document.write( "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? \r
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document.write( "A) y = b^a+3
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document.write( "B) 2a + 6 = 2
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document.write( "C) b = y
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document.write( "D) b = 2
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document.write( "E) b(2a + 6) = 2y\r
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document.write( "Explain too plz :) \n" );
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Algebra.Com's Answer #656171 by Boreal(15235)![]() ![]() You can put this solution on YOUR website! b^2a+6 = y^2, \n" ); document.write( "y= sqrt (b^(2a +6)), is y=b^(a+3) and that works. \n" ); document.write( "let a=3 and b=2. \n" ); document.write( "2^(12)=y^2 \n" ); document.write( "y=2^6=64 \n" ); document.write( "y=b^(a+3)=2^6=64 \n" ); document.write( "y=sqrt(b^2a+6)=[b^(2a+6)]^(1/2), and that is b^(a+3) ANSWER \n" ); document.write( "--------------------- \n" ); document.write( "2a+6=2. \n" ); document.write( "If b=y, this is true. If it is b^(2a+6)=b^2, then it works in all cases. \n" ); document.write( "Suppose, however, b=2 and a=3 Then 2^(12)=y^2, and y^2=4096 and y=64. \n" ); document.write( "--------------------- \n" ); document.write( "b(2a+6)=2y is a misunderstanding of raising to a power. b^2a+6 is raising to a power, not multiplying. \n" ); document.write( "-------------------- \n" ); document.write( "b=2. It can be, but they are all positive real numbers except 1. \n" ); document.write( "------------------- \n" ); document.write( "b=y only if 2a+6=2. Since they can be positive real numbers except 1, that isn't \"must\" \n" ); document.write( "-------------------- \n" ); document.write( " \n" ); document.write( " |