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.       

Tutorialspoint

Updated on: 10-Oct-2022

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