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)\"\" \"About 
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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
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