# Ask Uncle Colin: Approximating an embedded exponential

Dear Uncle Colin,

Help! My calculator is broken and I need to solve - or at least approximate - $0.1 = \frac{x}{e^x - 1}$! How would you do it?

-- Every $x$ Produces Outrageous Numbers, Exploring New Techniques

Hi, ExPONENT, and thanks for your message!

That’s a bit of a beast, and you’ll not get an exact solution using elementary functions. However, we can approximate it with a bit of rearrangement:

$e^x = 10x + 1$

Without a calculator (the Mathematical Ninja has been at mine, too), I’d say that when $x=4$, the left-hand side is about 55 and the right-hand side 41; when $x=3$, the left-hand side is about 20 and the right-hand side 31 - and as anyone with a passing knowledge of GCSE trial and improvement knows, that means there’s a solution between the two.

If $x=3.5$, the left hand side is about 32 (because $e^{0.7} \approx 2$) and the right-hand side is 36 - so we’re getting close.

At this point, I’d switch to a different rearrangement, $f(x) = x - \ln(10x+1)=0$, and apply Newton-Raphson. We’ll need the derivative, which is $f’(x) = 1 - \frac{10}{10x+1}$ ((It would be rude to use Leibniz notation here, don’t you think?)).

Starting with $x=3.5$, our next guess would be $3.5 - \frac{f(3.5)}{f’(3.5)}$. We can find $f(3.5) = 3.5-\ln(36) = 3.5 - 2\left(\ln(2)+ \ln(3)\right) \approx 3.5 - 2\left(0.693 + 1.099\right) = -0.084$ and $f’(3.5) = 1 - \frac{10}{36} = \frac{13}{18}$.

So, our next guess is $3.5 + 0.084\times \frac{18}{13}$. That works out to 3.616 or so, which is pretty close to the correct answer of 3.615.

I’m sure the Mathematical Ninja could get closer, but I don’t suppose he’d want to be bothered. Hope that helps!

- Uncle Colin