I am trying to prove that:
$x^{\frac{\ln(\ln(x))}{\ln(x)}} = \ln(x)$
My "solution":
$e^{\ln\left(x^{\frac{\ln(\ln(x))}{\ln(x)}}\right)} = e^{\frac{\ln(\ln(x))}{\ln(x)} \ln(x)} = e^{\ln(\ln(x))} = \ln(x)$
Is the first step valid, i.e is $\ln(x^{p(x)}) = p(x) \ln(x)$
How can I find out for myself?
$$x^{\frac{\ln(\ln(x))}{\ln(x)}}$$ $$=x^{\log_x(\ln(x))}$$ $$=\ln(x)^{\log_x(x)}$$ $$=\ln(x)$$
Given $\displaystyle x^{\frac{\ln(\ln x)}{\ln x}}\;,$ Now Let $\ln x= y\Rightarrow x=e^y$
So expression convert into $\displaystyle e^{\frac{y\ln y}{y}} = e^{\ln y} = y = \ln x$
Note that $\ln(x)^{p(x)} = p(x)\cdot \ln x\;,$ Where $x>0$
Answering the question. For $a>0$ we have by definition of logarithm that $$ a^b=e^{\ln(a^b)}. $$ and on the other hand $$ a^b=(e^{\ln a})^b=e^{b\ln a} $$ Comparing two expressions and noting that the exponential function is injective we get $$ \ln(a^b)=b\ln a. $$ Now apply it to $a=x$ and $b=p(x)$.
Let $y$ = $ln(x)$ So $x^{ln(y)/y}$ = $y$ Take ln both sides and simplify using the property: $ln(x^{ln(y)}/y)$ = $ln(y);ln(x)/y$. so $ln(y)ln(x)/y$ = $ln(y)$ We get $ln(y)y/y$ = $ln(y)$
