Question 124

If $$2^{x}=4^{y}=8^{z}$$ and xyz = 288, the value of $$\frac{1}{2x}$$ +$$ \frac{1}{4y} + \frac{1}{8z}$$ is

Solution

Expression : $$2^{x}=4^{y}=8^{z}$$

=> $$2^x = 2^{2y} = 2^{3z}$$

=> $$x = 2y = 3z = k$$ (let)

Now, $$xyz$$ = 288

=> $$k * \frac{k}{2} * \frac{k}{3}$$ = 288

=> $$k^3 = 12^3$$

=> $$k$$ = 12

=> $$x$$ = 12 , $$y$$ = 6 , $$z$$ = 4

To find : $$\frac{1}{2x}$$ +$$ \frac{1}{4y} + \frac{1}{8z}$$

= $$\frac{1}{24} + \frac{1}{24} + \frac{1}{32}$$

= $$\frac{11}{96}$$

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