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Solve the following equations:
\( 8^{x+1}=16^{y+2} \) and, \( \left(\frac{1}{2}\right)^{3+x}=\left(\frac{1}{4}\right)^{3 y} \)
Given:
\( 8^{x+1}=16^{y+2} \) and, \( \left(\frac{1}{2}\right)^{3+x}=\left(\frac{1}{4}\right)^{3 y} \)
To do:
We have to solve the given equations.
Solution:
We know that,
$(a^{m})^{n}=a^{m n}$
$a^{m} \times a^{n}=a^{m+n}$
$a^{m} \div a^{n}=a^{m-n}$
$a^{0}=1$
Therefore,
$8^{x+1}=16^{y+2}$
$(2^{3})^{x+1}=(2^{4})^{y+2}$
$\Rightarrow 2^{3 x+3}=2^{4 y+8}$
Comparing both sides, we get,
$3 x+3=4 y+8$
$\Rightarrow 3 x-4 y=8-3=5$.........(i)
$(\frac{1}{2})^{3+x}=(\frac{1}{4})^{3 y}$
$=[(\frac{1}{2})^{2}]^{3 y}$
$=(\frac{1}{2})^{6 y}$
Comparing both sides, we get,
$3+x=6 y$
$\Rightarrow x=6 y-3$
Substituting $x=6y-3$ in (i), we get,
$3(6y-3)-4y=5$
$18y-9-4y=5$
$14y=5+9=14$
$y=1$
$\Rightarrow x=6(1)-3=6-3=3$
The values of $x$ and $y$ are $3$ and $1$ respectively.