document.write( "Question 747802: log3(x+6)+log3(x-6)-log3x=2
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Algebra.Com's Answer #853761 by MathTherapy(10801)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "log3(x+6)+log3(x-6)-log3x=2\r\n" );
document.write( "solve the logarithmic equation\r\n" );
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document.write( "It's heart-wrenching to see how the other person (@MATHLOVER) makes this problem so long and time-consuming! Why??\r\n" );
document.write( "In my opinion, she should learn how to solve problems much more efficiently.\r\n" );
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document.write( "\"log+%283%2C+%28x+%2B+6%29%29+%2B+log+%283%2C+%28x+-+6%29%29+-+log+%283%2C+%28x%29%29+=+2\"\r\n" );
document.write( "This log equation starts with 3 log arguments: a1 (x + 6), a2 (x - 6), and a3 (x), the smallest being log argument\r\n" );
document.write( "a2, or x - 6. Argument \"a2\" \"tells\" one that x - 6 MUST be > 0. So, x - 6 > 0_____x > 6 \r\n" );
document.write( "We now have: \"log+%283%2C+%28x+%2B+6%29%29+%2B+log+%283%2C+%28x+-+6%29%29+-+log+%283%2C+%28x%29%29+=+2\", with x > 6           \r\n" );
document.write( "                       \"log+%283%2C+%28%28%28x+%2B+6%29%28x+-+6%29%29%2F%28x%29%29%29+=+2\" ---- Applying \"log+%28b%2C+%28a1%29%29+%2B+log+%28b%2C+%28a2%29%29+-+log+%28b%2C+%28a3%29%29\" = \"log+%28b%2C+%28%28a1%2Aa2%29%29%2Fa3%29\"\r\n" );
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document.write( "                               \"%28x+%2B+6%29%28x+-+6%29%2Fx+=+3%5E2\" --- Applying \"c+=+b%5Ed\", when: \"log+%28b%2C+%28c%29%29+=+d\"\r\n" );
document.write( "                               \"%28x+%2B+6%29%28x+-+6%29%2Fx+=+9\"\r\n" );
document.write( "                                  \"%28x%5E2+-+36%29%2Fx+=+9\"\r\n" );
document.write( "                                    \"x%5E2+-+36+=+9x\" --- Cross-multiplying\r\n" );
document.write( "                                \"x%5E2+-+9x+-+36+=+0\"\r\n" );
document.write( "                           (x - 12)(x + 3) = 0\r\n" );
document.write( "                            x - 12 = 0       OR      x + 3 = 0\r\n" );
document.write( "                                 x = 12      OR          x = - 3 \r\n" );
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document.write( "As stated above, x MUST be > 6, so ONLY x = 12 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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