Solving equation $\log_y(\log_y(x))= \log_n(x)$ for $n$

Question

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?

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score 7 · Answer 1 · answered Mar 05 '13 at 08:08

7

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 $\log_y(\log_y(x)) \neq \log_n(x)$

answered Mar 05 '13 at 08:08

mardat

366

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$x=y$ is a similarly nice example. – Mårten W Mar 05 '13 at 08:31

score 5 · Accepted Answer · answered Mar 05 '13 at 11:19

5

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

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

answered Mar 05 '13 at 11:19

Aang

14,672

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Bravo! I have once again witnessed the beauty of mathematics. – Archy Will He 何魏奇 Mar 05 '13 at 11:32

score 2 · Answer 3 · answered Mar 05 '13 at 08:22

2

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.

answered Mar 05 '13 at 08:22

lsp

4,745

i think my example is one reason of this – dato datuashvili Mar 05 '13 at 08:24
Yes, but you have to stick to the domain of the function as well. – lsp Mar 05 '13 at 10:33

score 1 · Answer 4 · edited Mar 05 '13 at 08:45

1

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.

edited Mar 05 '13 at 08:45

Stefan Hansen

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answered Mar 05 '13 at 08:18

dato datuashvili

9,194

Solving equation $\log_y(\log_y(x))= \log_n(x)$ for $n$

4 Answers4