If 2x = 4y = 8z and xyz = 288, then \(\frac{1}{2x}+\frac{1}{4y}+\frac

1.	If 2x = 4y = 8z and xyz = 288, then \(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}\) equals(a) \(\frac{11}{8}\) (b) \(\frac{11}{24}\)(c) \(\frac{11}{48}\)(d) \(\frac{11}{96}\)
Answer» (d) \(\frac{11}{96}\) 2^x = (2²)^y = (2³)^z ⇒ x = 2^y = 3^z Given xyz = 288 ⇒ \(x\times\frac{x}{2}\times\frac{x}{3} = 288\) ⇒ x³ = 6 × 288 ⇒ x³ = 1728 ⇒ x = \(\sqrt[3]{1728}=12\) ∴ y = \(\frac{12}{2}\) = 6 and = \(\frac{12}{3}\) = 4 ∴ \(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}\) = \(\frac{1}{24}+\frac{1}{24}+\frac{1}{32}\) = \(\frac{4+4+3}{96} = \frac{11}{96}.\)

If 2x = 4y = 8z and xyz = 288, then \(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}\) equals(a) \(\frac{11}{8}\) (b) \(\frac{11}{24}\)(c) \(\frac{11}{48}\)(d) \(\frac{11}{96}\)

Answer»

(d) \(\frac{11}{96}\)

2^x = (2²)^y = (2³)^z ⇒ x = 2^y = 3^z

Given xyz = 288 ⇒ \(x\times\frac{x}{2}\times\frac{x}{3} = 288\)

⇒ x³ = 6 × 288 ⇒ x³ = 1728

⇒ x = \(\sqrt[3]{1728}=12\)

∴ y = \(\frac{12}{2}\) = 6 and = \(\frac{12}{3}\) = 4

∴ \(\frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}\) = \(\frac{1}{24}+\frac{1}{24}+\frac{1}{32}\)

= \(\frac{4+4+3}{96} = \frac{11}{96}.\)