Why $f (x) = ln (x^{x})$ $f(x)=ln(x^x)$ is not the same as $g (x) = x \cdot ln (x)$ $g(x)=x·ln(x)$ ?

Question

Why $f (x) = ln (x^{x})$ $f(x)=ln(x^x)$ is not the same as $g (x) = x \cdot ln (x)$ $g(x)=x·ln(x)$ ?

There's a property in logaritms that says:
$k \cdot {log}_{b} (a) = {log}_{b} (a^{k})$ $k·log_b(a)=log_b(a^k)$

so $g (x) = x \cdot ln (x)$ $g(x)=x·ln(x)$ should be the same as $f (x) = ln (x^{x})$ $f(x)=ln(x^x)$ , but if we substitute some point we can easly see that it's not true, for example when $x = - 2$ $x=-2$ fuction $g (x)$ $g(x)$ has no solutions but fuction $f (x)$ $f(x)$ does:
$g (- 2) = - 2 ln (- 2) = U n d e f i n e d$ $g(-2)=-2ln(-2)=Undefi n ed$
$f (- 2) = ln ({(- 2)}^{- 2}) = ln (\frac{1}{4}) ≅ - 1.386$ $f(-2)=ln((-2)^(-2))=ln(1/4)~=-1.386$

Why is that?

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Answer 1 · 2017-11-22T10:42:02

$x > 0$ $x>0$

Explanation:

This is a very good question. So, the function $h (x) = ln x$ $h(x)=lnx$ is defined when $x > 0$ $x>0$

Having this in mind you cannot replace the value for $x = - 2$ $x=-2$
in any of the 2 functions. Having put for $x = - 2$ $x=-2$ in $f$ $f$ in the first place is a faulty thinking because for that value its undefined in either case.

For $x > 0$ $x>0$ you can see that you get the same values

example: ${ln x}^{k}$ $lnx^k$ , for $x = - 2, k = 2 n > 0 or k = 2 n + 1$ $x=-2 , k=2n>0 or k=2n+1$ $- - - \to$ $---->$ undefined

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Why $f (x) = ln (x^{x})$ $f(x)=ln(x^x)$ is not the same as $g (x) = x \cdot ln (x)$ $g(x)=x·ln(x)$ ?

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