Vtyjkyuk

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:

$Rightarrowlog_a(log_a (e^2)^a^2)$

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

$Rightarrowlog_a(2*(a^2)cdotlog_a (e))$ ... By log property $log x^a= a log x$

$Rightarrowlog_a(2cdot(a^2)cdot(frac1log_e(a)))$ ... By $log$ property

$Rightarrowlog_a(2cdot(a^2)) + log_a(frac1log_e(a))$ ...By $log$ property

$Rightarrow 2cdotlog_a(a^2)+log(frac1log_e(a))$

$Rightarrow 4 + log(frac1log_e(a))$

Couldn't solve beyond that

Trial 2:

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

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

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

First notice that your expression is
$$
log_a(nlog_a x)=log_a n+log_alog_ax
$$

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

You could observe that
$$
log_a2=fraclog 2log a
$$
and
$$
log_ae=frac1log a
$$
so
$$
log_alog_ae=log_afrac1log a=-log_alog a=-fracloglog alog a
$$
so you finally get
$$
2+fraclog2log a-fracloglog alog a
$$

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.

You were doing quite well but made a small error. Notice that you say:
$$log_a(2cdot(a^2)) + log_a(frac1log_e(a)) Rightarrow 2cdotlog_a(a^2)+log(frac1log_e(a))$$

But if we have logarithms we can't simply take the multiplication outside. Remember that
$log(acdot b)=log(a)+log(b)$. Thus we get:
$$log_a(2cdot(a^2)) + log_a(frac1log_e(a))=log_a(2)+log_a(a^2) + log_a(frac1log_e(a))$$
$$Rightarrow log_a(2)+2 - log_a(log_e(a))$$

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

$log_a(log_a x)=log_a(fracln x +icdot 2pi mln a)=fraclnfracsqrt(ln x^n)^2+(2pi m)^2ln a+i(arctanfrac2pi mln x^n+2pi k)ln a$, where $m,k in mathbbZ$. Applying that $x^n=e^4a$, and only taking the real solution ($m=k=0$):

$log_a(log_a x)=fraclnfrac2a^2ln aln a=2+fracln 2ln a-fracln(ln a)ln a$

If $m=4cdot a cdot l$, $l in mathbbZ-0$, and $k=-l$, we get additional solutions:

$log_a(log_a x)=fraclnfracsqrt(2a^2)^2+(8pi a cdot l)^2ln aln a$