Solve $\log_a(\log_a x^n)$ if $n=a^2$ ; $x= e^2$ 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, $a^{a^y}= x^n$
 A: $\log_a(\log_a x)=\log_a(\frac{\ln x +i\cdot 2\pi m}{\ln a})=\frac{\ln\frac{\sqrt{(\ln x^n)^2+(2\pi m)^2}}{\ln a}+i(\arctan\frac{2\pi m}{\ln x^n}+2\pi k)}{\ln a}$, 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:
$\log_a(\log_a x)=\frac{\ln\frac{\sqrt{(2a^2)^2+(8\pi a \cdot l)^2}}{\ln a}}{\ln a}$
A: 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. 
A: 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.


*

*The unadorned $\log$ symbol denotes the natural logarithm.

*I skipped over the checks for existence; the computations make sense for $a>1$ as the final formula shows.

