 aanpalendmw

2022-07-16

Prove symmetry of natural logarithm
Prove that $f\left(x\right)=\mathrm{ln}\sqrt{{x}^{2}+1}$ is symmetrical in $x=0$
$\mathrm{ln}\sqrt{\left(x-a{\right)}^{2}+1}=\mathrm{ln}\sqrt{\left(x+a{\right)}^{2}+1}$
$\sqrt{\left(x-a{\right)}^{2}+1}=\sqrt{\left(x+a{\right)}^{2}+1}$
$\left(x-a{\right)}^{2}+1=\left(x+a{\right)}^{2}+1$
${x}^{2}-2ax+{a}^{2}+1={x}^{2}+2ax+{a}^{2}+1$
$-2ax=2ax$
$-2ax=2ax$
$-x=x$?
I don't know what to do? Is this the proof or did I miss something? nuramaaji2000fh

If you can prove that your function is an even function that is
$f\left(-x\right)=f\left(x\right)$
then it is symmetrical about the $y$-axis or the line $x=0$. In your case we have
$f\left(-x\right)=\mathrm{ln}\left(\sqrt{\left(-x{\right)}^{2}+1}\right)=\mathrm{ln}\left(\sqrt{\left(x{\right)}^{2}+1}\right)=f\left(x\right)$
which proves the symmetry. Aleah Booth

If a function is symmetrical about a line $x=a$ then it can be said that:
$f\left(x\right)=f\left(2a-x\right)\mathrm{\forall }x\in \mathbb{R}$
$\mathrm{ln}\sqrt{{x}^{2}+1}=\mathrm{ln}\sqrt{\left(2a-x{\right)}^{2}+1}$
By using property $\mathrm{ln}{a}^{b}=b\mathrm{ln}a$
$\mathrm{ln}\left({x}^{2}+1\right)=\mathrm{ln}\left(\left(2a-x{\right)}^{2}+1\right)$
Since "$\mathrm{ln}$" is a one-one function:
${x}^{2}=\left(2a-x{\right)}^{2}$
Expanding and cancelling:
$4{a}^{2}=4ax$
It gives two solutions:
$a=0,x=a$
Since we want symmetry for all real points we take the solution line $x=a\left(=0\right)$
For this question there was an easy way: You could just have shown that
${x}^{2}=\left(-x{\right)}^{2}$
For this question there was an easy way: You could just have shown that
${x}^{2}=\left(-x{\right)}^{2}$
and the rest follows after.

Do you have a similar question?