9
$\begingroup$

When I was studying in high school, my teacher taught me about

log(a) + log(b) = log(ab)

log(a) - log(b) = log(a/b)

log(a) / log(b) = log$_b$(a)

Then, I asked my teacher "Why there is no formula log(a) * log(b) = (something) ?".

My teacher didn't know why ?

Now, I still don't know why there is no formula log(a) * log(b) = (something) ?

$\endgroup$
5
  • 3
    $\begingroup$ The formula is not particularly useful, but it is $\log a\cdot\log b=\log b^{\log a}=\log a^{\log b}$. If you like, you can rewrite the division formula as $\log b\cdot\log_b a=\log a$, which would be significantly more useful. $\endgroup$
    – abiessu
    Commented Apr 23, 2018 at 17:15
  • 1
    $\begingroup$ Why should there be such a formula? Just try to imagine by yourself what this formula could possibly look like. Also, when dealing with complex numbers the formulas you mentioned are not so straightforward. $\endgroup$
    – Marcel
    Commented Apr 23, 2018 at 17:18
  • 3
    $\begingroup$ You may just as well ask why is there no formula for $e^x + e^y$, for which there is $e^xe^y = e^{x+y}$. $\endgroup$ Commented Apr 23, 2018 at 17:19
  • 2
    $\begingroup$ Log turns multiplication into addition. It's for the inverse function of log (the exponential) that we would expect such an identity (turning addition into multiplication) and that is what we have: $e^{x+y} = e^x \cdot e^y$. $\endgroup$
    – Winther
    Commented Apr 23, 2018 at 17:19
  • $\begingroup$ For the record I think this is a good question. You can learn a lot in math by asking why something isn't $\endgroup$
    – pancini
    Commented Apr 23, 2018 at 17:21

2 Answers 2

7
$\begingroup$

For real numbers $x$ and $y$, the equation $(\log x)y = \log (x^y)$ holds.

Thus for real positive numbers $a$ and $b$, from letting $y = \log b$, it follows that $\log(a)\log(b) = \log(a^{\log b})$. Yes one can deduce that $\log a \log b$ is also $\log (b^{\log a})$.

These equations are not mentioned much, perhaps because they can easily be deduced from the other laws (and it doesn't seem all that interesting, at least to the generalist just learning this stuff).

$\endgroup$
2
$\begingroup$

There is currently no well-known function $\;f(x,y)\;$ such that $\;\log(x)\cdot\log(y)=\log(f(x,y)).\;$ That is, the function $\;f(x,y):=x^{\log(y)}=y^{\log(x)}\;$ has not been given a name yet, although it is a valid function. This situation may change at some future time. There are comparatively few named functions but new ones appear sometimes. One such example is the Lambert W function.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .