document.write( "Question 193653: log(x+1)+log(x-2)=1 \n" ); document.write( "

Algebra.Com's Answer #145349 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
In general:
\n" ); document.write( " $\"log%28a%29+%2B+log%28b%29+=+log%28ab%29\"$ so,
\n" ); document.write( " $\"log%28%28x%2B1%29%29%2Blog%28%28x-2%29%29=+log%28%28x%2B1%29%2A%28x-2%29%29\"$
\n" ); document.write( "also, $\"1+=+log%2810%29\"$
\n" ); document.write( " $\"log%28%28x%2B1%29%2A%28x-2%29%29+=+log%2810%29\"$
\n" ); document.write( "If the logs are equal, the antilogs are equal
\n" ); document.write( " $\"%28x%2B1%29%2A%28x-2%29+=+10\"$
\n" ); document.write( " $\"x%5E2+%2B+x+-+2x+-+2+-+10+=+0\"$
\n" ); document.write( " $\"x%5E2+-+x+-+12+=+0\"$
\n" ); document.write( " $\"%28x+-+4%29%2A%28x+%2B+3%29+=+0\"$
\n" ); document.write( " $\"x+=+4\"$
\n" ); document.write( " $\"x+=+-3\"$ this result doesn't work, becase it
\n" ); document.write( "gives me a $\"log%28-5%29\"$ and no log can give a
\n" ); document.write( "negative answer
\n" ); document.write( "check:
\n" ); document.write( " $\"x+=+4\"$
\n" ); document.write( " $\"log%28x%2B1%29%2Blog%28x-2%29=+1\"$
\n" ); document.write( " $\"log%285%29+%2B+log%282%29+=+1\"$
\n" ); document.write( " $\".6989700043+%2B+.3010299957+=+1\"$
\n" ); document.write( " $\".9999999997+=+1\"$ (inaccuracy is due to rounding off)
\n" ); document.write( "OK
\n" ); document.write( " \n" ); document.write( "

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