SOLUTION: Q: If a, b and y are positive real numbers with such that none of them is equal to 1 and b^2a+6 = y^2, which of these must be true? A) y = b^a+3 B) 2a + 6 = 2 C) b = y D)

Question 1041211: Q: If a, b and y are positive real numbers with such that none of them is equal to 1 and b^2a+6 = y^2, which of these must be true?
A) y = b^a+3
B) 2a + 6 = 2
C) b = y
D) b = 2
E) b(2a + 6) = 2y
Explain too plz :)
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
b^2a+6 = y^2,
y= sqrt (b^(2a +6)), is y=b^(a+3) and that works.
let a=3 and b=2.
2^(12)=y^2
y=2^6=64
y=b^(a+3)=2^6=64
y=sqrt(b^2a+6)=[b^(2a+6)]^(1/2), and that is b^(a+3) ANSWER
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2a+6=2.
If b=y, this is true. If it is b^(2a+6)=b^2, then it works in all cases.
Suppose, however, b=2 and a=3 Then 2^(12)=y^2, and y^2=4096 and y=64.
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b(2a+6)=2y is a misunderstanding of raising to a power. b^2a+6 is raising to a power, not multiplying.
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b=2. It can be, but they are all positive real numbers except 1.
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b=y only if 2a+6=2. Since they can be positive real numbers except 1, that isn't "must"
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