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Question
Answer the following:
If `log_2"a"/4 = log_2"b"/6 = log_2"c"/(3"k")` and a3b2c = 1 find the value of k
Solution
Let `log_2"a"/4 = log_2"b"/6 = log_2"c"/(3"k")` = R
∴ log2 a = 4R, log2 b = 6R, log2 c = 3kR
Now, a3b2c = 1
∴ log2 (a3b2c) = log2 1
∴ log2 a3 + log2 b2 + log2 c = 0
∴ 3 log2 a + 2 log2 b + log2 c = 0
∴ 3(4R) + 2(6R) +3kR = 0
∴ 12R + 12R + 3kR = 0
∴ 24R + 3kR = 0
∴ 3kR = – 24R
∴ k = – 8
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