Question

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(x)=ln(x^x)`, but if we substitute some point we can easly see that it's not true, for example when `x=-2` fuction `g(x)` has no solutions but fuction `f(x)` does:
`g(-2)=-2ln(-2)=Undefi n ed`
`f(-2)=ln((-2)^(-2))=ln(1/4)~=-1.386`

Why is that?

Answer 1

`x>0`

Explanation:

This is a very good question. So, the function `h(x)=lnx` 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: `lnx^k` , for `x=-2 , k=2n>0 or k=2n+1` `---->` undefined