Seeking an algebraic solution of this problem If $a$ is the solution of the equation $$xe^x=e^2$$
If $b$ is the solution of the equation $$x\ln x=e^2$$
Then what is $ab$?
I know the answer is $e^2$, using the symmetrical property of $\frac{e^2}{x}$,  $e^x$, and $\ln x$
But how can I do it algebraically?
 A: From your first equation, $a e^a = e^2$.  If you define $b = e^a$, then $a = \ln(b)$ so $b \ln(b) = e^a a = e^2$.  And then $ab = a e^a = e^2$.
A: If you're familiar with the Lambert $W$ function, it can make this exercise almost trivial. (Granted, it's far from the most elementary solution, arguably even overkill, but once you become familiar with the function, this exercise just begs for its use.) You can read up more on the function here, but it is essentially the inverse function to $f(x)=xe^x$. That is to say, if you know $xe^x = c$, then $x=W(c)$ (where $W$ is the Lambert $W$ function), and $W(xe^x) = W(x)e^{W(x)} = x$.
Another easily-verified property of the $W$ function is $W(x \ln x) = \ln x$. We make use of this later.
Thus, using the $W$ function on both equations, we see
$$xe^x = e^2 \overset{W(x)}{\implies} x = W(e^2)$$
$$x \ln x = e^2 \overset{W(x)}{\implies} \ln x = W(e^2) \implies x = e^{W(e^2)}$$
By these, if $a$ is the solution in the first equation and $b$ in the second, then $ab$ is given by
$$ab = W(e^2) e^{W(e^2)}$$
From the definition as stated earlier, $W(x)e^{W(x)} = x$. Thus, $ab=e^2$.
A: HINT:
$a$ is the (unique) solution of the equation 
$$F(a,\phi(a))= c$$
while $b$ is the (unique) solution of the equation
$$F( \phi^{-1}(b),b) = c$$
Then $b = \phi(a)$, $a=\phi^{-1}(b)$. Therefore
$$F(a,b) = c$$
