document.write( "Question 35350: Question goes like this....\r
\n" ); document.write( "\n" ); document.write( "Solve Equation.\r
\n" ); document.write( "\n" ); document.write( "log x + log (x+2) = log 3
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Algebra.Com's Answer #21462 by narayaba(40)\"\" \"About 
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log x + log (x+2) = log 3
\n" ); document.write( "log(a) + log(b) = log(a*b)
\n" ); document.write( "log(x) + log(x+2) = log(x*(x+2)) = log3
\n" ); document.write( "log(x*(x+2)) = log3
\n" ); document.write( "log on the right and left are of the same base
\n" ); document.write( "therefore x(x+2) = 3
\n" ); document.write( "x^2 +2x -3 = 0 factorising
\n" ); document.write( "(x+3)(x-1) = 0
\n" ); document.write( "x=-3 or x = 1\r
\n" ); document.write( "\n" ); document.write( "log of negative number is not possible hence x =1
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