Does loga/logb = log(a^(1/logb))? I know log(a^b)=blog(a). However, Wolfram Alpha tells me that (log(a))/(log(b)) does not equal log(a^(1)/(log(b))) Is Wolfram Alpha correct? If it is, why is it correct? I'm using base 10 logs instead of natural logs, although I doubt it makes a difference.

Marcelo Maxwell

Marcelo Maxwell

Answered question

2022-09-19

Does loga/logb = log(a^(1/logb))?
I know log ( a b ) = b log ( a )
However, Wolfram Alpha tells me that log ( a ) log ( b ) does not equal log ( a 1 log ( b ) )
Is Wolfram Alpha correct? If it is, why is it correct?
I'm using base 10 logs instead of natural logs, although I doubt it makes a difference.

Answer & Explanation

moderrockblog09

moderrockblog09

Beginner2022-09-20Added 7 answers

Actually you are correct.
However, you must take note that there are some constraints (which are pretty self-explanatory).
log ( a ( 1 log ( b ) ) ) = log a log b
As long as,
a > 0 , because you cannot take the logarithm of a negative number or 0
b 1   and   b > 0 , because then log b = 0 , which means that the fraction 1 log b would be undefined
Hope this helps you understand why.
Comment if you have any questions.
mydaruma25

mydaruma25

Beginner2022-09-21Added 2 answers

I think your issue might be that log ( x ) is a multivalued function if you allow complex outputs. This is due to the fact that e 2 π i n = 1 for n Z . The complex logarithm can give out different values as follows.
log ( z ) = ln | z | + i arg ( z + 2 π n )
So when you take one value for the left hand side of your original equation, you could take a completely different one for the right hand side (though I'm not entirely certain that this is the reason).

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