Question

plz dont copy paste and give full explanation. And dont scam or i will report​

Answer 1

Solution :-

${2}^{x} = {4}^{y} = {8}^{z} \\ {\mathtt{on \: comparing}} \\ {2}^{x} = {4}^{y} \\ {2}^{x} = {2}^{2y} \: : \: \: {\sf{on \: comparing \: exponentially}} \\ x=2y \\ y = \frac{x}{2} \\ \implies {2}^{x} = {8}^{z} \\ {2}^{x} = {2}^{3z} \: \: : \: {\sf{on \: comparing \: exponentially}} \\ x = 3z \\ z = \frac{x}{3}$

Substituting Values of x, y and z

$\frac{1}{2x} + \frac{1}{4y} + \frac{1}{6z} = \frac{24}{7} \: \: \: \: ( substitute \: values \: of \: x \: y \: and \: z)\\ \\ \frac{1}{2x} + \frac{1}{2x} + \frac{1}{2x} = \frac{24}{7} \\ \\ \frac{1 + 1 + 1}{2x} = \frac{24}{7} \\ \\ \frac{3}{2x} = \frac{24}{7} \\ \\ x = \frac{7}{16}$

$\red{ \tt{putting \: value \: of \: x \: in \: z = \frac{x}{3} }}$

$z = \frac{x}{3} \\ \\ z = \frac{7}{16} \times \frac{1}{3} = \frac{7}{48}$

HOPE IT HELPS YOU!✨✌️