document.write( "Question 299113: Solve: log(x-3) + log(x+5) =1\r
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document.write( "My teacher said to use the log of products to get an equivalent quadratic equation. I don't really follow him on how to do it, though. \n" );
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Algebra.Com's Answer #215031 by nerdybill(7384) You can put this solution on YOUR website! What he means is to apply a \"log rule\": \n" ); document.write( "log a + log b = log (ab) \n" ); document.write( ". \n" ); document.write( "Solve: log(x-3) + log(x+5) =1 \n" ); document.write( "log(x-3)(x+5) =1 \n" ); document.write( "(x-3)(x+5) =10^1 \n" ); document.write( "x^2+5x-3x-15 = 10 \n" ); document.write( "x^2+2x-15 = 10 \n" ); document.write( "x^2+2x-25 = 0 \n" ); document.write( "Solve by applying the quadratic formula. Doing so yields: \n" ); document.write( "x = {4.1, -6.1} \n" ); document.write( "Plug the above back into the original equation to check whether they are viable answers. The negative answer will not work (it's an extraneous solution) throw it out leaving: \n" ); document.write( "x = 4.1 \n" ); document.write( ". \n" ); document.write( "Details of quadratic formula to follow: \n" ); document.write( "
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