Question

pls prove it i beg you​ open the image and see the question

Answer 1

Step-by-step explanation:

Given,

$\sf \sf \bf :\longmapsto 2^x = 4^y = 8^z$

$\sf \bf :\longmapsto 2^x = (2^2)^y = (2^3)^z$

$\sf \bf :\longmapsto 2^x = 2^{2y} = 2^{3z}$

Bases are same thus powers are also same

$\sf \bf :\longmapsto x = 2y = 3z$

$\sf \bf :\longmapsto y = \dfrac{x}{2} , \ z=\dfrac{x}{3}$

Now putting these values in other equation:

$\sf \bf :\longmapsto \dfrac{1}{x} +\dfrac{1}{^2\cancel{4}\cfrac{x}{\cancel{2}}} +\dfrac{1}{4\cfrac{x}{3}}=4$

$\sf \bf :\longmapsto \dfrac{1}{x} + \dfrac{1}{2x} + \dfrac{3}{4x} = 4$

$\sf \bf :\longmapsto \dfrac{1}{x}\left(1 + \dfrac{1}{2} + \dfrac{3}{4} \right) = 4$

$\sf \bf :\longmapsto 4x = \dfrac{4+2 + 3}{4}$

$\boxed{\boxed{\sf \bf x = \dfrac{9}{16}}}$