# Prove that there are exactly two positive real numbers $x$ such that $e^x=3x$

I have a sketch:

How can I prove that there are exactly two positive real numbers $x$ such that $e^x=3x$ using the Mean Value Theorem/Rolle's Theorem? I'm not sure how to do this question as I can't see the significance of the gradient here... unless I'm missing something really obvious :S

If there are more than two solutions, then $f(x)=e^x-3x$ has at least three zeros. How many zeros must $f'(x)$ have?
To show there are at least two solutions, you might use the Intermediate Value Theorem.

• I've drawn a graph: puu.sh/cRC8l/75445a21c7.png If f(x) has at least 3 zeroes then f'(x) should have at least 2 zeroes..? Bit confused sorry, what's the difference between the IVT and MVT? – Jim Nov 15 '14 at 12:13
• Calculate $f'(x)$ - how many zeros does it actually have? IVT means that if $f(x)$ is positive at one place and negative at another, and it is continuous, then $f(x)=0$ somewhere in between. – Empy2 Nov 15 '14 at 12:40

Set $f(x)=e^x-3x$. Then $f'(x)=e^x-3$, so $f$ is decreasing in the interval $(-\infty,\log 3]$ and increasing in the interval $[\log 3,\infty)$.

Therefore $f$ can have at most two zeroes. (Why?)

Now $\lim_{x\to-\infty}f(x)=\infty$ and $\lim_{x\to\infty}f(x)=\infty$; moreover $$f(\log3)=3-3\log3=3(1-\log3)<0.\ \text{(Why?)}$$ Thus a zero exists in both the intervals above. (Why?)

consider f1(x)=e^x/x and the line f2(x)=3.

For f1(x), when x->0, f1->Inf and when x->Inf, f1->Inf

Also by differentating f1 you can find x=1, is a local minima. At x=1, f1=e<3.

So there lies a point between x=0 and x=1, and another between x=1 and x=Inf, that satisfies f1=f2. In other words e^x-3x=0.