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Question
Answer the following:
If `log"a"/(x + y - 2z) = log"b"/(y + z - 2x) = log"c"/(z + x - 2y)`, show that abc = 1
Solution
Let `log"a"/(x + y - 2z) = log"b"/(y + z - 2x) = log"c"/(z + x - 2y)` = k
∴ log a = k(x + y – 2z), log b = k(y + z – 2x), log c = k(z + x – 2y)
∴ log a + log b + log c = k(x + y – 2z) + k(y + z – 2x) + k(z + x – 2y)
∴ log abc = k(x + y – 2z + y + z – 2x + z + x – 2y) = 0
∴ log abc = log 1 ...[∵ log 1 = 0]
∴ abc = 1
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