document.write( "Question 132012: how can i solve \r
\n" ); document.write( "\n" ); document.write( "logX + logX + 1 = log12 ?
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Algebra.Com's Answer #96404 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
logX + logX + 1 = log12
\n" ); document.write( "2logX+1=log12
\n" ); document.write( "1=log12-2logx
\n" ); document.write( "1=log12-logx^2
\n" ); document.write( "\"1=log%2812%2Fx%5E2%29+\"\r
\n" ); document.write( "\n" ); document.write( "\"1=log%2810%2C12%2Fx%5E2%29\"\r
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\n" ); document.write( "\n" ); document.write( "By basic definition of logarithms:
\n" ); document.write( "\"10%5E1=+12%2Fx%5E2\"
\n" ); document.write( "\"10x%5E2+=+12\"
\n" ); document.write( "\"x%5E2=12%2F10=6%2F5\"\r
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\n" ); document.write( "\n" ); document.write( "Since x must be a positive number,
\n" ); document.write( "\"x=+sqrt%286%2F5%29\"
\n" ); document.write( "\"x=sqrt%2830%29%2F5\"\r
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\n" ); document.write( "\n" ); document.write( "R^2
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