Solutions of $y = kx$ and $y = 3\ln x + 3$ If $y = kx$ and $y = 3\ln x +3$ have exactly 2 solutions, then what is the range of $k$ ?  
By plotting the curves we can see that the range will be between $0$ and the value of $k$ at which $y = kx$ will be tangent to  $y = 3\ln x +3$. How do I find the upper limit of $k$ at which it will be a tangent?
 A: We want two roots for $f(x):=kx-3\ln x -3$, which $\to +\infty$ as $x\to 0^+$ and as $x\to +\infty$. Since $f'(x)=k-\frac{3}{x}$ vanishes only at $x=\frac{3}{k}$ where $f(\frac{3}{k})=3\ln\frac{k}{3}$ and $f''(\frac{3}{k}) =\frac{k^2}{3}$, there are two roots iff the global minimum $3\ln\frac{k}{3}$ is negative, i.e. iff $k\in (0,\,3)$.
A: You have correctly noted that $k=0$ is a critical value and correctly identified that the tangent is the critical point to find.
This requires us to solve $k=\frac{3}{x}$ and $kx=3\ln x +3$. This will give you $k=3$ and $x=1$ as follows:
$$\frac{3}{x}\times x=3\ln x + 3$$
$$3=3\ln x +3$$
$$0=3\ln x$$
$$0=\ln x$$
$$x=1$$
Hence $k=3$.
Note that there is only one solution. That means the solution space is divided into two at $k=3$. There will be a certain number of solutions for all $k>3$ and a certain number of solutions for all $0<k<3$.
Picking values of $k$ and plotting simple graphs shows that for $0<k<3$ we get two solutions and for $k>3$ we get zero solutions.
So the answer is $0<k<3$.
