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Do people good at math totally memorize these logarithm rules below? If so, are there good mnemonics for this? I'm bad at math and I only memorize these rules really vaguely by rote, thus when needed, I'm not sure if I remembered them correctly. So I always end up checking if they are true by testing them assigning numbers like 2 to m and n on the fly. Is this something smarter people would think is a stupidly inefficient process to go through? I wonder how other people deal with this?

  1. $\log_b (mn) = \log_b(m) + \log_b(n)$
  2. $\log_b(m/n) = \log_b(m) – \log_b(n)$
  3. $\log_b(m^n) = n \cdot \log_b(m)$
gt6989b
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stacko
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    Mathematicians rarely memorize anything. –  Jan 05 '16 at 20:35
  • I personally never memorized them directly, and instead rederive them each time I need them and feel unsure (but at this point I have learned them from practice) using properties of exponents which are more intuitive and the relationship between exponentiation and logarithms. Also, it would help make your post more readable if you were to visit this page to see how to type $\log_b(mn)=\log_b(m) + \log_b(n)$ – JMoravitz Jan 05 '16 at 20:36
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    @YagnaPatel I don't know how true that is. I certainly don't rederive $6\cdot 7$ every time it comes up. That being said, understanding how to derive the log rules from exponential laws is probably a good idea. – Hayden Jan 05 '16 at 20:37
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    People who use mathematics usually know them by heart. Whether they actively memorized them or grew used to them, they often do not know themselves. Mathematicians and people good at mathematics also remember the hypothesis under which these identities hold and existence conditions on $b,m,n$. –  Jan 05 '16 at 20:39
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    I doubt they don't memorize. I think you have to know quite a lot. E.g., you can't derive definitions; and deriving every theorem would make every math exam a certain failure due to lack of time – adjan Jan 05 '16 at 20:42
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    Logarithm transforms multiplication into addition. If you ever used a logarithmic ruler, you wouldn't have to remember these rules. Exponentiation does the reverse. – A.S. Jan 05 '16 at 20:46
  • In my case the rules seemed easy to remember by noticing that $\log(ab) = (\log a) @ (\log b),$ where operation $@$ is the "lower hierarchy level" version of operation $.$ (Have to cheat a little for $m^n.)$ Even so, I seem to recall having to work enough problems with logarithms in high school that you had to know the rules. Additionally, the relationships were incessantly used in trigonometry, since this was before calculators and so we used logarithm tables (and properties of logarithms) for all the numerical calculations involved in applied trig problems and for solving triangles. – Dave L. Renfro Jan 05 '16 at 21:28

4 Answers4

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You're asking questions and seeing if you can do things more efficiently -- that's what people who are good at math do.

I always forgot these when I first learned them. It might help to have a mnemonic. Anything that helps you remember is a mnemonic -- I find that getting an intuitive understanding or a feeling of why something is true helps me remember it.

Here's something that might help: think about $\log(n)$ as being the number of zeroes after $n$. (This is true if $n$ is a multiple of $10$ and the base of the log is $10$).

  1. $\log_b (mn) = \log_b(m) + \log_b(n)$

If I multiply a number with three zeroes ($1,000$) by a number with two zeroes ($100$), how many zeroes will the answer have? Five: $100,000$.

  1. $\log_b(m/n) = \log_b(m) – \log_b(n)$

Same deal as the previous, except with division. If I take $1,000$ and divide it by $100$, I get $10$. Three zeroes - two zeroes = one zero.

  1. $\log_b(m^n) = n \cdot \log_b(m)$

If my number gets raised to the $n$-th power, how many more zeroes will it have? $$100^3 = 100 \cdot 100 \cdot 100 = 1,000,000$$

so $2$ zeroes to the $3$rd power is six zeroes. That's $n$ times as many.

Of course these laws also work for numbers that are not multiples of $10$. In fact, $10$ wasn't special to the above. But starting from the simplest case makes it easier to remember.

EDIT: Anything that helps you remember a good mnemonic, just want to stress that point. Something like "$\log$ takes the power down and puts it in the front" works perfectly well. I like the above because it helps me remember why the laws are true.

Eli Rose
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The more you understand, the less you need to memorize. You just need to be confident that you'll be able to deduce all needed facts from a few basic ones.

In the case of logarithms, you may want to know the definition: $$ y = \log_b x \iff b^y = x $$ All properties of logarithms come from this definition and the corresponding properties of exponentiation.

If you get stuck when dealing with logarithms, you can always fall back to the definition.

That said, the next thing that you may want to remember is that logarithms transforms multiplication into addition. This is your property 1. Properties 2 and 3 follow directly from property 1. Try to deduce them. It'll give you a good feeling.

lhf
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All these rules can be easily derived from a few rules that you will inevitably memorize as you use them over and over to derive the more exotic properties of logarithms and exponentials. Say, for example, that you have memorized that $\exp(m) \exp(n) = \exp(m+n)$, $\exp(m)^n = \exp(mn)$, and $\log$ is the inverse of $\exp$. (Throughout all exponentials and logarithms are base $b$.) Now:

    1. If $m, n > 0$ then $\log(mn) = \log(\exp(\log(m)) \exp(\log(n))) = \log(\exp(\log(m) \log(n))) = \log(m) \log(n).$

  • 1.5. This is a property that you didn't ask about. It is that $\log(\frac{1}{n}) = -\log(n)$. We show this as follows. First, $1 = \exp(0) = \exp(x - x) = \exp(x) \exp(-x)$ so $\exp(-x) = \frac{1}{\exp(x)}$. Now $\frac{1}{n} = \frac{1}{\exp(\log(n))} = \exp(-\log(n))$ and so taking the logarithm of both sides, $\log(\frac{1}{n}) = -\log(n)$.

    1. $\log(m/n) = \log(m \cdot \frac{1}{n}) = \log(m) + \log(\frac{1}{n}) = \log(m) - \log(n)$ where we use properties 1 and 1.5.
    1. For natural number values of $n$ you can show this easily by applying rule 1 repeatedly $n-1$ times. In general, $\exp(n \log(m)) = \exp(\log(m))^n = m^n$ and taking the logarithm of both sides shows $\log(m^n) = n \log(m)$.
Reinstate Monica
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  • But isn't it circular - hence essentially bypassing learning anything of substance about $\log$? Log can be defined independently of $\exp$. – A.S. Jan 05 '16 at 21:28
  • @A.S. It's not circular if you start out defining $\exp$. But I absolutely agree that this is not the only way to start and ideally you would know multiple ways to define $\log$ and $\exp$ and how to derive each from the other. I didn't mean to elevate one such definition as the right one, but instead demonstrate that if you memorize a few properties you can easily derive many from them. – Reinstate Monica Jan 05 '16 at 23:07
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YES OF COURSE, people memorize those rules. As far as "how" goes I don't know how to answer - you just do it. One thing that ought to help is that it's all the same rule, really, because division is the inverse of multiplication and exponentiation is repeated multiplication. Maybe it will help you to think of it that way.

KipIngram
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