Question

Answer the following:

If `log"a"/(x + y - 2z) = log"b"/(y + z - 2x) = log"c"/(z + x - 2y)`, show that abc = 1

Answer 1

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