Ask Uncle Colin: A Mess of Logs

Dear Uncle Colin,

I have to show that \(-\frac{x}{2} = \ln (\sqrt{1+e^x} - \sqrt{e^x }) + \ln (\sqrt{1+e^{-x}} + 1)\). I can't get it anywhere near the right form!

- Proof Of It Not Coming - Any Reasonable Explanation?

Hi, POINCARE, and thanks for your message!

That's a bit of a mess - but with some careful book-keeping, it comes out ok!

Let's start by combining the logarithms to get \(\ln\br{\br{ \sqrt{1+e^x} - \sqrt{e^x }}\br{\sqrt{1+e^{-x}}+ 1}}\).

Now we have something inside the log-bracket we can expand to get: \(\sqrt{ \br{1+e^{x}} \br{1+e^{-x}} } + \sqrt{1+e^x} - \sqrt{ \br{e^x} \br{1+e^{-x}}} -\sqrt{e^x}\).

That can be expanded still further to get \(\sqrt{2 + e^x + e^{-x}} + \sqrt{1+e^x} - \sqrt{e^x+1} - \sqrt{e^x}\).

Those two middle terms are equal and opposite, so the log-bracket is now \(\sqrt{e^x + 2 + e^{-x}} - \sqrt{e^x}\).

I can rewrite that first square root by taking a 'factor' of \(\sqrt{e^x}\) out, to make the whole bracket \(\sqrt{\br{e^x}\br{1 + 2e^{-x} + e^{-2x}}} - \sqrt{e^x}\). Why do such a thing? It makes the second bracket in the first square root is \(\br{1+e^{-x}}^2\) - so the whole bracket is \(\sqrt{e^x} \br{1 + e^{-x}} - \sqrt{e^x}\).

Aha! That now becomes \(\sqrt{e^x} + e^{-\frac{1}{2}x} - \sqrt{e^x}\), or \(e^{-\frac{1}{2}x}\) - and it's all over bar the logging!

The entire right-hand side is \(\ln\br{ e^{-\frac{1}{2}x}}\), which is \(-\frac{1}{2}x\), as required. \(\blacksquare\)

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.


2 comments on “Ask Uncle Colin: A Mess of Logs

  • Mark Thornber

    If you take a factor of sqrt(e^x) from the first term the two brackets become a difference of two squares and it drops out in a couple of nice lines!

    • Colin

      That’s very neat! Thanks for sharing it 🙂

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