document.write( "Question 173580: Log6x+log6(x-5)=2
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Algebra.Com's Answer #854704 by MathTherapy(10858) $\"\"$ $\"About$

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document.write( "Log6x+log6(x-5)=2\r\n" );
document.write( "Check for inadmissible roots\r\n" );
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document.write( "Respondent, nycsub_teacher(90) provides the following solution,\r\n" );
document.write( "\"Log6x+log6(x-5)=2\r\n" );
document.write( "We write it like this:\r\n" );
document.write( "log_6(x/(x-5) = 2\r\n" );
document.write( "x/(x-5) = 6^2\r\n" );
document.write( "x/(x-5) = 36,\" and then asks, \"Can you finish now?\"\r\n" );
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document.write( "Whether or not the person can finish doesn't really matter, because the solution that'll be derived from the respondent's attempt, will be WRONG!!\r\n" );
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document.write( "Correct way: \r\n" );
document.write( "The smaller of the 2 log-variable arguments, x - 5 MUST be > 0, and so, x - 5 > 0____x > 5\r\n" );
document.write( "So, we get: , with x > 5\r\n" );
document.write( "                                  ---- Applying  = \r\n" );
document.write( "                                             -- Converting from LOGARITHMIC form to EXPONENTIAL form \r\n" );
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document.write( "                                 (x - 9)(x + 4) = 0 ---- FACTORIZING the trinomial\r\n" );
document.write( "                                  x - 9 = 0      or     x + 4 = 0 ---- Setting each factor equal to 0 (zero)\r\n" );
document.write( "                                        x = 9      or            x = - 4 (IGNORE)\r\n" );
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document.write( "As one of the 2 values, x = - 4, is NOT > 5, it is an INADMISSIBLE root, or an EXTRANEOUS solution.\r\n" );
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document.write( "Therefore, the SOLE ADMISSIBLE/VALID/ACCEPTABLE solution to this equation is x = 9.

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