If log2(3𝑥 + 𝑦) = 1 and log2 (

Question

If log2(3𝑥 + 𝑦) = 1 and log2 (

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).

Answer 2

To find the values of x and y, we will solve the given equations step-by-step.

Step 1: Use the first equation, log2(3𝑥 + 𝑦) = 1, to eliminate the logarithm.

Rewrite the equation in exponential form:
2^1 = 3𝑥 + 𝑦
2 = 3𝑥 + 𝑦

Step 2: Use the second equation, log2 (𝑥𝑦) = −2, to eliminate the logarithm.

Rewrite the equation in exponential form:
2^(-2) = 𝑥𝑦
1/4 = 𝑥𝑦

Step 3: Now we have the system of equations:
2 = 3𝑥 + 𝑦
1/4 = 𝑥𝑦

To solve this system of equations, we can use the method of substitution.

Step 4: Solve the second equation, 1/4 = 𝑥𝑦, for 𝑥 or 𝑦.
Let's solve for 𝑥:
1/4 = 𝑥𝑦
1/4𝑦 = 𝑥

Step 5: Substitute the value of 𝑥 in the first equation, 2 = 3𝑥 + 𝑦.

2 = 3(1/4𝑦) + 𝑦
2 = 3/4𝑦 + 𝑦

Step 6: Combine like terms:

Multiply the first term by 4/4 to get a common denominator:

2 = (3/4𝑦) * 4/4 + 𝑦
2 = 12/4𝑦 + 𝑦
2 = (12 + 4𝑦^2)/4𝑦
8𝑦 = 12 + 4𝑦^2

Step 7: Rearrange the equation to form a quadratic equation:

4𝑦^2 - 8𝑦 + 12 = 0

Step 8: Simplify the equation by dividing it by 4:

𝑦^2 - 2𝑦 + 3 = 0

Step 9: Solve the quadratic equation using the quadratic formula:

𝑦 = (-(-2) ± √((-2)^2 - 4(1)(3))) / (2(1))

𝑦 = (2 ± √(4 - 12)) / 2

𝑦 = (2 ± √(-8)) / 2

Since the value inside the square root is negative, there are no real solutions for 𝑦. Therefore, none of the given options (A, B, C, D) is correct.

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