Why does $\ln(x) = \epsilon x$ have 2 solutions? I was working on a problem involving perturbation methods and it asked me to sketch the graph  of $\ln(x) = \epsilon x$ and explain why it must have 2 solutions. Clearly there is a solution near $x=1$ which depends on the value of $\epsilon$, but I fail to see why there must be a solution near $x \rightarrow \infty$. It was my understanding that $\ln x$ has no horizontal asymptote and continues to grow indefinitely, where for really small values of $\epsilon, \epsilon x$ should grow incredibly slowly. How can I 'see' that there are two solutions?
Thanks!

Right after the first solution, you have that $\ln x>\epsilon x \Rightarrow \ln x-\epsilon x>0$. However, 
$$\lim_{x\to\infty}(\ln x-\epsilon x)=\lim_{x\to\infty}\ln\frac{x}{e^{\epsilon x}}\to-\infty
$$
since $\frac{x}{e^{\epsilon x}}\to 0$. (Assuming $\epsilon>0$)
Hence, $\ln x-\epsilon x$, must cross the $x$-axis a second time which gives your second solution.
