MitPY 2: Change of Base in Logarithms

This post is an offshoot of our offer of free maths help to anyone and everyone. Frequent commenter RF started the ball rolling, asking how one might teach (= understand) the change of base for logarithms:

\boldsymbol{\log_ax = \dfrac{\log_bx}{\log_ba}}

I’ll leave it as open as possible for teacher-commenters to discuss. I honestly don’t know how I’d teach it, and I have difficulty understanding it myself. But here are some preliminary thoughts.

Logarithms are intrinsically difficult because they are inverse things, implying that untangling any logarithmic statement also requires untangling the inverses, to get to the exponentials. This suggests any reasonable approach to the above identity must be grounded in some uninverted, exponential fact(s).

Pondering it quickly, I can see three ways that, together or separately, may lead to some understanding of the above identity:

  • First, multiply through to get rid of the fractions. (Never a bad bet.)
  • Second, think of a special case(s) that are easier to think of exponentially and to understand.
  • Third, give things names: if you want to understand, for example, what \boldsymbol{\log_ax} means then write \boldsymbol{\log_ax}= A. That then gives you the bones to be able to play around with the underlying exponential meaning.

I’ll leave it there, until others have commented.

UPDATE (25/03/20)

Thanks to everyone who has commented so far. I’ll look more carefully at the discussion later on, when I can breathe. But, as long as people are happily discussing things, I’ll take a back seat. After the conversation has about run its course, I’ll try to summarise up top the smartness in the various approaches.

Just a couple quick points:

  • I definitely should have included an experiment/special-cases dot point along the lines suggested by Storyteller in the comments.
  • For those who want to experiment with what LaTeX does and doesn’t work in comments, you can experiment here.

I can also edit comments (a power I plan to use only for niceness, rather than evil). So, I’ll fix up some of the TeX glitches in comments later today.

UPDATE (28/03/20)

Thanks to all the commenters below. Here’s an, um, “summary” of the discussion. (It’s intended to be slow and gentle, and could be further gentled for students/classes by the inclusion of numerical examples.)

Part 1 (Where log rules come from)

We want to make sense of the change of base formula

    \[\color{magenta}\boxed{\boldsymbol{\log_ax = \dfrac{\log_bx}{\log_ba}}}\]

That’s not obvious, since the inverseness of logarithms obscures everything. So, let’s forget about chasing the weird formula, and first get back to thinking about powers. Remember the fundamental meaning of logarithms:

    \[\color{red}\boxed{\boldsymbol{\log_b x = B \quad \Longleftrightarrow \quad  b^B  = x}}\]

That is, any logarithm equation is just an exponential/power equation, thought of from a different direction: what power of \boldsymbol{b} gives us \boldsymbol{x} (answer \boldsymbol{B}), rather than what does \boldsymbol{b} to the power of \boldsymbol{B} give us (answer \boldsymbol{x})?

Ultimately, as Terry emphasised, any log rule must come from a corresponding power rule via this equivalence. For example, note that

    \[\boldsymbol{b^D = a \quad \Longrightarrow \quad  b^{AD}  = a^A} \qquad \mbox{(because $\left(b^{D}\right)^A= b^{AD}).}\]

That simple power manipulation gives us the log rule

    \[\color{blue}\boxed{\boldsymbol{\log_b \negthinspace \left(a^A\right) = A\log_ba}}\]

(In words, the power needed to give us \boldsymbol{a^A} is \boldsymbol{A} times the power to give us \boldsymbol{a}.)


Part 2 (Experimentation)

We’ll give a more direct approach to the change of base rule in Part 3. First, we can experiment in the manner suggested by RF, and explored at length by Storyteller (aka Proust).

Let’s think about powers of 3. We have \boldsymbol{3^6 = 729} or, in log form, \boldsymbol{\log_3729 = 6}. But then, as powers of \boldsymbol{3^2}, we have \boldsymbol{\left(3^2\right)^3 = 729} or \boldsymbol{\log_{3^2}729 = 3}. We can summarise this in log form as

    \[\boldsymbol{\log_{3^2}729  =    \frac{\log_3 729}{2}\,.}\]

Notice that at its heart this calculation is just the blue power rule we proved above. And, critically, it is no coincidence that the 2 appears twice: on the left as a power and on the right as the denominator.

