absolute value of logarithm I have a problem with understand how function 2 <mrow cl

kokoszzm

kokoszzm

Answered question

2022-06-23

absolute value of logarithm
I have a problem with understand how function 2 | log 1 / 2 x | obtains values for the negative x ? I thought that there is the assumption that x>0 but wolframalpha shows chart that for negative x also obtains values.
I tried to do it in this way: 2 | log 1 / 2 x | for x ( 0 , 1 ) have formula y = 1 x and for x [ 1 , + ) equals y = x
But how it looks for the negative values?

Answer & Explanation

frethi38

frethi38

Beginner2022-06-24Added 16 answers

Disclaimer: I know nothing about this topic.
I think WolframAlpha is interpreting | log 1 / 2 x | as the modulus of a complex logarithm of x.
How can we find such for x negative? Let's start with natural logs:
e ln x = x so e i π 2 k i π + ln x = x so π i 2 k π i + ln x = ln ( x ), so ln x = ln ( x ) + π i + 2 k π i. The modulus of that is ( ln ( x ) ) 2 + ( π + 2 k π ) 2 , for whatever value of k is conventional. Raising 1/2 to that unwieldy power does not look likely to give anything pretty. That said, when |x| is very large, the value will be very close to −x=|x|.
Misael Matthews

Misael Matthews

Beginner2022-06-25Added 5 answers

log 1 / 2 x > 0 = log 1 / 2 1 x < 1
But since we're talking of a logarithmic function here it must be that x > 0 , so
Further hint:
log 1 / 2 x = log x log 1 2 = log x log 2 > 0

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