How do I take ln of a sum
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How do I take ln of a sum

[From: ] [author: ] [Date: 11-12-06] [Hit: ]
......
I have
y = (x^2 + 1)^(1/2) + e^C do i distribute ln
so,

ln(y) = ln(x^2 + 1)^(1/2) + ln(e^C)

this doesnt seem right to me

-
No, that's not right. We don't really have a way of dealing with logarithm of a sum.
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