document.write( "Question 734837: solve log3(x-3) + log3(x+2) = log3 6 \n" ); document.write( "

Algebra.Com's Answer #853762 by MathTherapy(10704) $\"\"$ $\"About$

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document.write( "solve log3(x-3) + log3(x+2) = log3 6\r\n" );
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document.write( "The other person's statement that if \"all log bases are the same (namely 3)\", then: x-3 + x+2 = 6\r\n" );
document.write( "                                                                                    2x - 1 = 6 and\r\n" );
document.write( "                                                                                    2x = 7 and x = 3.5, is a FALLACY!\r\n" );
document.write( "He couldn't be more WRONG!! If he'd checked, he would've seen that 3.5 doesn't: make sense\r\n" );
document.write( "                                                                                make the equation TRUE!\r\n" );
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document.write( "The smaller log argument, x - 3 \"tells\" one that x - 3 MUST be > 0. So, x - 3 > 0____x > 3\r\n" );
document.write( "We then get: , with X > 3\r\n" );
document.write( "                   ----- Applying  = \r\n" );
document.write( "                    (x - 3)(x + 2) = 6 ------ Applying c = d, if \r\n" );
document.write( "                          \r\n" );
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document.write( "                    (x - 4)(x + 3) = 0\r\n" );
document.write( "                     x - 4 = 0      OR     x + 3 = 0    \r\n" );
document.write( "                         x = 4      OR     x = - 3\r\n" );
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document.write( "As stated above, x MUST be > 3, so ONLY x = 4 is ACCEPTABLE. This makes x = - 3, an EXTRANEOUS solution to this\r\n" );
document.write( "equation, and is therefore IGNORED/REJECTED.

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