Question

Solve for $x$, if :

${\left(\sqrt{32}\right)}^x÷2^{y+1}=1$ and $8^y-{16}^{4-\frac{x}{2}}=0$

Answer 1

As given$\Rightarrow (\sqrt{32}{)}^x÷2^{y+1}=1$ and $8^y-{16}^{4-x/2}=0$

$\Rightarrow (\sqrt{32}{)}^x÷2^{y+1}=1$
$\Rightarrow {\left({\left(2^5\right)}^{\frac{1}{2}}\right)}^x÷2^{y+1}=1$
$\Rightarrow {\left(2^{\frac{5}{2}}\right)}^x÷2^{y+1}=1$
$\Rightarrow \frac{2^{\frac{5x}{2}}}{2^{y+1}}=2^0$
$\Rightarrow 2^{\frac{5x}{2}-(y+1)}=2^0$
$\Rightarrow \frac{5x}{2}-y-1=0$

$\Rightarrow 5x-2y=2$ ........ (1)

From$\Rightarrow 8^y-{16}^{4-\frac{x}{2}}=0$
$\Rightarrow {\left(2^3\right)}^y-{\left(4^2\right)}^{4-\frac{x}{2}}=0$
$\Rightarrow 2^{3y}-{\left({\left(2^2\right)}^2\right)}^{4-\frac{x}{2}}=0$
$\Rightarrow 2^{3y}-2^{16-2x}=0$
$\Rightarrow 2^{3y}=2^{16-2x}$
$\Rightarrow 3y=16-2x$

$\Rightarrow 2x+3y=16$ ........ (2)
Multiply eq1 by 3 and eq2 by 2
$\Rightarrow 15x-6y=6$ ......(3)
$\Rightarrow 4x+6y=32$ ...... (4)
Add eq3 and eq4
$\Rightarrow 19x=38\Rightarrow x=2$
Now put the value of x in eq1
$\Rightarrow 5\times 2-2y=2$
$\Rightarrow -2y=-8\Rightarrow y=4$