6
$\begingroup$

I'm just wondering, if I log a constant twice with the same base $y$,

$$\log_y(\log_y(x))= \log_n(x)$$

Can it be equivalent to logging the same constant with base $n$? If yes, what is variable $n$ equivalent to?

$\endgroup$
3
  • 1
    $\begingroup$ In light of your comment to Avatar, you might be interested in my post Exponential and Logarithmic Commutativity. $\endgroup$ Commented Mar 6, 2013 at 17:22
  • $\begingroup$ @DaveL.Renfro Your post is pretty inspirational. Thanks for sharing! I've learnt a new approach to solving log problems from it. When I have time I would definitely look into question 5 and 6 and try to find a solution for them. They look fun. $\endgroup$ Commented Mar 10, 2013 at 13:53
  • 1
    $\begingroup$ If you send me an email (I can't find your email address), I can send you a .pdf file of my original submitted "article". The .pdf version I have is a little more complete than the published version (because of space considerations), which in turn is longer than what I posted. Click on my name to find my email address. $\endgroup$ Commented Mar 11, 2013 at 17:02

4 Answers 4

7
$\begingroup$

No - take as example, $x = 1$. Then $\log_y(x) = 0$ and $\log_y(\log_y(x))$ is undefined. However as $x = 1$, $\log_n(x)$ is always equal to $0$, which means that $\log_y(\log_y(x)) \neq \log_n(x)$

$\endgroup$
1
  • 1
    $\begingroup$ $x=y$ is a similarly nice example. $\endgroup$
    – Mårten W
    Commented Mar 5, 2013 at 8:31
5
$\begingroup$

$$\log_y(\log_y(x))= \log_n(x)$$ $$\implies \log_y(\log_y(x))= \frac{\log_e(x)}{\log_e(n)}$$ $$\implies \log_e(n)=\frac{\log_e(x)}{\log_y(\log_y(x))}$$ $$\implies n=e^{\frac{\log_e(x)}{\log_y(\log_y(x))}}$$

Thus, for given $x,y$, if $e^{\frac{\log_e(x)}{\log_y(\log_y(x))}}$ is defined, then that is the value for $n$.

$\endgroup$
1
  • 1
    $\begingroup$ Bravo! I have once again witnessed the beauty of mathematics. $\endgroup$ Commented Mar 5, 2013 at 11:32
2
$\begingroup$

Yes you can do that with the initial conditions for a logarithm satisfying.

The conditions for log(x) to the base n are: x > 0, n > 0 and n != 1.

so you should be careful with the domain that you are choosing.

$\endgroup$
2
  • $\begingroup$ i think my example is one reason of this $\endgroup$ Commented Mar 5, 2013 at 8:24
  • $\begingroup$ Yes, but you have to stick to the domain of the function as well. $\endgroup$
    – lsp
    Commented Mar 5, 2013 at 10:33
1
$\begingroup$

suppose $x=4$ and $y=2$ then $\log_y(x)=2$ and $\log_2(2)=1$

which means that $\log_n(4)=1$ it means that $n=4$. I don't know if it is helpful for you.

$\endgroup$

You must log in to answer this question.

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