Prove symmetry of natural logarithm

Question

Prove that $f(x)=\ln\sqrt{x^2+1}$ is symmetrical in $x=0$.

$\ln\sqrt{(x-a)^2+1}=\ln\sqrt{(x+a)^2+1}$

$\sqrt{(x-a)^2+1}=\sqrt{(x+a)^2+1}$

$(x-a)^2+1=(x+a)^2+1$

$x^2-2ax+a^2+1=x^2+2ax+a^2+1$

$-2ax=2ax$

$-x=x$?

I don't know what to do? Is this the proof or did I miss something?

Answer 1

If you can prove that your function is an even function that is

$$ f(-x)=f(x) $$

then it is symmetrical about the $y$-axis or the line $x=0$. In your case we have

$$ f(-x) = \ln(\sqrt{(-x)^2+1}) = \ln(\sqrt{(x)^2+1}) = f(x) $$

which proves the symmetry.

Answer 2

If a function is symmetrical about a line $x=a$ then it can be said that: $$f(x)=f(2a-x)\forall x\in\mathbb R$$ $$\ln\sqrt{x^2+1}=\ln\sqrt{(2a-x)^2+1}$$ By using property $\ln a^b=b\ln a$ $$\ln(x^2+1)=\ln((2a-x)^2+1)$$ Since "$\ln$" is a one-one function: $$x^2=(2a-x)^2$$ Expanding and cancelling: $$4a^2=4ax$$ It gives two solutions: $$a=0,x=a$$ Since we want symmetry for all real points we take the solution line $x=a(=0)$

For this question there was an easy way: You could just have shown that $$x^2=(-x)^2$$ and the rest follows after.