Question

solve following question​

Answer 1

Answer:

d) 11/96

Explanation:

We will use this formula

$log a^b =blog a$

Let,

$2^x=4^y=8^z=k\\\\2^x=k\\Taking Log Both Side\\log 2^x=logk\\xlog2=logk\\x=logk/log2\\Similarly\\y=logk/log4=logk/log2^2\\y=logk/2log2\\Similarly\\z=logk/log8=logk/log2^3\\z=logk/3log2\\\\xyz=288\\logk/log2 * logk/2log2 * logk/3log2 =288\\(logk/log2)^3*1/6=288\\(logk/log2)^3=288*6=1728\\logk/log2=12\\log k=12log2$

Now substituting value in given equation

$1/2x=1/(2*(logk/log2))=log2/2logk=(log2)/(2*12log2)=1/24\\\\1/4y=1/(4*(logk/2log2))=2log2/4logk=(log2)/(2*12log2)=1/24\\\\1/8z=1/(8*(logk/3log2))=3log2/8logk=(3log2)/(8*12log2)=3/96\\\\1/2x+1/4y+1/8z=1/24+1/24+3/96=11/96$

Hope it helps :-)