I know that the inverse of $f(x)={10}^x$ is $f^{-1}\left(x\right)={\mathrm{log} x\ }$, geometrically this implies reflecting the original function across the line y=x. I am also aware that this can be demonstrated algebraically by finding the inverse.
However, I'm curious as to how we can prove that the inverse $f(x)={10}^x$ is, in fact, $f^{-1}\left(x\right)={\mathrm{log} x\ }$. Is there a method that can be used to demonstrate this? Any suggestions, methods, or explanations would be greatly appreciated, and I apologize if this is a silly question.
From $10^{x+y}=10^x10^y$ and $10^y>1$ for $y>0$ it is straightforward that $f$ is a strictly increasing function, and therefore has an inverse $f^{-1}$ which is well defined (on $\mathbb R^{+*}$).
As said in the comments depending on your definitions, there might be nothing to prove at all, but let's adopt the morphism point of view consisting in proving that if $f$ is an exponential (maps addition to multiplication) then $f^{-1}$ is a logarithm (maps multiplication to addition).
It suffice then to end the argument by identifying $10$ as the characteristic element of both functions ($10^1=10$ and $\log(10)=1$).
Since the inverse is well defined, let call $a=f^{-1}(x)$ and $b=f^{-1}(y)$
$f^{-1}(xy)=f^{-1}(f(a)f(b))=f^{-1}(f(a+b))=a+b=f^{-1}(x)+f^{-1}(y)$ as desired.