Prove that $\ln(x)=\frac{1}{x}$ has a unique solution The question is like:

Prove that the equation $\ln(x)=\frac{1}{x}$ for $x>0$ has a unique solution and explain why.

When it asks about the "unique solution" I try to find the exact value. Is it possible to find it, how would I solve it? Thanks.
 A: You don't need to find the unique solution, just to prove that it must be there somewhere.
It is probably easiest to think of it as finding a zero of the function $f(x)=\ln(x) - \frac1x$.
We know the function has at least one zero, because $f(1)=-1$ and $f(e)=1-\frac1e$ which is positive because $e>1$. Since $f$ is clearly continuous on $\mathbb R_+$, the intermediate value theorem says that $f(x)$ must be zero somewhere between $1$ and $e$.
On the other hand, the derivative of $f(x)$ is positive everywhere for $x>0$. (This is easy to see, even without computing it -- the logarithm is stricly increasing while $\frac1x$ is strictly decreasing, so both terms contribute positively to $f'(x)$). So if the function had two different zeroes, that would be a contradiction with Rolle's theorem (which you may know better as a special case of the Mean Value Theorem).
[Or, a different way to express the same reasoning without calculus: Suppose the graphs for $\ln x$ and $\frac1x$ intersected at two different points. The line joining those points would need to have positive slope becuase the logarithm is strictly increasing, and at the same time have a negative slope because $\frac1x$ is strictly decreasing. But it is absurd for the same line to have both positive and negative slope].
Combining these, there is at least one solution, and there cannot be more than one solution -- so the only possiblity is that there is exactly one solution.

The actual solution is $x=e^{W(1)}\approx 1.76322$, where $W$ is Lambert's W function. It cannot be written in a nicer form than that, and you're almost certainly not expected to know about the $W$ function and find this expression that the level where this problem would be posed.
A: You can prove the uniqueness of the solution elementarily using derivatives. If $x>0$, then $f(x)=x\ln(x)$ is continuous everywhere, $f(x)$ is negative when $x\in(0,1)$, non-negative and strictly increasing when $x\in[1,+\infty)$ (because if $x\in[1,+\infty)$, then $f'(x)=\ln x+1\ge 1$) with $\lim_{x\to +\infty}f(x)=+\infty$. You can find the solution too:
You're solving $\frac{1}{x}e^{\frac{1}{x}}=1$ with $x>0$. Taking Lambert W function on both sides gives $\frac{1}{x}=W(1)$, so $x=\frac{1}{W(1)}$ is the unique solution. 
It can be re-written as $x=e^{W(1)}$, because $\frac{1}{W(1)}=e^{W(1)}\iff 1=W(1)e^{W(1)}$, which is true.
A: $f(x)=x\log x$ is an increasing function on $\left(\frac{1}{e},+\infty\right)$, since $f'(x)=1+\log x>0$ over such interval. $f(x)$ is negative over $\left(0,1\right)$ and $f(e)=e$, hence $f$ attains the value $1$ at some point of the interval $\left(1,e\right)$ and nowhere else.
