Is there a way to use a calculator for logarithmic form equations that aren't base 10 or base e?

I just find this really hard to believe and quite unacceptable that the only way to do this is plugging in numbers to find an answer.

Also, it seems this function is probably available because I see a lot of sample problems on tutorials that suggest you do the problems "without a calculator".

I understand the exponential equivalent of logarithmic form but I don't understand why they would say to do these problems without a calculator if calculators don't even do that stuff.


3 Answers 3


To quote Wikipedia.

$\log_b a = {\log_d a \over \log_d b}$ This identity is useful to evaluate logarithms on calculators. For instance, most calculators have buttons for $\ln$ and for $\log_{10}$, but not for $\log_2$. To find $\log_2 3$, one could calculate $\frac{\log_{10} 3}{\log_{10} 2}$ (or $\frac{\ln 3}{\ln 2}$, which yields the same result).

So to calculate $\log_b a$ Just do $\frac{\ln a}{\ln b}$.

I suspect the reason they do this has to do with a Stack as the internal data structure of the calculator and Prefix / Postfix notation.

When you do $\ln a$ your first type $a$ and then $\ln$ So it primarily takes $1$ operand.

  • 1
    $\begingroup$ Awesome, +1 for the extra info. I guess it's just not needed but it seems like it should've been included as part of the natural display type OS versions but maybe because the version I bought was only $12. $\endgroup$
    – mchid
    Commented Oct 8, 2015 at 22:29
  • 1
    $\begingroup$ Welcome to MathSE. Nice answer. Here is a tutorial on how to typeset mathematics on this site. $\endgroup$ Commented Oct 8, 2015 at 22:30
  • $\begingroup$ I think it depends on OS to OS, I have seen variants of the exponentiation operator. Will research more on this. $\endgroup$ Commented Oct 8, 2015 at 22:31

Below you will find an example of the kind of calculator they DON'T want you to use. You will notice the $\log_ \square \square $ button just below the "ON" button. This allows you to calculate things like $\log_9 (27) $ directly.

The problem with this calculator is that it allows you to get "correct answers" without having gained any understanding of what logarithms are. This is not an issue if the goal is to create machines that can mindlessly churn out answers, but I guess that the goal in your case is to gain a deeper understanding than that.

So, let's consider $\log_9 (27) $ for amoment. The calculator below will tell you that the answer is $1.5$, but why is that?

The key (in my opinion) is to remember that logarithms are powers.

Asking "What is the logarithm of $27$?" is equivalent to asking "What power gives the result $27$?" and of course depends on us having an agreed base.

Here I have stated that I want the logarithm of $27$ with the base equal to $9$. This means that I want to solve the equation:


The first reaction may be that this is not obvious. After all, the only easy ways to express $27$ as a power are these:


$$27=729^\frac 12$$


and none of these has $9$ as the base.

However, we do know that $3=9^\frac 12$, so we can rewrite the third of those expressions as:

$$27=\left(9^\frac 12 \right)^3=9^\frac 32$$

Now in answer to the question "What is the logarith of 27 (with base 9)" we can ask ourselves "What power of 9 is 27?" and confidently state "$\frac 32$" or "$1.5$"

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  • $\begingroup$ As a Casio user, I feel it should be noted that I have accepted the other answer only because I feel the other answer applies to more users. This is the exact answer for most all NATURAL-V.A.P.M. devices, thanks! $\endgroup$
    – mchid
    Commented Nov 4, 2015 at 19:27

The simplest method, which any calculator can implement, is to use the identity $\log_a b=\ln b/\ln a$.


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