document.write( "Question 613455: How do you find log(x+21)+logx=2? \n" ); document.write( "
Algebra.Com's Answer #854655 by MathTherapy(10858)\"\" \"About 
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document.write( "How do you find log(x+21)+logx=2?\r\n" );
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document.write( "Respondent @radh(108) went along a very long path, and arrived at a PARTIALLY-CORRECT solution set. \r\n" );
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document.write( "The smaller of the 2 logarithmic variable-arguments, x, MUST be > 0, so x > 0\r\n" );
document.write( "We then get: log (x + 21) + log (x) = 2, with x > 0\r\n" );
document.write( "  log x(x + 21) = 2 ---- Applying \"log+%28b%2C+%28c%29%29+%2B+log+%28b%2C+%28d%29%29\" = \"log+%28b%2C+%28cd%29%29\"\r\n" );
document.write( "\"log+%28%28x%5E2+%2B+21x%29%29+=+2\"\r\n" );
document.write( "          \"x%5E2+%2B+21x+=+10%5E2\" ---- Converting to EXPONENTIAL form\r\n" );
document.write( "          \"x%5E2+%2B+21x+=+100\"\r\n" );
document.write( " \"x%5E2+%2B+21x+-+100+=+0\"\r\n" );
document.write( "(x - 4)(x + 25) = 0 --- Factoring above TRINOMIAL\r\n" );
document.write( "  x - 4 = 0     or       x + 25 = 0\r\n" );
document.write( "        x = 4     or                x = - 25 (IGNORE)\r\n" );
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document.write( "The value, 4, for x, is > 0, but - 25 is NOT! Therefore, ONLY x = 4 is valid/ACCEPTABLE, while - 25 is an EXTRANEOUS solution.
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