document.write( "Question 59509: 
Solve for x
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document.write( "(log3x)2 + log3x2 + 1 = 0 \n" );
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Algebra.Com's Answer #40848 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "(log3x)2 + log3x2 + 1 = 0\r\n" );
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document.write( "On the second term use the rule\r\n" );
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document.write( "logban = n·logba\r\n" );
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document.write( "(log3x)2 + 2·log3x + 1 = 0\r\n" );
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document.write( "Let the letter w, or any letter other\r\n" );
document.write( "than x, be such that\r\n" );
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document.write( "w = log3x\r\n" );
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document.write( "Then the above equation becomes\r\n" );
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document.write( "        (w)2 + 2·w + 1 = 0\r\n" );
document.write( "           w2 + 2w + 1 = 0   \r\n" );
document.write( "Factoring\r\n" );
document.write( "        (w + 1)(w + 1) = 0\r\n" );
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document.write( "So the solutions w = -1 and w = -1 are \r\n" );
document.write( "equal.\r\n" );
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document.write( "Now since w = log3x, then w = -1 becomes\r\n" );
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document.write( "         log3x = -1\r\n" );
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document.write( "Now use the rule of logarithms that says:\r\n" );
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document.write( "logba = c can be rewritten as a = bc \r\n" );
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document.write( "to rewrite log3x = -1 as\r\n" );
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document.write( "           x = 3-1\r\n" );
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document.write( "or         x = 1/3\r\n" );
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document.write( "Edwin
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