Find out if $\log(3x + \sqrt{9x^2 + 1})$ is even or odd

Question

I'm trying to find if this function is odd or even :

$f(x) = \log(3x + \sqrt{9x^2 + 1})$

I know that it's an odd one because if I try $f(x) + f(-x) = 0$ it shows that it's odd. But I want to know how to figure problems like this? I used this method http://www.purplemath.com/modules/fcnnot3.htm ( plug -x to check ) but its not working here.

Answer 1

We have $$\begin{align}f(x)+f(-x)&=\ln(3x+\sqrt{9x^2+1})+\ln(-3x+\sqrt{9x^2+1})\\&=\ln\left((3x++\sqrt{9x^2+1})(-3x+\sqrt{9x^2+1})\right)=\ln 1=0.\end{align}$$

Answer 2

Hint \begin{align*}f(x)&=\log{(3x+\sqrt{9x^2+1})}=\log{\left(\dfrac{(\sqrt{9x^2+1}-3x)(\sqrt{9x^2+1}+3x)}{\sqrt{9x^2+1}-3x}\right)}\\ &=\log{\dfrac{1}{\sqrt{9x^2+1}-3x}}=-\log{(\sqrt{(-3x)^2+1}-3x)}\\ &=-f(-x) \end{align*}