Question

If log (x+y/3)=1/2(log+logy) then find the value of x/y+y/x

Answer 1

Given, log(x+y/3) = 1/2(logx+logy)

           = 2(logx+y/3) = log(x) + log(y)

          = 2(log(x+y)/3) = log (xy)

          = ((x+y)/3^2) = xy

          = (x+y)^2 = 9xy

          = x^2 + y^2 = 7xy

          Divide both sides with xy, we get


         = x^2/xy + y^2/xy = 7xy/xy

         = (x/y) + (y/x) = 7.


Hope this helps!

Answer 2

hello users .....

we have given that :
log {(x+y)/3}=1/2(log+logy)

we have to find :
 x / y + y / x = ?

solution:-
we know that:
$n log m = log m^{n}$
and 
log mn = log m + log n 

Here ;
 log {(x+y)/3}=1/2(logx+logy) 

=> 2 log {(x+y)/3}=(logx+logy)
 
=> log {(x+y)/3}² = log xy 

=> {(x+y)/3}² = xy 

=> (x + y)² / 3² = xy 

=> (x + y)² = 9 xy 

=> x² + y² + 2xy = 9 xy 

=> x² + y² =9 xy - 2 xy = 7 xy 

now;
divide both side by xy , we get 
x² / xy + y² / xy = 7 xy / xy

=> x / y + y / x = 7 answer 

✮✮ hope it helps ✮✮