Question

tell me answer in details

Answer 1

Hi...☺

Here is your answer...✌

${2}^{x} = {4}^{y} = {8}^{z} \\ \\ {2}^{x} = {2}^{2y} = {2}^{3z} \\ \\ = > x = 2y = 3z \: \: \: \: \: \: .....(1) \\ \\ now \\ \\ \frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z} = \frac{24}{7} \\ \\ \frac{1}{2 \times (3z)} + \frac{1}{2 \times ( 3z)} + \frac{1}{6z} = \frac{24}{7} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: |from \: (1)| \\ \frac{1}{6z} + \frac{1}{6z} + \frac{1}{6z} = \frac{24}{7} \\ \\ \frac{3}{6 z} = \frac{24}{7} \\ \\ \frac{1}{2z} = \frac{24}{7} \\ \\ \frac{1}{z} = \frac{24 \times 2}{7} \\ \\ \frac{1}{z} = \frac{48}{7} \\ \\ = > z = \frac{7}{48}$

HENCE,

Option (C) 7/48 is the correct answer