Question

if   2 raise x  =  3  raise  y   =  6 raise minus  z  then   1 upon  x  plus 1  upon y  plus 1 upon z  is  equal  to   ?

Answer 1

$2^x=3^y=6^{-z}=2^{-z}3^{-z}$
$log_{2} 3^{y}=x \\ ylog_{2}3=x ...(1)\\2^{x}=6^{-z}=2^{-z}3^{-z} \\ 2^{x+z}=3^{-z} \\ log_{2}3^{-z}=x+z \\ -zlog_{2}3=x+z \\ log_{2}3=-1- \frac{x}{z} \\ \frac{1}{z}= \frac{-1}{x}- \frac{log_{2}3}{x} \\ \frac{1}{x}+ \frac{1}{y}+ \frac{1}{z}= \frac{1}{x}+ \frac{1}{y}- \frac{1}{x}- \frac{log_{2}3}{x} \\ = \frac{1}{y}- \frac{log_{2}3}{x}$
from (1)
$= \frac{1}{y}- \frac{1}{y}=0$