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 $$z$$.

Note by Swapnil Das
3 years, 1 month ago

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We have $$2^{x} = 2^{2y} = 2^{3z} \Longrightarrow x = 2y = 3z \Longrightarrow y = \dfrac{x}{2}, z = \dfrac{x}{3}.$$

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

- 3 years, 1 month ago

@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.

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Thank you Sir. I am greatly obliged to get your compliment.

- 3 years, 1 month ago

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

- 3 years, 1 month ago

You're welcome, Swapnil. :)

- 3 years, 1 month ago

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

- 3 years, 1 month ago

Yes! Thank U! Please post the solution.

- 3 years, 1 month ago