We now wave the Mathologer magic wand (perhaps after more experimentation). We replace the 3 by \boldsymbol{b}, the 2 by \boldsymbol{B} and (with much more trepidation) the 729 by \boldsymbol{x}. With fingers crossed, that gives

    \[\color{blue}\boxed{\boldsymbol{\log_{b^B}x  =    \frac{\log_b x}{B}}}\]

This is a change of base rule for logs we can actually understand: it tells us that if \boldsymbol{b^B} is the base then we need (1/\boldsymbol{B})th the power than if \boldsymbol{b} were the base. And, again, this is just the blue power rule, but written in log form.

Lastly, let’s write \boldsymbol{b^B =a}. Then \boldsymbol{B = \log_ba}, and our blue log rule now takes the form

    \[\color{magenta}\boxed{\boldsymbol{\log_{a}x  =    \frac{\log_b x}{ \log_ba}\,.}}\]

This is exactly the magenta log rule we’re after, and we’ve kind of semi-proved it. The gaps are justifying that:

(i) Any \boldsymbol{x} can be written in terms of the bases \boldsymbol{b} and \boldsymbol{b^B};

(ii) Any \boldsymbol{a} can be written in terms of the base \boldsymbol{b};

(iii) The blue power rule holds in this more general context.

That is, the proper justification of the change of base rule requires a deeper exploration of the real numbers and is, therefore, pretty much outside the school world.

Part 3 (More direct “proof”)

This is essentially the proof given by Franz, Glen and Anonymous, but framed more like SRK’s argument.

The change of base rule for logarithms has to do with quantities written with different bases. So, let’s ask that question directly. Suppose we have something with base \boldsymbol{a} and we want to write it as something with base \boldsymbol{b}. That is, if we have

    \[\boldsymbol{x =a^A\,,}\]

how can we rewrite

    \[\boldsymbol{x =b^B\,?}\]

(The question can be made more concrete by specifying to common numerical bases. So, one can ask how to rewrite 10^X, or even 10^6, as 2^Y or e^Z. Even more concretely, we can be back in the experimental world of Part 2.)

Well, what power of \boldsymbol{b} do we need to get \boldsymbol{a^A}? The answer to that must be \boldsymbol{B = \log_b \left(a^A\right)}, by the very definition of logarithms. But then our blue power rule tells us

    \[\color{magenta}\boxed{\boldsymbol{b^B = a^A \quad \Longrightarrow \quad  B  = A\log_ba} }\]

This magenta power rule tells us how to change from one base to another, so it is really all we need to know. In fact, we don’t even need to that much: in this case, it is better to remember the technique rather than the formula. But, let’s go one step further.

Write \boldsymbol{x} for the common quantity \boldsymbol{x = a^A = b^B}. Then \boldsymbol{A = \log_ax} and \boldsymbol{B = \log_bx}, and our magenta power rule becomes

    \[\color{magenta}\boxed{ \boldsymbol{\phantom{A^A}\hspace{-7\jot}\log_bx   =   \log_ba\cdot \log_ax  \phantom{A^A}\hspace{-6\jot} }}}\]

This is exactly the magenta log rule we’re after, with the fraction multiplied out.

As in Part 2, the proof assumes that logs and the blue power rule work for general real numbers, not just for 3^6 and the like.

33 Replies to “MitPY 2: Change of Base in Logarithms”

  1. I used to do it by solving the equation a^m = x in two different ways, first by using \log_a and then by using \log_b. Of course you first need to prove that log(a^m) = m \log a$.

    That was in the days before they abolished logarithms from the German curriculum. Now they’re explaining the meaning of exponential growth on prime time TV (and get it wrong).

    1. Oh, God. Thanks, Franz, and feel free to send some PoSWW coronavirus material. (And, sorry for the delay. Your comment was stuck in spam for some reason.)

  2. I was wondering if some link could be made to sequences and series by fixing the log base (say 5) and then having the arguments follow a geometric sequence: 5, 25, 125, 625 and have students calculate these (by hand). Then, fix the argument at, say, 5 and vary the bases, 5, 25, 125, 625 and again have students investigate.

    To prove the pattern (yes, any VCAA spies reading this, I said a naughty word) you would need the change of base rule, surely?

    Nothing more than a thought at this point.

    1. Thanks, RF. Someone emailed me this morning with good suggestions somewhat along your lines. Sort of experimental, which I definitely should have included as a dot point. I’m pushing them to post the approach as a comment here.

  3. I really like this question. Because the answering of it brings into question ordering issues of when and how to teach index/log laws, as well as problems of rigor and special cases.

    I don’t have a good answer myself — looking forward to reading what others have to say!

  4. The change of base formula is typically taught in Methods Unit 3, so there should be some solid ground of exponentials and logarithms to build on.

    I like to start with the definition of a log:

    a^A = x \Leftrightarrow A = \log_a (x), where the base a is defined for a > 0 and a \neq 1.

