View Full Version : a petite observation about logarithms
homology
Nov15-05, 09:07 AM
Here's something cute:
Consider the graph of ln(x^2) and then consider the graph of 2ln(x), missing anything?
I was momentarily caught off guard by this until I realized that when we derive the property: ln(x^a)=aln(x), we choose the positive root.
Has anyone ever run into a situation where it was better to say that ln(x^2)=2ln|x| ?
Lonewolf
Nov15-05, 09:53 AM
Well, consider the case x < 0. Then ln(x^2) is real, but 2ln(x) has a non-zero imaginary part, so clearly they can't be equal.
HallsofIvy
Nov15-05, 10:16 AM
Yes, that's a very good point.
Solving the equation ln(x^2)= 0 is NOT the same as solving 2ln(x)= 0 and, yes, it is better to write ln(x^2)= 2ln(|x|).
Yes, that's a very good point.
Solving the equation ln(x^2)= 0 is NOT the same as solving 2ln(x)= 0 and, yes, it is better to write ln(x^2)= 2ln(|x|).
On most computers it is not better to write ln(x^2) instead of 2ln(|x|) in a computer program.
HallsofIvy
Nov20-05, 09:19 PM
On most computers it is not better to write ln(x^2) instead of 2ln(|x|) in a computer program.
I'm sorry, what does that have to do with my response? My point was NOT that it was better to write ln(x^2) rather than 2ln(|x|) but rather that it was better to write ln(x^2)= 2ln(|x|) rather than ln(x^2)= 2ln(x).
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