Question

Solve for x, if :

${\left(\sqrt{27}\right)}^x÷3^{y+4}=1$ and $8^{4-\frac{x}{3}}-{16}^y=0$

• A. 2
• B. 4
• C. 8
• D. 6

Answer 1

The correct option is B 4

$\Rightarrow (\sqrt{27}{)}^x÷3^{y+4}=1$ and $\Rightarrow 8^{4-\frac{x}{3}}-{16}^y=0$
$\Rightarrow (\sqrt{27}{)}^x÷3^{y+4}=1$
$\Rightarrow {\left({\left(3^3\right)}^{\frac{1}{2}}\right)}^x÷3^{y+4}=1$
$\Rightarrow {\left(3^{\frac{3}{2}}\right)}^x÷3^{y+4}=1$
$\Rightarrow \frac{3^{\frac{3x}{2}}}{3^{y+4}}=3^0$
$\Rightarrow 3^{\frac{3x}{2}-(y+4)}=3^0$
$\Rightarrow \frac{3x}{2}-y-4=0\Rightarrow 3x-2y=8eq1$
$\Rightarrow 8^{4-\frac{x}{3}}-{16}^y=0$
$\Rightarrow {\left(2^3\right)}^{4-\frac{x}{3}}-{\left(2^4\right)}^y=0$
$\Rightarrow 2^{12-x}=2^{4y}$
$\Rightarrow 12-x=4y$
$\Rightarrow x+4y=12eq2$

$\Rightarrow 3x-2y=8eq1$
$\Rightarrow x+4y=12eq2$
multiply eq2 by 3
$\Rightarrow 3x+12y=36eq3$

substract eq2 and eq3
$\Rightarrow -14y=-28\Rightarrow y=2$
Now put the value of y in eq1
$\Rightarrow 3x-2\times 2=8$
$\Rightarrow 3x=8+4\Rightarrow 3x=12\Rightarrow x=4$