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If ${2^{2x - y}} = 32$ and ${2^{x + y}} = 16$ then ${x^2} + {y^2}$ is equal to
A. $9$
B. $10$
C. $11$
D. $13$

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Last updated date: 17th Jun 2024
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Answer
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Hint: In this problem, first we will multiply the given equations. Then, we will use the law of exponents which is given by ${a^m} \times {a^n} = {a^{m + n}}$. After simplification, we will get the value of $x$. Then, we will find the value of $y$. Then, we will get the required value.

Complete step-by-step answer:
In this problem, we have two equations. Let us write the given equations.
${2^{2x - y}} = 32 \cdots \cdots \left( 1 \right)$ and ${2^{x + y}} = 16 \cdots \cdots \left( 2 \right)$
Now we will multiply the equation $\left( 1 \right)$ and $\left( 2 \right)$. Therefore, we get
$
  \left( {{2^{2x - y}}} \right) \times \left( {{2^{x + y}}} \right) = 32 \times 16 \\
   \Rightarrow \left( {{2^{2x - y}}} \right) \times \left( {{2^{x + y}}} \right) = {2^5} \times {2^4} \cdots \cdots \left( 3 \right) \\
 $
Now we will use the law of exponents which is given by ${a^m} \times {a^n} = {a^{m + n}}$ on both sides of equation $\left( 3 \right)$. Therefore, we get
$
  {2^{2x - y + x + y}} = {2^{5 + 4}} \\
   \Rightarrow {2^{3x}} = {2^9} \\
   \Rightarrow 3x = 9\quad \left[ {\because {a^x} = {a^y} \Rightarrow x = y} \right] \\
   \Rightarrow x = \dfrac{9}{3} \\
   \Rightarrow x = 3 \\
 $
Now we will substitute the value $x = 3$ in either equation $\left( 1 \right)$ or equation $\left( 2 \right)$ to find the value of $y$. Let us substitute the value $x = 3$ in equation $\left( 1 \right)$. Therefore, we get
$
  {2^{2\left( 3 \right) - y}} = 32 \\
   \Rightarrow {2^{6 - y}} = {2^5} \\
   \Rightarrow 6 - y = 5\quad \left[ {\because {a^x} = {a^y} \Rightarrow x = y} \right] \\
   \Rightarrow y = 6 - 5 \\
   \Rightarrow y = 1 \\
 $
Now we have $x = 3$ and $y = 1$. Therefore, ${x^2} + {y^2} = {\left( 3 \right)^2} + {\left( 1 \right)^2} = 9 + 1 = 10$.
Therefore, if ${2^{2x - y}} = 32$ and ${2^{x + y}} = 16$ then ${x^2} + {y^2}$ is equal to $10$.
So, option B is the right answer


Note: We can also solve this problem by using different methods. In one of the methods, we will use one property which is given by ${a^x} = {a^y} \Rightarrow x = y$. After using this property in both given equations, we will get two linear equations in $x$ and $y$. Then, we will use a simple elimination method to find the values of $x$ and $y$. Then, we will get the required value.