document.write( "Question 455693: solve log(x+20)-log(x+2)=logx \n" ); document.write( "

Algebra.Com's Answer #312847 by robertb(5830) $\"\"$ $\"About$

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$\"log%28%28%28x%2B20%29%2F%28x%2B2%29%29%29+=+logx\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"%28x%2B20%29%2F%28x%2B2%29+=+x\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"x%2B+20+=+x%28x%2B2%29\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"x%2B20+=+x%5E2+%2B+2x\"$ \r
\n" ); document.write( "\n" ); document.write( "==> $\"0+=+x%5E2+%2B+x++-+20\"$ \r
\n" ); document.write( "\n" ); document.write( "==> 0 = (x +5)(x - 4) ==> x = -5, 4.\r
\n" ); document.write( "\n" ); document.write( "-5 does not satisfy the original equation. Hence the final answer is x = 4. \n" ); document.write( "

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