verifying property of inverse function of ln without using the fact that inverse is $e^x$ Let $\phi(x) = \ln^{-1}(x)$. We want to prove that:
$\phi(x + y) = \phi(x)\phi(y)$ for all $x, y \in \mathbb{R}$. We cannot use the fact that $\phi(x) = e^x$, just the fact that $\ln$ has an inverse (that is $\phi$) (which I was asked to prove in an earlier problem).
I thought about starting from the identity $\ln(xy) = \ln(x) + \ln(y)$ and taking the inverse of both sides, but that left me with $xy = \phi(\ln(x) + \ln(y))$, and I didn't know where to go from there.
 A: Since you know that $\ln(xy) = \ln x + \ln y$, just notice that
$$\ln \big(\phi(x) \phi(y)\big) = \ln \phi(x) + \ln \phi(y)  = x + y = \ln \phi(x + y)$$
where we have used that $\phi$ and $\ln$ are inverse functions. Now conclude that $\phi(x) \phi(y) = \phi(x + y)$ due to the injectivity of $\ln$.
A: You were very close.  As a hint: we know that
$$
\ln[\phi(x)\phi(y)]=\ln[\phi(x)]+\ln[\phi(y)]
$$
A: Use the fact that:
$$\ln (\phi(x) \phi(y)\big) = \ln \phi(x) + \ln \phi(y)  = x + y = \ln \phi(x + y)$$
A: Since you are assuming that $\ln$ is invertible, it is surjective. So given $x,y\in\mathbb R$, there exist $a,b\in\mathbb R_+$ with $x=\ln a$, $y\ln b$. Then
$$
\phi(x+y)=\phi(\ln a+\ln b)=\phi(\ln ab)=ab=\phi(x)\phi(y). 
$$
A: First, you need a definition of $\ln$.  Without that, of course you can prove nothing about $\ln$.  I like the defnintion $\ln x = \int_1^x \frac{dt}{t}$.  Then we can prove
$$
\int_1^{xy}\frac{dt}{t} =
\int_1^{x}\frac{dt}{t}+\int_x^{xy}\frac{dt}{t}
= \int_1^{x}\frac{dt}{t} + \int_1^{y}\frac{dt}{t}
$$
using a simple substitution in the second integral.  
added
OK, second part, duplicating some other answers.  Suppose $\phi$ is a bijection (from set $A$ to set $B$) satisfying $\phi(xy)=\phi(x)+\phi(y)$.  Let $\psi$ be its inverse (from set $B$ to set $A$).  We want to prove $\psi(u+v) = \psi(u)\psi(v)$.  Now
$$
\phi\big(\psi(u)\psi(v)\big)\big)
= \phi\big(\psi(u)\big) + \phi\big(\psi(v)\big)\big)
= u + v = \phi\big(\psi(u+v)\big)
$$
But $\phi$ is injective, so this implies $\psi(u)\psi(v)=\psi(u+v)$.
