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?

1 Answer
Nov 22, 2017

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