Question 1041211
 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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