Question

log(x+3)=2logx+logy​

Answer 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 it will help u...

Answer 2

Step-by-step explanation:

log(x+3)=logx^2+logy

log(x+3)=log(x^2*y)

x+3=x^2y

x-x^2y=-3

x(1-xy)=-3