SOLUTION: Prove: 2/(log_8(a))-4/(log_2(a)) = 1/(log_4(a))

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Prove: 2/(log_8(a))-4/(log_2(a)) = 1/(log_4(a))      Log On


   

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) About Me  (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
2%2Flog_8%288%5Ek%29+-+4%2Flog_2%288%5Ek%29+=+1%2Flog_4%288%5Ek%29
We can easily adjust the denominators using rules of exponent, and the fraction becomes
2%2FK+-+4%2F3K+=+1%2F%283%2F2%29k. 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) About Me  (Show Source):
You can put this solution on YOUR website!
Prove: 2/(log_8(a))-4/(log_2(a)) = 1/(log_4(a))
log%288%2Ca%29+=+log%282%2Ca%5E3%29+=+3log%282%2Ca%29
log%284%2Ca%29+=+log%282%2Ca%5E2%29+=+2log%282%2Ca%29
---------------------------
2%2Flog%282%2Ca%5E3%29+-+4%2Flog%282%2Ca%29+=+1%2Flog%282%2Ca%5E2%29 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.