SOLUTION: ok.well im having a lot of troulble with logs so i have a question.here is my problem...(16 to the 3/4 power - 16 to the 1/4 power)to the -3/2power.
i have no clue on what to do.p
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Exponential-and-logarithmic-functions
-> SOLUTION: ok.well im having a lot of troulble with logs so i have a question.here is my problem...(16 to the 3/4 power - 16 to the 1/4 power)to the -3/2power.
i have no clue on what to do.p
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Question 188836: ok.well im having a lot of troulble with logs so i have a question.here is my problem...(16 to the 3/4 power - 16 to the 1/4 power)to the -3/2power.
i have no clue on what to do.please help! Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! ok.well im having a lot of troulble with logs so i have a question.here is my problem...(16 to the 3/4 power - 16 to the 1/4 power)to the -3/2power.
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...(16 to the 3/4 power - 16 to the 1/4 power)to the -3/2power.
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Take it one step at a time
16 to the 3/4 power
= (16^(1/4))^3
16^(1/4) = (2^4)^1/4 = 2
then, 2^3 = 8
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The 1st term is 8.
16^(1/4) is 2 (still).
The 2nd term is 2
So now it's:
(8 - 2)^(-3/2)
= 6^(-3/2) at least it's shorter
= 1/(6^(3/2))
= 1/6 * 1/sqrt(6)
= 1/6 * sqrt(6)/6
=
