One day in math class, we were working on homework. It involved calculus, and looking at $e^x$ and $e^x/2$. From a small section of the graph, it looked like if you chose any point on $e^x$, the normal (the perpendicular of the tangent) would be perpendicular to $e^x/2$ as well. For any point.
Which was completely stupid I found out later, because their rates of growth are different so their normals can't be perpendicular to each other everywhere.
So then I thought, hmmm... how about $e^x$ and $e^x-1$? Getting all the formulas and plugging in is easy. The simplification part is horrid, and that's what I need help with. Here's what I've done/tried:
$g(x) = e^x,\ g'(x)=e^x$ and $h(x) = e^x-1,\ h'(x)=e^x$.
The point slope of a line is $y-f(a)=f'(a)(x-a)$, where $a$ is the point of 'tangency' to a curve $f(x)$.
So now we have $y-e^a=e^a(x-a)$ and $y-e^b+1=e^b(x-b)$ for some a for $g(x)\ \mathrm{at}\ a$ and $h(x)\ \mathrm{at}\ b$, respectively. We can set these lines equal to each other (by solving for y) to find what a and b are:
$e^a(x-a+1)=e^b(x-b+1)-1$
Now can someone explain how to continue simplifying this to solve for $a$ or $b$?