Question

Solve for x and y:

`8^(3 - x/2) - 2^(2y) = 0` and `(sqrt(32))^x ÷ 2^y = 4`

Answer 1

Given the system of equations:

`8^(3 - x/2) - 2^(2y) = 0` and `(sqrt(32))^x ÷ 2^y = 4`

Step-wise calculation:

1. Rewrite the first equation:

`8^(3 - x/2) = 2^(2y)`

Since (8 = 2³), substitute:

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

Simplify the left side:

`2^(3 xx (3 - x/2)) = 2^(2y)`

`2^(9 - (3x)/2) = 2^(2y)`

Equate exponents of 2:

`9 - (3x)/2 = 2y`

2. Rewrite the second equation:

`(sqrt(32))^x/(2^y) = 4`

Note that `sqrt(32) = sqrt(2^5)`

`sqrt(32) = 2^(5/2)`

And 4 = 2²,

So substitute:

`((2^(5/2))^x)/2^y = 2^2`

Simplify numerator:

`2^((5x)/2)/2^y = 2^2`

`2^((5x)/2 - y) = 2^2`

Equate exponents of 2:

`(5x)/2 - y = 2`

3. Now, solve the system:

`9 - (3x)/2 = 2y`

`(5x)/2 - y = 2`

From the first, express (y):

`2y = 9 - (3x)/2`

⇒ `y = (9 - (3x)/2)/2`

⇒ `y = 9/2 - (3x)/4`

Substitute into the second:

`(5x)/2 - (9/2 - (3x)/4) = 2`

Simplify:

`(5x)/2 - 9/2 + (3x)/4 = 2`

Multiply everything by 4 to clear denominators:

`4 xx (5x)/2 - 4 xx 9/2 + 4 xx (3x)/4 = 4 xx 2`

10x – 18 + 3x = 8

13x – 18 = 8

13x = 26

x = 2

Now find (y):

`y = 9/2 - (3 xx 2)/4`

`y = 9/2 - 6/4`

`y = 9/2 - 3/2`

`y = 6/2`

y = 3