Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd

Question

I know that a function is odd when $$f(-x) = -f(x)$$ Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$

Then the function is odd!

I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.

However, I would want to prove directly that $$f(-x) = -f(x)$$ In other words, I want to solve $$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$ and to come at the end to this: $$-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$$

This was my approach: $$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$ $$\ln\left(-x+\sqrt{x^2 + 4}\right) - \ln2$$ $$\ln\left(\frac{-x+\sqrt{x^2 + 4}}{2}\right)$$ $$\ln\left(\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)^{-1}\right)$$ $$-\ln\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)$$

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Here I got stuck. I want to get to $-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$ but if I use $\ln\left(\frac ab\right) = \ln a - \ln b$ then I will get back to $f(-x)$ and not to $-f(x)$.

Any help?

Answer 1

Hint: Rationalise the denominator of the fraction inside the ln.

I.e. multiply top and bottom by $x+\sqrt{x^2+4}$.

Answer 2

Hint:

$$\frac{2}{\sqrt{x^2+4}-x}=\frac{2(\sqrt{x^2+4}+x)}4$$

Answer 3

$$\left(x+\sqrt{x^2 + 4}\right)\left(-x+\sqrt{x^2 + 4}\right)=4=2\cdot2$$ so that, taking the logarithm,

$$f(x)+f(-x)=0.$$

Answer 4

A faster approach. If we set $f(x)=\log\left(\frac{x+\sqrt{x^2+4}}{2}\right)$ and $x=2\sinh\theta$ we have $$f(2\sinh\theta) = \log\left(\frac{2\sinh\theta+2\cosh\theta}{2}\right) = \log(e^\theta) = \theta $$ and since $\theta\mapsto 2\sinh\theta$ is a bijective odd function from $\mathbb{R}$ to $\mathbb{R}$, $$ f(-x) = -f(x) $$ readily follows. That also implies $f(x) = \text{arcsinh}\left(\frac{x}{2}\right)$.

Answer 5

It simply means \begin{align} &\ln(-x+\sqrt{x^2+4})-\ln2=-\ln(x+\sqrt{x^2+4})+\ln2\\ \iff &\ln(x+\sqrt{x^2+4})+\ln(-x+\sqrt{x^2+4})=2\ln 2\\ \iff&\ln[(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)]=\ln4\\ \iff &(\sqrt{x^2+4}+x)(\sqrt{x^2+4}-x)=4. \end{align}