log(2)16 = 4
log(2)(1/2) = -1
=>
(log2)16/(log2(1/2)) = -4
2007-01-07 18:31:21
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answer #1
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answered by Patrick M 2
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Looks complicated but isn't. Usually best to re-write expression in an order which looks simpler to you: example,
16 (log2) / (1/2) (log2) =
16 (log2) / (0.5) (log2) =
16 / (0.5) = 32
So the answer is D) 32.
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If the notation (log2)16 means log of 16 to the basis of 2, then the previous 2 answers yielding -4 is correct. Working in reverse from those answers by factoring :
16 = 2^4 so (log2)16=4;
1/2 = 2(^-1) so (log2)(1/2)=-1
therefore
(log2)16/(log2)(1/2) = 4/(-1) = -4.
Don't clutch up at all those mixed up expressions. Re-write the expressions in a simpler way for you to read.
2007-01-08 02:36:16
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answer #2
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answered by kyq 2
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(log2)(16)/(log2)(1/2)
This is ambiguous. Do you mean:
(log2)(16)/{(log2)(1/2)}
or
{(log2)(16)/(log2)}(1/2)
Either way, start by dividing numerator and denominator by log2.
I'll take the first one.
(log2)(16)/{(log2)(1/2)}
= 16/(1/2) = 32
The answer is D).
I should add one more. Does log2 mean
The (log of 2) or (log to the base 2)?
If it means log to the base 2 then the answer is:
(log2)16/(log2)(1/2)
4/-1 = -4
The answer is A).
Please state your questions clearly in the future. We shouldn't have to guess at what the question is supposed to be.
2007-01-08 03:05:24
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answer #3
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answered by Northstar 7
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Helmut, Patrick M and Chetan all have the correct answer. However, some of these people here seem to be reading your problem incorrectly.
log (base 2) 16 = 4, because (2)^4 = 16.
log (base 2) 1/2 = -1, because (2)^-1 = 2^(0-1) = 2^0/2^1 = 1/2.
Therefore, log (base 2) 16 / log (base 2) 1/2 = 4 / -1 = -4.
2007-01-08 02:59:49
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answer #4
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answered by MathBioMajor 7
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(log2)16/(log2)(1/2) =
4/-1 =
-4
2007-01-08 02:30:54
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answer #5
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answered by Helmut 7
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answer is D. 32
2007-01-08 02:28:40
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answer #6
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answered by rishi 2
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Correct answer is -4.
2007-01-08 02:32:32
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answer #7
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answered by Anonymous
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a) listen in class
b) get tutoring
c) use a calculator
d) all of the above
the answer is d
2007-01-08 02:27:36
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answer #8
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answered by wtfitsnguyen 2
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