Why is $\ln(x^x)=x\ln(x)$ valid? I know that $\ln(x^k)=k\ln(x)$ for any constant $k$, but why is $\ln(x^x)=x\ln(x)$. The exponent $x$ is not constant.
 A: The general rule for logarithms is $\log(a^b)=b\log(a)$ for any real numbers $a$ and $b$ (as long as $a$ is positive).  In particular, it holds when $a=b=x$ (assuming, again, that $x$ is positive).
A: As $x$ is probably not an integer, $x^x$ is defined as :
$$x^x = e^{x\ln(x)}$$
Hence, taking the logarithm give you $\ln{x^x}=x\ln(x)$
A: For all $y > 0$, $\ln y$ is by definition the power that $e$ must be raised by to give the value $y$. So,
\begin{equation}
e^{\ln y} = y.
\end{equation}
In particular it is true for any $x$ such that $x^x > 0$. So substituting $x^x$ for $y$,
\begin{equation}
e^{\ln x^x} = x^x.
\end{equation}
But we also have from the exponent laws and the definition of $\ln$ that,
\begin{equation}
e^{x \ln x} = (e^{\ln x})^x = x^x
\end{equation}
Comparing the left and right hand sides of the above two equations it follows that $\ln x^x = x \ln x$.
A: For any positive real number $x$, $x\ln x = \ln (x^x)$.  This is a statement about very many "constants" $x$.  It means $3\ln 3 = \ln (3^3)$, $4\ln 4 = \ln (4^4)$, etc.  The only difference between this and $k\ln x = \ln (x^k)$ is the latter is allowed to have $k$ be different from $x$.
A: another way to think about it, for positive real $x,y$: 
$$
\ln y = \log_x y \cdot \ln x \tag{1}
$$
and, again by definition
$$
\log_x x^x= x
$$
A: For every $x>0$ we have: $$e^{x\ln x}=(e^{\ln x})^x=x^x=e^{\ln x^x}$$
Next to that function $x\mapsto e^x$ is injective, so we are allowed to conclude: $$x\ln x=\ln x^x$$
