# Why f(x)=ln(x^x) is not the same as g(x)=x·ln(x)?

## There's a property in logaritms that says: k·log_b(a)=log_b(a^k) so g(x)=x·ln(x) should be the same as $f \left(x\right) = \ln \left({x}^{x}\right)$, but if we substitute some point we can easly see that it's not true, for example when $x = - 2$ fuction $g \left(x\right)$ has no solutions but fuction $f \left(x\right)$ does: $g \left(- 2\right) = - 2 \ln \left(- 2\right) = U n \mathrm{de} f i n e d$ $f \left(- 2\right) = \ln \left({\left(- 2\right)}^{- 2}\right) = \ln \left(\frac{1}{4}\right) \cong - 1.386$ Why is that?

Nov 22, 2017

$x > 0$

#### Explanation:

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

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

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

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