Question

Solve for $x$ and $y$, if: $(\sqrt{32}{)}^x÷2^{y+1}=1$ and $8^y-{14}^{4-\frac{x}{2}}=0$

Answer 1

Finding the values of $x$ and $y$
Given, $(\sqrt{32}{)}^x÷2^{y+1}=1$
${\left[\right(32{)}^{\frac{1}{2}}]}^x\times \frac{1}{2^{y+1}}=1$ $[\therefore a÷b=a\times \frac{1}{b}]$
$\Rightarrow {\left(\right(2^5\left){)}^{\frac{1}{2}}\right)}^x\times \frac{1}{2^{y+1}}=1$
$\Rightarrow 2^{5\times \frac{1}{2}\times x}=1\times 2^{y+1}$ $[\therefore ((a^m{)}^n{)}^p=a^{mnp}]$
$\Rightarrow 2^{\frac{5}{2}x}=2^{y+1}$
So, $\frac{5}{2}x=y+1$
$\Rightarrow \frac{x}{2}=\frac{y+1}{5}$ ... (i)
Now,
$\Rightarrow 8^y-{16}^{4-\frac{x}{2}}=0$
Substitute the value of $\frac{x}{2}$ from (i), we get,
$\Rightarrow (2^3{)}^y=(2^4{)}^{4-\frac{y+1}{5}}$
$\Rightarrow 2^{3y}=2^{4(4-\frac{y+1}{5})}$ $\left[\right(a^m{)}^n=a^{m\times n}]$
$\Rightarrow 2^{3y}=2^{16-\frac{4y+4}{5}}$
$\Rightarrow 3y=16-\frac{4y+4}{5}$
$\Rightarrow 3y+\frac{4y+4}{5}=16$
$\Rightarrow \frac{5\times 3y+4y+4}{5}=16$
$\Rightarrow 15y+4y+4=16\times 5$
$\Rightarrow 19y=80-4\Rightarrow y=\frac{76}{19}$
Substitute, the value of be in (i), we get,
$\frac{x}{2}=\frac{1+\frac{76}{19}}{5}=2\times \frac{5}{5}=2\times 1=2$
Hence, the value of $x$ and $y$ are $2$ and $\frac{76}{19}$ respectively.