Question involving index laws; find the value of $x$ Here is the following question, I am currently working on:

$$x^x = 2^{2048}$$
Find the value of $x$.

I decided to start by letting $x = 2^n$, where $n \in \mathbb{R}$.
Then I used index laws:

*

*${2^n}^{(2^n)}=2^{2048}$

*$2^n = 2^{(2^{11-n})}$
Finally acquiring:

*

*$n=2^{11-n}$

*$n^{\frac{1}{11-n}} = 2$
That is where I have been able to get so far. Anyone has ideas to progress further from here?
 A: Forewarning: This solution is probably overkill and far from intended, especially considering the answer itself ($256$) is a power of $2$. This is simply the first approach that comes to mind when handling stuff involving $x^x$ or $xe^x$.

Overkill Solution Using Lambert $W$:
Abstract things a bit to the equation $x^x = a$; $2^{2048}$ is simply a red herring and we can handle that specific case later.
Then we know that
$$x^x = e^{\ln(x^x)} = e^{x \ln(x)} = a \implies x \ln(x) = \ln(a)$$
By a property of the Lambert $W$ function, applying it to both sides yields
$$W(x \ln(x)) = \ln(x) = W(\ln(a))$$
and so, by a further property,
$$x = e^{W(\ln(a))} = \frac{\ln(a)}{W(\ln(a))}$$
For $a=2^{2048}$ this becomes
$$x = \frac{2048 \ln(2)}{W(2048 \ln(2))}$$
$W(2048 \ln(2))$ is hard to work with, but we note the property we used before, $W(x \ln(x)) = \ln(x)$. If we can somehow rebalance $2048 \ln(2)$ so that the same thing is outside and inside the logarithm, we're golden. Luckily, since $2048$ is a power of $2$ ($2^{11}$) this is easy: just reallocate one power of $2$ at a time back inside:
$$\begin{align*}
2048 \ln(2) 
&= 1024 \ln(4) \\
&= 512 \ln(16) \\
&= 256 \ln(256)
\end{align*}$$
Hence:
$$x = \frac{2048 \ln(2)}{\ln(256)} = \frac{2048 \ln(2)}{8 \ln(2)} = 256$$
A: Hint
You properly arrived at the equation
$$n=2^{11-n}$$
Take logarithms
$$\log_2(n)=11-n \implies n+\log_2(n)=11\implies n <11$$  There is a very small range to explore.
