document.write( "Question 457945: Solve log base 2 (x)+ log base 2 (2x+1)=0 \n" ); document.write( "

Algebra.Com's Answer #314112 by robertb(5830) $\"\"$ $\"About$

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$\"log%282%2C+x%29+%2B+log%282%2C+2x%2B1%29+=+0\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"log%282%2C+x%282x%2B1%29%29+=+0\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"x%282x%2B1%29+=+2%5E0\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"2x%5E2+%2B+x+-+1+=+0\"$ \r
\n" ); document.write( "\n" ); document.write( "==> (2x-1)(x+1) = 0 ==> x = 1/2, -1.\r
\n" ); document.write( "\n" ); document.write( "Reject x = -1 (will not satisfy the original equation).\r
\n" ); document.write( "\n" ); document.write( "Hence the final answer is x = 1/2. \n" ); document.write( "

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