Showing that $e^x=3x$ has exactly 2 solutions in $\mathbb{R}$ Show, that $e^x=3x$ has exactly 2 solutions in $\mathbb{R}$
Let $f(x) = e^x-3x$. $f$ is oviously  continuous.
Consider $f(0)= e^0-3\cdot 0 = 1 > 0,$ $f(1)= e^1-3\cdot 1 = e-3 < 0$, $f(10) = e^10-3 \cdot 10 >0$
$\Rightarrow$ $f$ has, according to the Intermediate value theorem,  at leat $2$ solutions in $\mathbb{R}$. But how can i show that $f$ has exactly $2$ solutions?

Usually in these types of problems one uses the Mean Value Theorem (or more specifically Rolle's Theorem) to show that a function can have at most some number of zeroes.  
In your case you want to show $f(x) = e^x-3x$ has at most 2 zeroes.  So pretend you have 3, say at $x=a < b < c$.  Use Rolle's theorem to show that $f'$ must have a zero between $a$ and $b$, and a zero between $b$ and $c$.  So $f'$ has at least 2 zeroes.  But using Rolle's theorem again this means $f''$ has a zero.  Now show this is a contradiction...
