document.write( "Question 867265: Solve
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document.write( "a)log(base 2)(x + 1)+ log(base 2)(x-1) = 3
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document.write( "b)log(base 2)(x^2-1) = 3
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document.write( "c)explain why the equations in part a and b are not equivalent
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document.write( "Can you please help me out? Thanks so much in advance:) \n" );
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Algebra.Com's Answer #522918 by edjones(8007)![]() ![]() You can put this solution on YOUR website! a) \n" ); document.write( "log(base 2)(x + 1)+ log(base 2)(x-1) = 3 \n" ); document.write( "2^log(base 2)(x + 1)*2^log(base 2)(x-1)=2^3 \n" ); document.write( "(x+1)(x-1)=8 \n" ); document.write( "x^2=9 \n" ); document.write( "x=3 \n" ); document.write( ". \n" ); document.write( "b) \n" ); document.write( "log(base 2)(x^2-1) = 3 \n" ); document.write( "2^log(base 2)(x^2-1) =2^3 \n" ); document.write( "x^2-1=8 \n" ); document.write( "x^2=9 \n" ); document.write( "x=3, x=-3 \n" ); document.write( ". \n" ); document.write( "c) x can't equal -1 in a) because a log cannot equal 0 or be negative. \n" ); document.write( ". \n" ); document.write( "Ed\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |