Question

Solve the following system of equations.
$2{\log }_{2}x-3^y=15,3^y.{\log }_{2}x=2{\log }_{2}x+3^{y+1}.$

Answer 1

$2{\log }_{2}x-3^y=15$

$\Rightarrow 2{\log }_{2}x=15+3^y$

$\Rightarrow {\log }_{2}x=\frac{15+3^y}{2}$(equation $1$)

$3^y\cdot {\log }_{2}x=2{\log }_{2}x+3^{y+1}$

$\Rightarrow (3^y-2){\log }_{2}x=3^y\left(3\right)$

$\Rightarrow {\log }_{2}x=\frac{3^{y+1}}{3^y-2}$(equation $2$)

From equation $1$ and $2$

$\frac{3^{y+1}}{3^y-2}=\frac{15+3^y}{2}$

$\Rightarrow 2\cdot 3^{y+1}=(15+3^y)(3^y-2)$

$\Rightarrow 6\cdot 3^y=(3y{)}^2+13\cdot 3^y-30$

$\Rightarrow (3^y{)}^2+7\cdot 3^y-30=0$

$\Rightarrow (3^y+10)(3^y-3)=0$

$\therefore 3^y=3,3^y=-10$

$\therefore y=1,3^{-10}$

From equation $1$

${\log }_{2}x=\frac{15+3^y}{2}$

So,${\log }_{2}x=\frac{15+3}{2}$

$\Rightarrow x=2^9$

$(x,y):(29,1)$

and ${\log }_{2}x=\frac{15-10}{2}$

$\Rightarrow x=2^{5/2}$

$(x,y):(3^{-10},2^{5/2})$