solve for $x$: $\frac{e^{x/80} - 1}{e^{x/4} - 1} = \epsilon$

Question

I am working on a scaling problem in biology. I came across the following equation:

$$\frac{e^{x/80} - 1}{e^{x/4} - 1} = \epsilon.$$

According to wolfram, left-hand side does not have a root, but the function seems to become arbitrarily small. I was wondering if there is some standard way that, given any $\epsilon > 0$, I can find some $x \in \mathbb{R}$ such that this equation is true.

Answer 1

If the question is about finding an approximation to the root, we can construct the Puiseux series solution for $u - 1 = \epsilon (u^{20} - 1)$. The first-order approximation comes from $u = \epsilon u^{20}$, giving $$x \sim -\frac {80} {19} \ln \epsilon, \quad \epsilon \to 0^+.$$

Answer 2

If you are interested in the limit, equaling the expression of the function to an arbitrarily small $\varepsilon$ isn't the way to do so. But, simply enough it can be found by using L'Hospital's rule, as stated in the comments by Ian :

$$\lim_{x \to \infty} \frac{e^{x/80} - 1}{e^{x/4} - 1} \stackrel{\frac{\infty}{\infty} D.L.H.}{=} \lim_{x \to \infty} \frac{\frac{1}{80}e^{x/80}}{\frac{1}{4}e^{x/4}} = 0 $$

Now, for showing that there exists an $x \in \mathbb R^+$ such that the equation is fullfilled, one may proceed by calculus theorems and manipulations of by numerical methods. The exact computation of it is rather complex by hand.

To answer to your comment, the $x$ for which the limit becomes zero "is" $\infty$. This is exactly what we observe in the graph, as it asymptotically reaches zero.