2007-10-12 09:08:07 · 5 answers · asked by Robert G 2 in Science & Mathematics ➔ Mathematics
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 · answer #1 · answered by Astral Walker 7 · 1⤊ 0⤋
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 · answer #2 · answered by Francisco Orduña 2 · 0⤊ 0⤋
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 · answer #3 · answered by Como 7 · 0⤊ 0⤋
No.
- log (n + 1) = log[(n + 1)^-1]
2007-10-12 09:11:58 · answer #4 · answered by Amit Y 5 · 1⤊ 0⤋
yes
2007-10-12 09:10:49 · answer #5 · answered by Here's Johnny! 3 · 0⤊ 2⤋