Ask Uncle Colin: Approximating an embedded exponential

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to and Uncle Colin will do what he can.

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}$1.

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


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

  1. It would be rude to use Leibniz notation here, don't you think? []


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