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

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if \(\frac {log_2\, a} {4} = \frac {log_2\,b}{6} = \frac {log_2\, c}{3k} \)and a3 b2 c = 1, find the value of k.

log2 a/4 = log2 b/6 = log2 c/ 3k

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∴ 12x + 12x + 3kx = 0

∴ 3kx = -24x

∴ k = -8

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