    Students should have access to all the log rules (taught in mainstream Yr 10 maths (in theory) and taught again in Yr 11 Methods Unit 2), so you can take the log to the base of b of both sides of a^A = x. So:

    a^A = x

    \Rightarrow A = \log_a (x). …. (1)

    a^A = x

    \Rightarrow \log_b (a^A) = \log_b (x)

    \Rightarrow A \log_b (a) = \log_b (x)

    \displaystyle \Rightarrow A = \frac{\log_b (x)}{\log_b (a)}. …. (2)

    Now equate (1) and (2) to get the formula.

    1. OK, I don’t accept this “definition” of a log, but it is largely irrelevant (I was taught to ln(x) as the area under a 1/t graph from 1 to x) from which the exponent is then defined as the inverse of a log… all irrelevant for my purposes.

      The rest of what you write is valuable and helpful.

      So thanks.

      1. Hi RF.

        Re: The definition of log you were taught. Indeed. Same here.

        There are several definitions of log, arguably some ‘better’ than others. I probably should have said:

        “I like to start with the definition of a log used in Mathematical Methods and earlier years:”

        since the integral definition is beyond Maths Methods at the point in time when logarithms are taught and would also require an element of ‘undoing’ what has been taught in previous years.

        However, given your non-acceptance of the definition of a log that I used, I’m curious what definition you would use in Year 10 Maths or Maths Methods Unit 2 (I’m guessing not the integral definition, which doesn’t leave much wriggle room lol!).

        So before any discussion on how to teach the change of base formula, a discussion on how to teach the concept of a logarithm should be had ….?

        It should be noted that many functions met in secondary school (including the inverse trigonometric functions) are more usefully defined via integrals (but typically are not). And most of the so-called ‘advanced functions’ met at university are defined either by integrals or differential equations.

  5. OK, so I think I like this short proof, but I would really discuss in detail each of the steps, the hypotheses involved, and the whole idea of “proof”.

    I wouldn’t call it a rule either — it is a theorem!

    \aligned &x = x \\ \Leftrightarrow &a^{\log_a x} = b^{\log_b x} \\ \Leftrightarrow &\Big(b^{\log_b a }\Big)^{\log_a x} = b^{\log_b x} \\ \Leftrightarrow &b^{\log_b a \log_a x} = b^{\log_b x}\\ \Leftrightarrow &\log_b a \log_a x} = \log_b x \,. \endaligned

    (Edited by Marty to correct the LateX)

    1. Shit. Can someone help me out with the LaTeX? I can’t get it to compile for some reason…. is there a way to include pictures? Because I can just compile it locally and paste a picture in :/.

      1. Such language! Glen, did you enter it as I suggested above? Worked for me. If you google “entering latex in WordPress comments” someone has WordPress sandbox post for experimenting. Will look for the link later. (Still waking up …)

        Definitely don’t do pictures! At worst I’ll edit your comment later and fix the Tex.

        1. Oh I’m an idiot, I can’t edit my comment anymore :/. Sorry Marty. (But yes I googled it and it seems as though I could have just used the aligned environment within dollar signs.)

  6. Assuming:

    1) the learner is comfortable with fractions

    2) the learner has a clean way to think about (interpret) what log_b(x) means – e.g. “to what do I raise b to get x” and so appreciates log_3(3^2) = 2 as more than a rule

    3) the learner has blindly accepted that an index can be something other than an integer

    4) we want to take an approach where we “create” the critter.

    The story:

    Once upon a time, I needed to determine log_3(10).
    I figured it is 2 and a bit, but what is the bit?

    I know I can figure out log_10(11), because I have log_10 tables, or a scientific calculator, depending on the era in which this story is set.
    (Oh dear, but if the story is set now, I might also have a machine that can … – best ignore this!)

    Mmm. Can I convert all loggie things with nasty bases into log_10(x)

    Make life easier for a minute, stick with simple stuff.
    I realise that:
    log_3(9) = 2 and 2 = log_10(100) and so

    log_3(3^2) = log_10(10^2) and so

    I can write logs in base 3 as something equal in value but as a log in base 10.

    But how, when things get ugly?
    Let’s try two simple things, that are not equal (and get all log statements for numbers):

    log_10(10^4) is twice log_3(3^2)
    2*log_3(9) = log_10(10^4)

    log_3(9) = log_10(10^4)/ 2 = log_10(10^4)/ log_10(10^2) (= 2)

    Mmmm …how does that help?

