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If `xa n dy`are real numbers such that `2log(2y-3x)=logx+logy` ,then find `x/y`. |
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Answer» `2log(2y-3x)=logx+logy` `log(2y-3x)^2=logxy` `(2y-3x)^2=xy` `y^2[2-3x/y]^2=xy` `4+9(x/y)^2-12x/y=x/y` Let `x/y=t` `4+9t^2-12t-t=0` `9t^2-13t+4=0` `9t^2-9t-4t+4=0` `9t(t-1)-4(t-1)=0` `(t-1)(9t-4)=0` `t=1,4/9` `x/y=1,4/9`. |
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