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

Answer 1

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