document.write( "Question 473707: I need help solving for x in the following equation log(x+3) + log(x-1) = log 5 \n" ); document.write( "
Algebra.Com's Answer #324920 by Theo(13342)![]() ![]() You can put this solution on YOUR website! equation is: \n" ); document.write( "log(x+3) + log(x-1) = log(5) \n" ); document.write( "since log(x) + log(y) = log(x*y), this equation becomes: \n" ); document.write( "log((x+3)*(x-1)) = log(5) \n" ); document.write( "this will be true if (x+3)*(x-1) = 5 \n" ); document.write( "multiply the factors together to get: \n" ); document.write( "x^2 + 3x - x - 3 = 5 \n" ); document.write( "this becomes x^2 + 2x - 8 = 0 \n" ); document.write( "this can be factored to get: \n" ); document.write( "(x+4)*(x-2) = 0 \n" ); document.write( "this leads to: \n" ); document.write( "x = -4 or x = 2. \n" ); document.write( "when x = -4, the original equation becomes: \n" ); document.write( "log(-1) + log(-5) = log(5) \n" ); document.write( "those logs don't work because you can't get the log of a negative number. \n" ); document.write( "however, if you use the laws of logarithms to combine the 2 logs on the left to one log, then you get: \n" ); document.write( "log((-1)*(-5)) = log(5) which becomes log(5) = log(5) which is true. \n" ); document.write( "when x = 2, the original equation becomes: \n" ); document.write( "log(5) + log(1) = log(5) \n" ); document.write( "combine the logs on the left hand side and you get: \n" ); document.write( "log(5*1) = log(5) which becomes log(5) = log(5) which is true. \n" ); document.write( "the solution to this equation is: \n" ); document.write( "x = -4 or x = 2. \n" ); document.write( "there is that funny little wrinkle where: \n" ); document.write( "log(-1) + log(-5) = log(5) is not valid, but log((-1)*(-5)) = log(5) is valid.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |