document.write( "Question 255027: Hi :)
\n" ); document.write( "I’m new to logarithms and not sure if I’m on the right track with this one. Hoping you can help?... thanks heaps
\n" ); document.write( "log₃x + log₃(x + 2) = 1 (find x)
\n" ); document.write( "My attempt:
\n" ); document.write( "> log₃((x )(x+2)) = 1
\n" ); document.write( "> log₃(x²+2x) = 1
\n" ); document.write( "> x² + 2x = 3¹
\n" ); document.write( "> x² + 2x – 3 = 0
\n" ); document.write( "> (x - 1)(x + 3) = 0
\n" ); document.write( "> so x = +1 & -3
\n" ); document.write( "Is that right???
\n" ); document.write( "thanks again \n" ); document.write( "

Algebra.Com's Answer #187345 by nerdybill(7384) $\"\"$ $\"About$

You can put this solution on YOUR website!
log₃x + log₃(x + 2) = 1 (find x)
\n" ); document.write( "My attempt:
\n" ); document.write( "> log₃((x )(x+2)) = 1
\n" ); document.write( "> log₃(x²+2x) = 1
\n" ); document.write( "> x² + 2x = 3¹
\n" ); document.write( "> x² + 2x – 3 = 0
\n" ); document.write( "> (x - 1)(x + 3) = 0
\n" ); document.write( "> so x = +1 & -3
\n" ); document.write( ".
\n" ); document.write( "So far, so good!
\n" ); document.write( "At this point, you must \"test\" for \"extraneous\" solutions.
\n" ); document.write( "Notice that if x was equal to -3
\n" ); document.write( "log₃(x + 2) = log₃(-3 + 2) = log₃(-1)
\n" ); document.write( "You cannot take a log of a negative number --
\n" ); document.write( "therefore, -3 is an extraneous solution -- throw it out leaving:
\n" ); document.write( "x = 1
\n" ); document.write( "as your solution \n" ); document.write( "

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