Question

Solve the following system of equations.
${\log }_{4}x-{\log }_{x}y=\frac{7}{6},xy=16$

Answer 1

${\log }_{4}x-{\log }_{2}y=\frac{7}{6}xy=16$

$\frac{\log x}{\log 4}-\frac{\log y}{\log x}=\frac{7}{6}y=\left(\frac{16}{x}\right)\to 1$

$\frac{{\left(\log x\right)}^2-\left(\log 4\right)\left(\log y\right)}{\left(\log 4\right)\left(\log x\right)}=\frac{7}{6}$

$6{\left(\log x\right)}^2-6\left(\log 4\right)\left(\log y\right)=7\left(\log 4\right)\left(\log x\right)\to 2$

From $1$ and $2$

$6{\left(\log x\right)}^2-6\left(\log 4\right)\left(\log \left(\frac{16}{x}\right)\right)=7\left(\log 4\right)\left(\log x\right)$

$6{\left(\log x\right)}^2-6\log 4(2\log 4-\log x)=7\left(\log 4\right)\left(\log x\right)$

$6{\left(\log x\right)}^2-1\left(\log 4\right)\left(\log 4\right)+6\left(\log 4\right)\left(\log x\right)-7\left(\log x\right)\left(\log 4\right)=0$

$6{\left(\log x\right)}^2-\left(\log 4\right)\left(\log x\right)-12{\left(\log 4\right)}^2=0$

$(2\log x-3\log 4)(3\log x+4\log 4)=0$

$\log x=\frac{3}{2}\left(\log 4\right)\log x=-\frac{4}{3}\log 4$

$\log x=\log {\left(4\right)}^{3/2}\log x=\log {\left(4\right)}^{-4/3}$

$x={\left(4\right)}^{3/2}x={\left(4\right)}^{-4/3}$

$x=8$

From $1$

$y=\frac{16}{x}y=\frac{16}{8}=2$

$y=\frac{16}{{\left(4\right)}^{-4/3}}=\frac{4^2}{{\left(4\right)}^{4/3}}=4^{(2+\frac{4}{3})}={\left(4\right)}^{\left(\frac{10}{3}\right)}$

$(x,4):(8,2),(4^{-4/3},4^{10/3})$