document.write( "Question 1156244: Log3(1-x)=log3(x+16-x^2) \n" ); document.write( "
Algebra.Com's Answer #778940 by Edwin McCravy(20060) $\"\"$ $\"About$
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document.write( "We can't tell whether you mean the 3's to be the base of the logarithm\r\n" );
document.write( "like this:\r\n" );
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document.write( "or whether you mean the 3's to be multipliers and the logarithms understood\r\n" );
document.write( "to be common logs with understood base 10, like this:\r\n" );
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document.write( "If it's the first way then we drop the logs and get this:\r\n" );
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document.write( "If it's the second way then we drop the logs and get this:\r\n" );
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document.write( "But if it's that way we divide both sides by 3 and get\r\n" );
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document.write( "So, luckily here it doesn't matter which you meant but in other\r\n" );
document.write( "logarithm problems it would make a difference.  So be careful.\r\n" );
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document.write( "Get 0 on the right and descending order on the left:\r\n" );
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document.write( "But we must check in the original equation for extraneous solutions.\r\n" );
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document.write( "If it was the first way, we check to see if 5 is a solution:\r\n" );
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document.write( "It looks like it checks but it doesn't because logs cannot be taken of\r\n" );
document.write( "negative numbers in real number mathematics.  So 5 is extraneous.\r\n" );
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document.write( "If it was the first way, we check to see if -3 is a solution:\r\n" );
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document.write( "That checks.  So x = -3 is the only solution.\r\n" );
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document.write( "If you meant the other way, it still would be only x = -3.\r\n" );
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document.write( "Edwin
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