SOLUTION: Prove: 2/(log_8(a))-4/(log_2(a)) = 1/(log_4(a))
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Question 252137: Prove: 2/(log_8(a))-4/(log_2(a)) = 1/(log_4(a))
Found 2 solutions by drk, Alan3354:
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Okay, got it . . . The bases are 8, 2, and 4 respectively. Since the bases have a common multiple of 8, we can express a in terms of these bases. Let a = 8^K, where k is greater or equal to 1. Now we get
We can easily adjust the denominators using rules of exponent, and the fraction becomes
. Now if you solve the left side, we get
(6 - 4) / 3K
2/3K.
Thinking about a reciprocal, we get 1/[(3/2)K] which is now the right side.
I hope that helps.
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Prove: 2/(log_8(a))-4/(log_2(a)) = 1/(log_4(a))
---------------------------
I think this is what you mean.
All logs are base 2.
2/(3log(a)) - 4/log(a) = 1/(2log(a))
The LCD is 6log(a)
4/(6log(a)) - 24/(6log(a)) = 3/(6log(a))
It's not equal.
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