2007-06-12 08:18:34 · 5 answers · asked by Dani 1 in Science & Mathematics ➔ Mathematics
log[2] (1/32) =
log[2] (32^-1) =
-1 * log[2] (32) =
-log[2] (2^5) =
-5 log[2](2) =
-5
2007-06-12 08:22:43 · answer #1 · answered by Anonymous · 2⤊ 0⤋
1/32 can be written as 2^(-5)
If we replace this in the question then we will get
log2(2^(-5))
=> (-5)log2(2)
=>-5 (as log of any number to the same is always one, except for 0 and negative numbers)
there fore -5 is the answer
2007-06-12 15:32:45 · answer #2 · answered by anksbaj 1 · 0⤊ 0⤋
2^5 is 32, so log2(1/2^5)is
-5
2007-06-12 15:23:16 · answer #3 · answered by Anonymous · 1⤊ 1⤋
log2(1/32) = - log2(32) = - log2(2^5) = -5
2007-06-12 15:25:05 · answer #4 · answered by Steiner 7 · 1⤊ 0⤋
Let log base 2 be referred to as "log" in following answer:-
log (1 / 32)
= log (1 / 2^5)
= log(2^(-5))
= (-5).log 2
= - 5
2007-06-12 17:43:51 · answer #5 · answered by Como 7 · 0⤊ 0⤋