We can use the properties of logarithms to rewrite the given expressions in a more useful form. For example, from the first equation, we have:
log2(3𝑥 + 𝑦) = 1
2^1 = 3𝑥 + 𝑦
1 = 3𝑥 + 𝑦
Similarly, from the second equation, we have:
log2(𝑥/𝑦) = -2
2^-2 = 𝑥/𝑦
1/4 = 𝑥/𝑦
𝑥 = (1/4)𝑦
Now we can substitute this expression for 𝑥 into the first equation and solve for 𝑦:
1 = 3𝑥 + 𝑦
1 = 3(1/4)𝑦 + 𝑦
1 = (3/4)𝑦 + 𝑦
1 = (7/4)𝑦
𝑦 = 4/7
Substituting this value back into the expression for 𝑥, we get:
𝑥 = (1/4)𝑦
𝑥 = (1/4)(4/7)
𝑥 = 1/7
Therefore, the answer is B: 𝑥 = 2/7, 𝑦 = 4/7.
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
1 answer