Please post a solution to the given problem. Any help would be appreciated. Thanks!

If \(2^{x}\) = \(4^{y}\) = \(8^{z}\) and \(xyz = 288\) , find the values of \(x\), \(y\) and \(z\).

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## Comments

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TopNewestWe have \(2^{x} = 2^{2y} = 2^{3z} \Longrightarrow x = 2y = 3z \Longrightarrow y = \dfrac{x}{2}, z = \dfrac{x}{3}.\)

Thus \(xyz = x*\dfrac{x}{2} * \dfrac{x}{3} = \dfrac{x^{3}}{6} = 288 \Longrightarrow x^{3} = 1728 \Longrightarrow x = 12, y = 6, z = 4.\)

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@Brian Charlesworth Sir you are quick enough to post the solution. As I am not good at LaTex , I could not post the solution quick. Thank you for posting the solution.

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You had the correct answer first. 3 minutes after the question was posted; that's fast! :)

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I convey my heartfelt thanks to you respected Sir! @Brian Charlesworth

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You're welcome, Swapnil. :)

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Is the answer like this: x = 12, y=6, z=4

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Yes! Thank U! Please post the solution.

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