# 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

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.

#### Share

• ##### 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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