Inverse of logarithmic function I have this problem:
$f:(0,\infty)\rightarrow \mathbb R$,
$$f(x)=x+1+\ln x $$
$g$ is the reverse of $f$.
I tried to solve the problem like this
$y=f(x) ; x+\ln x=y-1$ and I got stuck 
The problem is: 
$g(2) = $?
Also g'(2) =?
Anyone got any idea?
 A: You cannot solve for $x$ as a function of $y$, as you are attempting.  There is no formula for $f^{-1}$ in terms of common functions like $e^x$, etc.  
The question amounts to solving for $x$ in the equation $2 = x + 1 + \ln x$. 
The simplest way to do this is to plug in some whole numbers in for $x$ until you find one that outputs $2$. This happens to succeed here, but only because the problem was constructed with this in mind. 
If the solution here were not as easy to guess, you could also graph $f$ and try to estimate visually the input that gives the output $2$.
If you are familiar with numerical methods like Newton's method, you can also obtain a solution to $2 = x + 1 + \ln x$ by guessing an initial value and then computing successively better approximations until you reach a desired amount of precision.
A: The function is invertible, because
$$
f'(x)=1+\frac{1}{x}>0
$$
so $f$ is strictly increasing. Since
$$
\lim_{x\to0}f(x)=-\infty,
\qquad
\lim_{x\to\infty}f(x)=\infty
$$
the range of $f$ is $\mathbb{R}$, so $g$ is defined over $\mathbb{R}$ and has $(0,\infty)$ as its range.
Saying $g(2)=x$ is the same as saying $f(x)=2$, which means
$$
x+1+\ln x=2
$$
that is,
$$
x+\ln x=1
$$
We know that a solution exists and is unique; guessing it should be easy.
There is no “explicit” expression for $g$ in terms of “elementary” functions (polynomials, radicals, exponential, logarithm, trigonometric functions) and rational expressions thereof.
A: You are supposed to solve $x+1+\ln x = 2$. Even though you cannot solve this generally (unless you use something like Lambert W function), for $f(x)=2$ you can actually guess the correct solution quite easily (try some simple values of $x$ that simplify $\ln x$). 
Also even if you guess the solution, you still need to do more work by showing that is the only solution. You can show that by showing that the function is monotonic. But that should already been shown since you are taking inverse if the function, which does not make sense if it is not monotonic.
