Question

Solve the following system of equations.
${\log }_2(x-y)=5-{\log }_2(x+y),\frac{\log x-\log 4}{\log y-\log 3}=1.$

Answer 1

${\log }_2(x-y)=5-{\log }_2(x+4)$

$\frac{\log x-\log 4}{\log y-\log 3}=1$

${\log }_2(x-y)=5{\log }_2(x+4)$

${\log }_2(x-4)+{\log }_2(x+y)=5$

${\log }_2(x^2-y^2)=5$

$x^2-y^2=2^5$

$x^2-y^2=32\to 1$

$\frac{\log x-\log 4}{\log y-\log 3}=1$

$\log x-\log 4=\log y-\log 3$

$\log x-\log y=\log 4-\log 3$

$\log \left(\frac{x}{y}\right)=\log \left(\frac{4}{3}\right)$

$x=\left(\frac{4}{3}y\right)\to 2$

From $1$ and $2$

${\left(\frac{4}{3}\right)}^2-y^2=32$

$7y^2=32\times 9$

$y=\pm 12\sqrt{\frac{2}{7}};x=\pm 16\sqrt{\frac{2}{7}}$

$(x,y):(16\sqrt{\frac{2}{7}},12\sqrt{\frac{2}{7}}),(-16\sqrt{\frac{2}{7}},-12\sqrt{\frac{2}{7}})$