Determining whether equation $\ln x=\frac{1}{3}x$ has at least 3 solutions I'm trying to say whether
$$\ln x=\frac{1}{3}x$$
Has at least 2 solutions, and after that the problem asks us to tell if the third solution exists...
So I managed to find 1 solution: $f(x)=\ln x - \frac{1}{3} x$, $f(e)>0, f(1)=-\frac{1}{3} < 0$, so by the mean value theorem we have $f(y)=0$ for some $y\in [1,e]$.
What to do about the other two?
UPDATE: I haven't covered derivatives yet in my course...
 A: Hint: Find first derivative and second derivative. Analyze the signs. What do they tell you?
A: If you look at the graph of the given equation, the plot of ln(x) bends towards X-axis after the second point of intersection while the plot of $\frac13x$ continues straight above. So, there won't be a third solution of the equation.
A: Change to $x = e^{x/3}$, and then to
$e^x = 3 x$ 
sketch the graph of $e^x$ and of $3 x$. Observe that:
$e^x$ is always increasing in $x$, and always curving upwards. 
$3 x$ (a straight line) is always increasing in $x$, does not curve.
So, starting from small $x$, if your straight line crosses $e^x$, then it must cross back again (because the $e^x$ keeps on curving upwards).
You cannot get more than 2 crossings because $e^x$ only curves upwards (you would need e.g. an "s" shape to get 3 crossings of a straight line).
The only other possibility would be that the 2 lines touch but do not cross, giving 1 solution.
Let's see now: 
at $x=0$, $e^x=1$, $3 x = 0$   so $e^x > 3 x$  
at $x=1$, $e^x = 2.71...$, $3 x = 3$  so $e^x < 3 x$  
and for $x \to \infty$, $e^x > 3 x$  
So we get exactly 2 crossings.
