If log2(3𝑥 + 𝑦) = 1 and log2 (

Question

𝑥
𝑦
) = −2, find
the values 𝑜𝑓 𝑥 𝑎𝑛𝑑 𝑦
A. 𝑥 = 8
5 , 𝑦 = −32
B. 𝑥 = 2
7 , 𝑦 = 8
7
C. 𝑥 = −2, 𝑦 = 8
D. 𝑥 = 1, 𝑦 = 4

Answer 1

We can rewrite the first equation as 2 = log2(3x+y) and the second equation as -2 = log2(x/y). Using the properties of logarithms, we can rewrite the second equation as 2 = log2(y/x).

Now we can set up a system of equations:

2 = log2(3x+y)
2 = log2(y/x)

Using the inverse property of logarithms, we can rewrite the first equation as 2^2 = 3x+y, or 4 = 3x+y.

Using the change of base formula, we can rewrite the second equation as log2(y) - log2(x) = log2(y/x) = 2.

We can then use substitution to solve for y:

log2(y) - log2((4-y)/3) = 2

log2(y) - log2(4-y) + log2(3) = 2

log2(y(4-y)) = log2(12)

y(4-y) = 12

4y - y^2 = 12

y^2 -4y + 12 = 0

Using the quadratic formula, we get y = 2 +/- 2i√2

Therefore, there are no real solutions for y.

If we go back to the equation 4 = 3x+y, we can solve for x:

4 = 3x+y

4 = 3x + (2 + 2i√2)

2 = 3x + i√8

2/3 = x + i√8/3

Therefore, x = 2/3 and y = 2 + 2i√2.

So the answer is none of the above (E).