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My brother challenged me to solve this problem. Trying since 2 days. I came up with $a^{a^y}= x^n$ assuming $y$ is $\log_a(\log_a x^n)$. There's no solution available on net as well. If someone can solve it, it would be of great help! Thanks

Trial 1:

$\Rightarrow\log_a(\log_a (e^2)^{a^2})$

$\Rightarrow\log_a(\log_a \exp(2\cdot a^2))$ ...By $\exp$ property

$\Rightarrow\log_a(2*(a^2)\cdot\log_a (e))$ ... By log property $\log x^a= a \log x$

$\Rightarrow\log_a(2\cdot(a^2)\cdot(\frac{1}{\log_e(a)}))$ ... By $\log$ property

$\Rightarrow\log_a(2\cdot(a^2)) + \log_a(\frac{1}{\log_e(a)})$ ...By $\log$ property

$\Rightarrow 2\cdot\log_a(a^2)+\log(\frac1{\log_e(a)})$

$\Rightarrow 4 + \log(\frac{1}{\log_e(a)})$

Couldn't solve beyond that

Trial 2:

Considering $y=\log_a(\log_a (x^n))$

$\Rightarrow a^y= \log_a (x^n)$

so, $a^{a^y}= x^n$

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First notice that your expression is $$ \log_a(n\log_a x)=\log_a n+\log_a\log_ax $$

For $n=a^2$ you have $\log_an=\log_a(a^2)=2$; for $x=e^2$, $$ \log_a\log_a x=\log_a(2\log_a e)=\log_a2+\log_a\log_ae $$

You could observe that $$ \log_a2=\frac{\log 2}{\log a} $$ and $$ \log_ae=\frac{1}{\log a} $$ so $$ \log_a\log_ae=\log_a\frac{1}{\log a}=-\log_a\log a=-\frac{\log\log a}{\log a} $$ so you finally get $$ 2+\frac{\log2}{\log a}-\frac{\log\log a}{\log a} $$

Notes.

  1. The unadorned $\log$ symbol denotes the natural logarithm.
  2. I skipped over the checks for existence; the computations make sense for $a>1$ as the final formula shows.
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  • $\begingroup$ You made a sligth mistake, your $\frac{1}{\log a}$ should be $\frac{\log 2}{\log a}$, you did not copy it correctly. $\endgroup$ – Jan Aug 13 '18 at 9:16
  • $\begingroup$ @Jan Yes, fixed. $\endgroup$ – egreg Aug 13 '18 at 9:22
  • $\begingroup$ This is the closest solution I could see by far! Thank you @egreg $\endgroup$ – Sayli_ambure Aug 13 '18 at 9:23
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You were doing quite well but made a small error. Notice that you say: $$\log_a(2\cdot(a^2)) + \log_a(\frac{1}{\log_e(a)}) \Rightarrow 2\cdot\log_a(a^2)+\log(\frac1{\log_e(a)})$$

But if we have logarithms we can't simply take the multiplication outside. Remember that $\log(a\cdot b)=\log(a)+\log(b)$. Thus we get: $$\log_a(2\cdot(a^2)) + \log_a(\frac{1}{\log_e(a)})=\log_a(2)+\log_a(a^2) + \log_a(\frac{1}{\log_e(a)})$$ $$\Rightarrow \log_a(2)+2 - \log_a(\log_e(a))$$

I do not think we can easily simply further at this point.

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  • $\begingroup$ Thank you for pointing out the error! :) $\endgroup$ – Sayli_ambure Aug 13 '18 at 8:55
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$\log_a(\log_a x)=\log_a(\frac{\ln x +i\cdot 2\pi m}{\ln a})=\frac{\ln\frac{\sqrt{(\ln x^n)^2+(2\pi m)^2}}{\ln a}+i(\arctan\frac{2\pi m}{\ln x^n}+2\pi k)}{\ln a}$, where $m,k \in \mathbb{Z}$. Applying that $x^n=e^{4a}$, and only taking the real solution ($m=k=0$):

$\log_a(\log_a x)=\frac{\ln\frac{2a^2}{\ln a}}{\ln a}=2+\frac{\ln 2}{\ln a}-\frac{\ln(\ln a)}{\ln a}$

If $m=4\cdot a \cdot l$, $l \in \mathbb{Z}-\{0\}$, and $k=-l$, we get additional solutions:

$\log_a(\log_a x)=\frac{\ln\frac{\sqrt{(2a^2)^2+(8\pi a \cdot l)^2}}{\ln a}}{\ln a}$

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  • $\begingroup$ Have you tried giving different values to $a$ to see if you always get the same numerical result? I don't think that the $a$ can be simplified. $\endgroup$ – TheAverageHijano Aug 13 '18 at 9:00

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