Question

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

Answer 1

Hi,

log [ ( x + y )/3] = 1/2 ( logx + logy)

2× log[ ( x + y )/3 ] = log x + log y

log [ ( x + y ) / 3 ]^2 = log xy

{ since i ) n log a = log a^n

ii ) log a + log b = log ab }

Remove log bothsides,

[ ( x + y ) / 3 ]^2 = xy

( x + y )^2 / 3^2 = xy

x^2 + y^2 + 2xy = 9xy

x^2 + y^2 = 9xy - 2xy

x^2 + y^2 = 7xy

Divide each term with xy

x^2 / xy + y^2 / xy = 7xy / xy

x / y + y / x = 7

I hope this help you.

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Answer 2

Answer:

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

2logx+y/3=logx+logy

log(x+y)²/3²=logxy

logx²+y²/9=logxy

Cancel the logs on both sides

x²+y²/9=xy

x²+y²+2xy=9xy

x²+y²=9xy-2xy

x²+y²=7xy

Divide the xy on all

x²/xy+y²/xy=7xy/xy

x/y+y/x=7

So proved