document.write( "Question 22137: log2x+log2(x-6)=4 \n" ); document.write( "

Algebra.Com's Answer #854133 by MathTherapy(10801) $\"\"$ $\"About$

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document.write( "log2x+log2(x-6)=4\r\n" );
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document.write( "The solution, (x = 8, or x = - 2) by the other person who responded, is PARTIALLY WRONG!!\r\n" );
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document.write( "The SMALLER log argument, x - 6, MUST be > 0. So, x - 6 > 0 ===> x > 6.\r\n" );
document.write( "We then have: , with x > 6.\r\n" );
document.write( "                                          ----- Applying  = \r\n" );
document.write( "                                                   --- Converting to EXPONENTIAL form <=== Note that the other person has  i/o ,\r\n" );
document.write( "                                                                                                                                                       but both have the same value, 16\r\n" );
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document.write( "                                       (x - 8)(x + 2) = 0 \r\n" );
document.write( "                                                      x - 8 = 0     OR    x + 2 = 0 ---- Setting each FACTOR equal to 0\r\n" );
document.write( "                                                            x = 8     OR           x = - 2\r\n" );
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document.write( "The x-value, 8, is > 6, but - 2 is NOT. This makes - 2 an EXTRANEOUS solution!! So, x = 8 is the only VALID/ACCEPTABLE solution!!

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