If log2 a/4 = log2 b/6 = log2 c/3k and a^3b^2c = 1, find the value of k.

Question

If  \(\cfrac{log_2a}{4}\) = \(\cfrac{log_2b}{6}\) = \(\cfrac{log_2c}{3k}\)

log₂ a/4 = log₂ b/6 = log₂ c/3k and a³b²c = 1,

Answer 1

Let  \(\cfrac{log_2a}{4}\) = \(\cfrac{log_2b}{6}\) = \(\cfrac{log_2c}{3k}\) = x

\(\therefore\) log₂ a = 4x, log₂ b = 6x, log₂ c = 3k.x    .....(i)

Also, a³ b² c = 1

Taking log to the base 2 throughout, we get

log₂ (a³ b² c) = log₂ 1