Ask Uncle Colin: Finding a curve with an asymptote

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,

I'm looking for the equation of a curve that goes through the points $\left(10, \frac{1}{64}\right)$ and $\left(100, \frac{1}{32} \right)$, as -- as $x$ gets large -- approaches 1. How do I go about it?

Always Silence Your Mobile Phone Typing Out Tricky Equations

Good advice, ASYMPTOTE! And a great question.

There are infinitely many curves that fit the bill, but the one that jumps out at me is of the form $y = 1 - Ae^{-kx}$. From here, it's 'simply' a case of working out $A$ and $k$.

We have two equations:

$\frac{1}{64} = 1 - Ae^{-10k}$

$\frac{1}{32} = 1 - Ae^{-100k}$

A little rearrangement to make them (slightly) nicer gives:

$\frac{64}{63} = Ae^{10k}$

$\frac{32}{31} = Ae^{100k}$

Dividing these gives

$\frac{32 \times 63}{31 \times 64} = e^{90k}$

With a quick cancel and a logarithm:

$\frac{1}{90}\ln\left(\frac{63}{62}\right) = k$.

We can substitute this back into one of the earlier equations to find $A$:

$\frac{64}{63} = A e^{-\frac{1}{9} \ln\left(\frac{63}{62}\right)}$

Simplify the exponential:

$\frac{64}{63} = A \left(\frac{63}{62}\right)^{-\frac 19}$

And Bob's your uncle:

$\frac{64}{63} \left(\frac{63}{62}\right)^{\frac 19}= A $

Here's what it looks like (click to see details):

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.


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