Question

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

• A. $\frac{11}{12}$
• B. $\frac{11}{96}$
• C. $\frac{29}{96}$
• D. none of these

Answer 1

Answer: (B)

$\frac{11}{96}$

$\begin{array}{c}Giventhat\\ 2^x=4^y=8^z--\left(1\right)\\ and,\\ xyz=288---\left(2\right)\\ u\sin geq{u}^n\left(1\right)weget\\ 2^x={\left(2^2\right)}^y={\left(2^3\right)}^z\\ \Rightarrow 2^x=2^{2y}=2^{3z}\left[\therefore {\left(a^m\right)}^n=a^{mn}\right]\\ \Rightarrow x=2y=3z\\ \Rightarrow x=2yandx=3z\\ \Rightarrow x=2yandx=3z\\ \therefore y=\frac{x}{2}andz=\frac{x}{3}\\ Now,subsitute\left(y=\frac{x}{2}\right)\&\left(z=\frac{x}{3}\right)ineq{u}^n\left(2\right)\\ Wehave\\ x\left(\frac{x}{2}\right)\left(\frac{x}{3}\right)=288\\ \Rightarrow x^3=288\times 6\\ \Rightarrow x^3=1728\therefore x=12\\ \therefore y=\frac{12}{6}=6\&z=\frac{12}{3}=4\\ \therefore \frac{1}{2x}+\frac{1}{4y}+\frac{1}{8z}\\ =\frac{1}{2\times 12}+\frac{1}{4\times 6}+\frac{1}{8\times 4}\\ =\frac{1}{24}+\frac{1}{24}+\frac{1}{32}=\frac{11}{96}\\ Hence,optionBiscorrect.\end{array}$