So the question is as follows:
Let $f(x) = \int_{0}^x \frac{x}{2\sqrt{t}}dt$. Suppose $f(f(f(...f(f(a))...)))$ (done $2013$ times) $= 2^{2013}$. Find the real-valued solution of $a$
Now, for my solution:
Integrating $f(x)$, we get that:
$$f(x) = \left[x\sqrt{t}\right]_{0}^x = x\sqrt{x} = x^{3/2}$$
Now, by finding a pattern by continuously performing $f(f(a))$ $2013$ times, we get that:
$$x^{\left(3/2\right)^{2013}} = 2^{2013}$$
However, I was unable to determine an answer. In their solution it stated that:
$$x^{\left(3/2\right)^{2013}} = x^{\left(3/2\right)*{2013}}$$
It didn't say (which is the process I did):
$$x^{\left(3/2\right)^{2013}} = x^{\left(3^{2013}/2^{2013}\right)}$$
Now, which form is correct? Does the parenthesis change the way to evaluate the exponents?
Any help would be appreciated.