verifying property of inverse function of ln without using the fact that inverse is $e^x$

Question

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.

Answer 1

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 $\ln$.

Answer 2

You were very close. As a hint: we know that $$ \ln[\phi(x)\phi(y)]=\ln[\phi(x)]+\ln[\phi(y)] $$

Answer 3

Use the fact that:

$$\ln (\phi(x) \phi(y)\big) = \ln \phi(x) + \ln \phi(y) = x + y = \ln \phi(x + y)$$

Answer 4

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 $$ \phi(x+y)=\phi(\ln a+\ln b)=\phi(\ln ab)=ab=\phi(x)\phi(y). $$

Answer 5

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 $\psi(u)\psi(v)=\psi(u+v)$.