Question

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

Answer 1

Hint:
It is in question that If ${2^{2x - y}} = 32$ and ${2^{x + y}} = 16$ .
Then what is the value of ${x^2} + {y^2}$ .
We can write 32 as ${2^5}$ and 16 as ${2^4}$ .
First, we will find the values of x and y by comparing the powers of ${2^{2x - y}} = {2^5}$ and ${2^{x + y}} = {2^4}$.
Thus, after finding the values of x and y, find the required value of ${x^2} + {y^2}$.

Complete step by step solution:
It is in question that If ${2^{2x - y}} = 32$ and ${2^{x + y}} = 16$.
Then what is the value of ${x^2} + {y^2}$.
Since, ${2^{2x - y}} = 32$
We can also write 32 as ${2^5}$
 $\Rightarrow {2^{2x - y}} = {2^5}$
As the denominator same, so we can compare the powers,
 $\Rightarrow 2x - y = 5$ (I)
Now, ${2^{x + y}} = 16$
We can also write 16 as ${2^4}$
 $\Rightarrow {2^{x + y}} = {2^4}$
Since, the denominator is so we can compare the powers,
 $\Rightarrow x + y = 4$ (II)
Now, add equation (I) and equation (II), we get,
 $\Rightarrow 2x - y + x + y = 5 + 4$
 $\Rightarrow 3x = 9$
 $\Rightarrow x = \dfrac{9}{3}$
 $\Rightarrow x = 3$
Now, substitute the value of x in equation (I)
 $\Rightarrow 2x - y = 5$
 $\Rightarrow 2\left( 3 \right) - y = 5$
 $\Rightarrow 6 - y = 5$
 $\Rightarrow y = 6 - 5$
 $\Rightarrow y = 1$
Thus, value of $x = 3$ and $y = 1$
Now,
 ${x^2} + {y^2} = {3^2} + {1^2}$
 ${x^2} + {y^2} = 9 + 1$
 ${x^2} + {y^2} = 10$

Hence, the value of ${x^2} + {y^2} = 10$

Note: The above question can be solved by using another method i.e. by logarithm.
It is in question that If ${2^{2x - y}} = 32$ and ${2^{x + y}} = 16$ .
Then what is the value of ${x^2} + {y^2}$ .
Since, ${2^{2x - y}} = 32$
We can also write 32 as ${2^5}$
 $\Rightarrow {2^{2x - y}} = {2^5}$
Now, take log on both sides of the equation.
 $\Rightarrow \log {2^{2x - y}} = \log {2^5}$
Using property of logarithm $\log {b^a} = a\log b$ on the above equation
 $\Rightarrow 2x - y\log 2 = 5\log 2$
 $\Rightarrow 2x - y = 5$ (I)
Now, ${2^{x + y}} = 16$
We can also write 16 as ${2^4}$
 $\Rightarrow {2^{x + y}} = {2^4}$
Now, take log both sides of the equation.
 $\Rightarrow \log {2^{x + y}} = \log {2^4}$
Using property of logarithm $\log {b^a} = a\log b$ on the above equation, we get,
 $\Rightarrow x + y\log 2 = 4\log 2$
 $\Rightarrow x + y = 4$ (II)
Now, add equation (I) and equation (II), we get,
 $\Rightarrow 2x - y + x + y = 5 + 4$
 $\Rightarrow 3x = 9$
 $\Rightarrow x = \dfrac{9}{3}$
 $\Rightarrow x = 3$
Now, substitute the value of x in equation (I)
 $\Rightarrow 2x - y = 5$
 $\Rightarrow 2\left( 3 \right) - y = 5$
 $\Rightarrow 6 - y = 5$
 $\Rightarrow y = 6 - 5$
 $\Rightarrow y = 1$
Thus, value of $x = 3$ and $y = 1$
Now,
 ${x^2} + {y^2} = {3^2} + {1^2}$
 ${x^2} + {y^2} = 9 + 1$
 ${x^2} + {y^2} = 10$
Hence, the value of ${x^2} + {y^2} = 10$.