document.write( "Question 105760: Solve log(x-1)- log6 = log(x-2)- logx \n" ); document.write( "

Algebra.Com's Answer #854648 by MathTherapy(10858) $\"\"$ $\"About$

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document.write( "Solve log(x-1)- log6 = log(x-2)- logx\r\n" );
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document.write( "@HyperBrain(694)'s answer, below, is WRONG, and DOUBLE-WRONG for the negative value of x, since x MUST BE > 2.\r\n" );
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document.write( "The smallest log variable-argument, x - 2 MUST BE > 0, so x > 2. We then get:\r\n" );
document.write( ", with x > 2.\r\n" );
document.write( "           ----- Applying  = \r\n" );
document.write( "                      ----- Applying e = f, since  = \r\n" );
document.write( "                  x(x - 1) = 6(x - 2) --- Cross-multiplying\r\n" );
document.write( "                     \r\n" );
document.write( "      \r\n" );
document.write( "            \r\n" );
document.write( "          (x - 4)(x - 3) = 0\r\n" );
document.write( "           x - 4 = 0       or       x - 3 = 0 ---- Setting each factor equal to 0\r\n" );
document.write( "                 x = 4       or            x = 3\r\n" );
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document.write( "Both values of x are VALID/ACCEPTABLE, since both are > 2.

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