Question

Solve the following equations :

iv) $8^{x+1}={16}^{y+2}\text{and}{\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{4}\right)}^{3y}$

Answer 1

Given : $8^{x+1}={16}^{y+2}$

$\Rightarrow (2^3{)}^{x+1}=(2^4{)}^{y+2}$

$\Rightarrow 2^{3x+3}=2^{4y+8}\{\because (a^m{)}^n=a^{m\times n}\}$

On comparing both the sides, we get

$\Rightarrow 3x+3=4y+8$

$\Rightarrow 3x-4y=5........\left(i\right)$

Also, ${\left(\frac{1}{2}\right)}^{3+x}={\left(\frac{1}{4}\right)}^{3y}$

$\Rightarrow \frac{1}{2^{3+x}}=\frac{1}{(2^2{)}^{3y}}\{\because {\left(\frac{1}{a}\right)}^m=\frac{1}{a^m}\}$

$\Rightarrow 2^{3+x}=2^{6y}$

On comparing both the sides, we get

$3+x=6y$

$x=6y-3......\left(ii\right)$

Putting value of $x$ in equation ($i)$ we get,

$3(6y-3)-4y=5$

$\Rightarrow 18y-9-4y=5$

$\Rightarrow 14y=14$

$\Rightarrow y=1$

Putting value of $y$ in equation $\left(ii\right)$ we get,

$x=6\times 1-3$

$\Rightarrow x=3$

Hence, $x=3\text{and}y=1$