Question

Solve for x.

log₂ x + log₄ x + log₁₆ x = `21/4`

Answer 1

log₂ x + log₄ x + log₁₆ x = `21/4`

∴ `logx/log2 + logx/log4 + logx/log16 = 21/4`

∴ `(logx)[1/log2 + 1/log4 + 1/log16] = 21/4`

∴ `(logx)[1/log2 + 1/log2^2 + 1/log2^4] = 21/4`

∴ `(logx) [1/log2 + 1/(2log2) + 1/(4log2)] = 21/4`

∴ `(logx/(log2)) [ 1 + 1/2 + 1/4] = 21/4`

∴ `(logx/(log2)) [(4 + 2 + 1)/4] = 21/4`

∴ `7/4.(logx/log2) = 21/4`

∴ `logx/log2` = 3

∴ log x = 3 log 2 = log2³

∴ log x = log 8

∴ x = 8