I have the graphs of $y = F(x) = e^x$, $y = G(x) = ln (x)$, and $L : y = x$ drawn.
Of course, $y = F(x)$ and $y = G(x)$ are inverses to each other and therefore they are symmetric about $L$.
Let $P(p, q)$ be a point on $y = G(x)$. Through $P$, I drop a perpendicular to $L$, cutting it at $M$, and $y = F(x)$ at $H(h, k)$ such that $M$ is the midpoint of $HP$.
I know that $p = k$ and $h = q$ is a fact. However, I want to prove that via the equations:-
(1) $k = e^h$
(2) $q = log_e (p)$
(3) $h + p = k + q$
Using (1) and (2) to eliminate p and k from (3), I end up with $h + e^q = e^h + q$. A trivial solution is $h = q$.
My questions are (1) is that solution unique? And (2) how can I prove that in a more vigorous manner?
PS Please ignore the circle.
The equation $h+e^q=e^h+q$ is equivalent to $e^q-q=e^h-h$, or equivalently $f(q)=f(h)$ where $f(x) = e^x-x$. And it is easy to check that $f$ is increasing for $x>0$ (since $f'(x) = e^x-1>0$), hence one-to-one—so that $f(q)=f(h)$ implies $q=h$. (Indeed, if you allow $h$ or $q$ to be negative then the solution is no longer unique.)