No. Here is how you can figure it out.
b*log(a) = log(a^b) which is the general rule of logarithms.
So
-log(n+1) = -1*log(n+1) = log[(n+1)^-1] = log[1/(n+1)]
And in general,
log[1/(n+1)] <> [log(n+1)]^-1
For example, if n=0
log[1/(n+1)] = log [1] = 0
[log(n+1)]^-1 = log[1]^-1 = 1/0
2007-10-12 09:19:40
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answer #1
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answered by Astral Walker 7
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I think that is not true:
A true statement is: -log(n+1)=log[(n+1)^-1]
Or the same thing in other manner:
-log(n+1)=log[1/(n+1)]
2007-10-12 09:31:29
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answer #2
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answered by Francisco Orduña 2
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n + 1 is but a number.
For n + 1 > 0 :-
LHS
- log (n + 1)
RHS
log (n + 1)^(-1) = (-1) log (n + 1) = - log (n + 1)
Statement is true
2007-10-16 04:32:54
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answer #3
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answered by Como 7
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No.
- log (n + 1) = log[(n + 1)^-1]
2007-10-12 09:11:58
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answer #4
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answered by Amit Y 5
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yes
2007-10-12 09:10:49
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answer #5
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answered by Here's Johnny! 3
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