Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Solve the system of equations: $$\begin{cases}\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy} \\\left(\sqrt{3}+xy\right)^{\log_2x}+\dfrac{x}{\left(\sqrt{3}-xy\right)^{\log_2y}}= 1+\dfrac{x}{y}\end{cases}$$

My try: $\dfrac{x^2+xy+y^2}{x^2+y^2+1}=\dfrac{1}{xy}\\\Leftrightarrow xy\left(x^2+y^2\right)+\left(xy\right)^2= x^2+y^2+1\\\Leftrightarrow\left(xy-1\right)\left(x^2+y^2+xy+1\right)=0\\\Leftrightarrow xy=1\,\,\,\mbox{(because:}\,\, x^2+y^2+xy+1>0)$

And here I'm stuck!

share|improve this question
    
if i answered your question then please mark it so! –  Atul Gangwar Nov 10 '13 at 6:33

1 Answer 1

up vote 1 down vote accepted

Subsitute $xy=1 , y=1/x,and \, log y=-logx$ :-

$log(3^{1/2}+1)^{logx}+ x*log(3^{1/2}-1)^{logx}=1+x^2$

then take Log at both sides of your equation then your 2nd equation becomes

$logx*log(3^{1/2}+1)+logx+ logx*log(3^{1/2}-1)=2logx$

$log(3^{1/2}+1)+ log(3^{1/2}-1)=1$

$log((3^{1/2}+1)(3^{1/2}-1))=1$

which is always true!

share|improve this answer
    
of course i have leap frogged some steps. –  Atul Gangwar Nov 9 '13 at 17:45

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.