations that actually involve finding a number, not just rearranging things. But I’m working on it.

  • Change of base can be fairly confusing I agree. The problem is that it’s usefulness can be hard to understand, and the formula seems more or less arbitrary.

    A possible example that might help is

    3^x = 5000

    On the one hand, we know the answer to that is log_3(5000) (log base 3 of 5000). However, I have no idea what that number is. What we could do is do a change of base

    log_3(5000) = log_10(5000) / log_10(3)

    Now, it’s much easier to figure out what this is.

    log_10(5000) = log_10(5 * 1000) = log_10(5) + log_10(1000).
    log_10(1000) = 3 (10^3=1000)
    log_10(5) is about 0.7 .

    then the last term,
    log_10(3) is about 0.5

    So that means that log_3(5000) is about 3.7/0.5 = 7.4. Plugging the number into google I find 7.75 . Not too shabby. So, while I had no idea what log_3(5000) is, by doing a change of base I was able to get a pretty good idea.

    A few hints on the change of base formula itself that might help
    1) Remember that it’s called the “Change of base formula” for a reason. The “base” in both of the final logs is the same. If we want to change a log_3 into a log_5, then both of our final terms will also be log_5. That’s why we’re doing it in the first place
    2) We know that log_10(5000) > log_10(3), because 5000>3. We also know that log_b(a) is greater than 1 if a>b, and less than 1 if b>a. Why is this? Lets go back to the definition of the log.
    a^x = b.
    If b>a, we’re going to need to make a bigger, i.e. raise it to a power greater than 1. If b3. Then, we know that it has to be
    log_10(5000)/log_10(3), because (from hint 1) log_10(5000)>log_10(3)

    If we did it the other way
    log_10(3)/log_10(5000), we can see that this number is less than 1. However, we know that log_3(5000)>1, so that must be wrong!

    Anyway, I hope that helps and it isn’t just gobbledygook.

  • That is very helpful, thanks. So were the various videos, etc. I watched while studying. Unfortunately on the last test, I screwed it all up anyway. Sigh.

  • With math there are three stages. The first is learning the vocabulary of the subject, the second is learning how to solve the problems, and the third is really understanding what is going on in the broader concept. The problem is that the tests are always before that third step

  • The usefulness of the change of base formula shouldn’t be that hard to understand. A typical scientific calculator gives you logs in exactly two bases: base 10 (log) and base e (ln). If you want to compute a log to any other base, you’ll need to use the change of base formula to get it. (And this was even more true in the days before calculators, where you might have only a precalculated table for logs in only one base to work with).

    So – let’s say we want to compute log_4(64). That is, we want to find x such that 4^x = 64. Or maybe that’s our starting equation, and we recognize that solving it is equivalent to finding log_4(64). (In this case, of course, we might recognize the numbers involved and realize that x=3, but in general, we’ll need to compute it.)

    Ok, so we don’t have a way of computing log_4(64) directly on our calculator, but we do have an equation there, and we can do the same thing to both sides. So let’s take log_10 of both sides: log_10(4^x) = log_10(64). Now we can use the property of logs that says log(a^b) = b*log(a). So apply that to the left hand side here: log_10(4^x) = x*log_10(4) = log_10(64). Both these last two logs are constants that we can calculate on our calculator. Now we have a simple linear equation in x that we can solve by dividing both sides by log_10(4): x = log_10(64)/log_10(4). That’s the change of base formula!

    Plugging in the results from the calculator, x = 1.806179974/0.602059991 = 3. That’s the answer, log_4(64). We could also have calculated the answer using ln instead of log_10: ln(64)/ln(4) = 4.158883083/1.386294361 = 3. It doesn’t matter which base we calculate the original logs in, as long as we use the same base for both numerator and denominator.

    While 10 and e are the two most common bases for logs (that’s why they have their own keys on the calculator), logs to specific other bases show up in various problems, so it’s useful to remember the change of base formula. For example, how many rounds do you need for a single-elimination tournament with n entrants? It’s log_2(n), rounded up to the next integer – which on the calculator is log(n)/log(2). So for 150 entrants, you would need log_10(150)/log_10(2) = 2.176091259/0.301029995 = 7.22881869, which rounds up to 8 rounds. That’s useful to know if you are the tournament director trying to draw up a schedule!

    So to summarize: to find the log of something to an arbitrary base on a scientific calculator, take the log (or ln) of the thing you want to find the log of, and divide by the log (or ln) of the base you want to use. That’s the change of base formula!

    (Also, as before, if you want to post problems here that you couldn’t figure out or had to guess on, your audience might be able to offer explanations.)

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