Ask Uncle Colin: Disguised quadratics

Dear Uncle Colin,

I have to solve $5^x = 6 - 5^{1-x}$ - I understand it’s going to end up as a quadratic, but I can’t see how!

- Explain, Uncle Colin, Like I Demand!

Hi, EUCLID and thanks for your message!

The key thing here is to spot that you can split up the last term as either $5 \times 5^{-x}$ or (equivalently) as $\frac{5}{5^x}$. If you then let $y = 5^x$, you have:

$y = 6 - \frac{5}{y}$

Multiplying through by $y$ and rearranging gives you $y^2 - 6y + 5=0$.

That factorises nicely as $(y-5)(y-1)=0$, so $y=5$ or $y=1$.

However, we made $y$ up! We need to solve finally $5^x = 5$ or $5^x=1$, giving us $x=1$ or $x=0$.

Hope that helps!

- Uncle Colin

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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2 comments on “Ask Uncle Colin: Disguised quadratics

  • Barney Maunder-Taylor

    “Disguised quadratic” or “Hidden quadratic”? Both seem to be used in equal measures. A student of mine a few years back used to call them “weird quadratics”,which I love!

  • Colin

    Disguised and hidden are both terribly dull. What about ‘masquerade’?

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