document.write( "Question 260514: log(x+3)=1-logx \n" ); document.write( "

Algebra.Com's Answer #191890 by drk(1908) $\"\"$ $\"About$

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here is the question
\n" ); document.write( " $\"log%28x%2B3%29=1-logx\"$
\n" ); document.write( "step 12 - add log(x) to both side to get
\n" ); document.write( " $\"log%28x%2B3%29+%2B+log%28x%29+=+1\"$
\n" ); document.write( "use rules of logs to get
\n" ); document.write( " $\"%28log%28x%5E2%2B3x%29%29+=+1\"$
\n" ); document.write( "rewrite as exponents as
\n" ); document.write( " $\"10%5E1+=+x%5E2+%2B+3x\"$
\n" ); document.write( "set = 0 and solve for x as
\n" ); document.write( " $\"x%5E2+%2B+3x+-10+=+0\"$
\n" ); document.write( "factor
\n" ); document.write( " $\"%28x%2B5%29%28x-2%29+=+0\"$
\n" ); document.write( "solve
\n" ); document.write( "x+5 = 0
\n" ); document.write( "x = -5
\n" ); document.write( "x-2 = 0
\n" ); document.write( "x=2
\n" ); document.write( "we can't use x = -5, so the only answer is x = 2. \n" ); document.write( "

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