Question

Solve the following systems of equations.
$log(x^2+y^2)-1=log13,log(x+y)-log(x-y)=3log2$

Answer 1

Solve $log(x^2+y^2)-1=log\mathrm{13...}\left(1\right)$

$log(x+y)-log(x-y)=3log2...\left(2\right)$

by eq.(1) $log(x^2+y^2)-lo{g}_{10}^{10}=log13$

$(\because 1=lo{g}_{10}^{10})$

$log(x^2+y^2)=log13+log10$

$log(x^2+y^2)=log(13\times 10)$ $(\because logAB=logA+logB)$

$\Rightarrow x^2+y^2=\mathrm{130...}\left(3\right)$

by Eq.(2)

$log(x+y)=lo{g}_2^3+log(x-y)$

$log(x+y)=log8(x-y)$

$\Rightarrow x+y=8(x-y)$

$7x=9y...\left(4\right)$

by Eq.(3) & (4)

${\left(\frac{9y}{7}\right)}^2+y^2=130$

$\frac{130}{49}{y}^2=130$

$y^2=49$

$y=\pm 7$

put in Eq.(4)

$7x=\pm 9\times 7$

$x=\pm 9$

sol. $x=\pm 9,y=\pm 7$