document.write( "Question 70987: I need to solve this equation on a worksheet:\r
\n" ); document.write( "\n" ); document.write( "log X +log (X + 2) = log 3\r
\n" ); document.write( "\n" ); document.write( "Thank you!!!
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Algebra.Com's Answer #50709 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
\"log+X+%2Blog+%28X+%2B+2%29+=+log+3\"
\n" ); document.write( "\"log%28x%5E2%2B2x%29=log+3\"Combine logs using basic identity log(x)+log(y)=log(x*y)
\n" ); document.write( "\"cross%2810%29%5E%28log%28x%5E2%2B2x%29%29=cross%2810%29%5E%28log+3%29\"Cancel out logs by making the base 10 with the equation as a raised power (which undoes the logs with base 10)
\n" ); document.write( "\"x%5E2%2B2x=3\"Now you have an equation in which you can solve for x
\n" ); document.write( "\"%28x-1%29%28x%2B3%29=0\"Get everything to one side and factor it, then solve for x
\n" ); document.write( "\"x=1\" and \"x=-3\" are your solutions. But wait, we're not done yet. We need to check these answers
\n" ); document.write( "Check:
\n" ); document.write( "Let x=1
\n" ); document.write( "\"log+%281%29%2B+log+%283%29=log+3\"
\n" ); document.write( "\"log+%281%2A3%29+=+log+3\"
\n" ); document.write( "\"log+3=+log+3\" Answer works
\n" ); document.write( "Let x=-3
\n" ); document.write( "\"log+%28%28-3%29%29+%2B+log+%28%28-3%2B2%29%29=log+3\" You cannot take the log of a negative number (if you want a real number) so this solution does not work.
\n" ); document.write( "So the the only solution that works is x=1
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