document.write( "Question 972538: Solve for x. The only solutions of the form
what is b?\r
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document.write( "What I did: ln((x-3)(x+2))
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document.write( " ln(x^2 -x-6)
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document.write( " a= 1 b = -1 c = -6 \r
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document.write( "that would come out to \r
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document.write( "Where did I go wrong? \n" );
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Algebra.Com's Answer #594836 by Boreal(15235) You can put this solution on YOUR website! You need to convert the problem to a quadratic form. You had an ln(quadratic form). One thing to do is to graph it, and just see what it looks like.\r \n" ); document.write( "\n" ); document.write( "The way I like to deal with ln is raise everything to the e power. Then you have a quadratic =1, and you can deal with it accordingly.\r \n" ); document.write( "\n" ); document.write( "Raise everything to the e power \n" ); document.write( "(x-3)(x+2)=1\r \n" ); document.write( "\n" ); document.write( "x^2-x-7=0 \n" ); document.write( "[1 +/- sqrt (1+28)]/2 \r \n" ); document.write( "\n" ); document.write( "x=3.19 and -2.19\r \n" ); document.write( "\n" ); document.write( "b=28\r \n" ); document.write( "\n" ); document.write( "substitute in ln (0.19)(5.19) is very close to 1, and ln (1)=0\r \n" ); document.write( "\n" ); document.write( "ln (-5.19)((-0.19) is the same,\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |