Unravelling the Mystery of logn Identities

[quasar987]

I don't get where these two identities come from:

$$(logn)^{logn} = n^{log(logn)}$$

and

$$(logn)^{log(logn)} = e^{(log(logn))^2}$$

 

[Muzza]

I can only think of this roundabout way to show the first:

log(n)^log(n) = x
log(log(n)^log(n)) = log(x)
log(n) * log(log(n)) = log(x)
10^(log(n) * log(log(n))) = 10^log(x)
(10^log(n))^log(log(n)) = x
n^log(log(n)) = x

So log(n)^log(n) = n^log(log(n)).

 

[NateTG]

If you're using the natural logarithm, its usually better to use $\ln$ or $\log_e$ rather than $\log$ which can be interpreted in other ways (for example as $\log_{10}$ or as a log with unspecified base) depending on context.

The identities are similar:

$$n^{\ln(\ln(n))}=\left(e^{\ln(n)}\right)^{\ln(\ln(n))}=e^{\ln(n) \times \ln(\ln(n))}=e^{\ln(\ln(n)) \times \ln(n)}=\left(e^{\ln(\ln(n))}\right)^{\ln(n)}=\left(\ln(n)\right)^{\ln(n)}$$

$$e^{\left(\ln(\ln(n))\right)^2}=e^{\ln(\ln(n)) \times \ln(\ln(n))}=\left(e ^{\ln(\ln(n))}\right)^{\ln(\ln(n))}=\left(\ln(n)\right)^{\ln(\ln(n))}$$

 

[quasar987]

Oh, I see! Well thanks a bunch !