Question

If 2^x+y = 2^2x-Y = √8 then the respective values of x and y are_​

Answer 1

Answer:

x=1and y=1/2

Step-by-step explanation:

explained in the pic

Answer 2

Answer:

The value of x = 1 and the value of y = 1/2

Step-by-step explanation:

Given,

$2^{x+y} = 2^{2x-y} = \sqrt{8}$

To find,

The values of x and y

Recall the concept,

If$a^m = a^n$, then m = n  ---------------------(A)

$\sqrt{a} = a^{1/2}$ --------------(B)

Solution:

We have $\sqrt{8} = 8^{1/2} = 2^{3X1/2} = 2^{3/2}$ (by the identity (B))

Since $2^{x+y} =\sqrt{8}$ ⇒ $2^{x+y} = 2^{3/2},$

Then by equation (A), we get

x+y = 3/2 -------------(1)

Again

Again since$2^{2x-y} = \sqrt{8}$ ⇒$2^{2x-y} = 2^{3/2}$, then by equation (A) we get

2x - y = 3/2 ----------(2)

Adding equation (1) and (2) we get

3x = 3/2 +3/2

3x = 6/2 = 3

x = 1

Substitute the value of x = 1 in equation (1) we get

x+ y = 3/2

1+y = 3/2

y = 3/2 -1

y = 1/2

The value of x = 1 and the value of y = 1/2

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