    Then I just noticed that log_10(10^4)/ log_10(10^2) = log_10^2(10^4) = 2 – that is peculiar!

    So is log_10(10^10)/ log_10(10^5) = log_10^5(10^10)
    Yes – as they both = 2!

    So is log_10(10^8)/ log_10(10^3) = log_10^3(10^8)
    Yes – as they both = 8/3

    So what am I saying?

    log_10(x)/ log_10(a) = log_a(x)

    OMG – really, how bizarre?!
    Test a few more cases …

    Hang on, can 10 can be anything? I.e. log_b(x)/ log_b(a) = log_a(x)

    Could that work?

    Suppose x = b^5 and a = b^4 then the LHS = 5/4

    The RHS = log_(b^4) (b^5) and so to what do I raise b^4 to get b^5?
    5/4, so we are all good, in this case.

    Now suppose x = b^p and a = b^q then the LHS = p/q

    The RHS = log_(b^q) (b^p) and so to what do I raise b^q to get b^p?
    p/q, so we are all good, in this case.

    So we are all good for all x and a that can be written as b^?, where ? is a member of …

    So maybe “necessity is the mother of invention” rings true once again and some mother created it so they needed just 1 (or 2) sets of log tables (or a scientific calculator).
    But later on, some other mother invented calculators with more log buttons than just log_10 and log_e and so maybe we do not need this thing now; or do we?

    The End.

        1. OK. Will update if I every find how to answer in the affirmative.

          But I’m currently “teaching” via video conference, writing a major assessment and playing an online Mahjong tournament, so it might be a while.

          Multi-tasking was never my strong suit.

  7. I feel like my approach has been to dodge the issue rather than confront it (so arguably this isn’t an answer to your question), but FWIW:

    For year 10-ish students (who at this stage aren’t quite ready for a more abstract treatment) I don’t explicitly teach the change of base rule. That’s because at this stage generally all that is required is knowing how to use a scientific calculator (which only has \log_{10} or \log_e functions) to find approximate solutions to equations like 2^x = 5. So I just encourage them to solve such equations by taking \log_{10} of both sides then using a log law to get x\log_{10}2 = \log_{10}5.

    I’ll have to mess around with the approaches mentioned above by RF and Storyteller though. Seem interesting, although my institutionalisation makes me worry about time-pressure / opportunity costs / etc.

    1. Thanks, SRK. I think your approach is very good, and I think your time pressure / opportunity cost point is doubly important. First, it is important as a practical issue: given the change of base rule just sits there in the curriculum, time may well be better spent digging deep on other topics.

      But I think your point is even more important at a second, conceptual level: is it maybe better, just on principle, to do without the change of base rule? It seems the main useful point is to note that all exponential functions are essentially the same. That is, it is important to realise that one can always write \boldsymbol{a^x} as \boldsymbol{b^y}. But, one can see that both specifically and generally as you (SRK) have done, directly and without an explicit change of base formula.

      1. Yeah, that second point seems especially useful. It’s something I lean on when introducing \frac{d}{dx}\left(e^x\right) = e^x. (Something like: (i) some handwaving that the derivative of an exponential function is an exponential function, ie. \frac{d}{dx}\left(b^x\right) = Ab^x, (ii) wouldn’t it be nice if there was a value of b such that A = 1? (iii) ask students to trust me that the solution exists and is unique).

  8. As an alternative to the story above:


    1) the learner is comfortable with fractions,

    2) the learner has a clean way to think about what \log_b(x) means – e.g. “to what do I raise b to get x” and so appreciates \log_3(3^2) = 2 as more than a rule

    3) the learner is comfortable that a number can be represented in many different exponential forms (e.g. 256 = 16^2 = 4^4 = 167772=16^\frac{1}{3} = ...)

    3) the learner has blindly accepted that an index can be something other than an integer

    4) we choose to take this critter as a given, and then try to make sense of it.

    So, here goes.

    \log_a{x}=\frac{\log_b{x}}{\log_b{a}} , meaning I can write a single log, in a given base, as a particular quotient of logs in another base.

    Why would that be true?

    What is the most basic thing I know about logs?
    \log_2{8} = 3 since the LHS is saying to what do I raise 2 to make 8, or more informatively written: \log_2{2^3} = 3.

    So where \log_a{x}=\frac{\log_b{x}}{\log_b{a}} is concerned, to see if we can make more sense of it, make x=b^m, a=b^n and we have:


    LHS = \frac{m}{n}

    RHS = \frac{m}{n}


    1. I like this one.

      It isn’t really lower-level, actually, these steps can be made perfectly rigorous, by just being careful about where these numbers live.

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