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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