Hi Friends please help me with this question if u give correct answer í will give you 50 points and Brainliest
Attachments:
Answers
Answered by
1
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.
Pls mark me as the brainliest.
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.
Pls mark me as the brainliest.
Answered by
1
hey mate...
here is your answer ..
Attachments:
Similar questions