Proving the Property of Logarithms: Examples with Exponents

[Nano-Passion]

Edit: I answered my own question, I guess this thread serves no purpose so mods, you can delete this.

y = x^2
ln y = ln x^2
ln y = 2 ln x

Can we do the same thing with:

$$y = x^{2/x}$$
$$ln y = ln x^{2/x}$$
$$ln y = \frac{2}{x} x$$

Would that be correct? I just want to make sure because I used this technique for differentiation.

 

[iRaid]

Doesn't: $$ln y = \frac{2}{x} x$$ just make it 2 :p

But from what I remember about logs, yes you can do that..

 

[gb7nash]

The OP might have already figured this out, but $ln x^{2/x} = \frac{2 ln(x)}{x}$, not $\frac{2}{x}x$. Typo maybe?

 

[Nano-Passion]

The OP might have already figured this out, but $ln x^{2/x} = \frac{2 ln(x)}{x}$, not $\frac{2}{x}x$. Typo maybe?

Yep, it was a typo.