Question 613455
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How do you find log(x+21)+logx=2?
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Respondent @radh(108) went along a very long path, and arrived at a PARTIALLY-CORRECT solution set. 

The smaller of the 2 logarithmic variable-arguments, x, MUST be > 0, so x > 0
We then get: log (x + 21) + log (x) = 2, with x > 0
  log x(x + 21) = 2 ---- Applying {{{log (b, (c)) + log (b, (d))}}} = {{{log (b, (cd))}}}
{{{log ((x^2 + 21x)) = 2}}}
          {{{x^2 + 21x = 10^2}}} ---- Converting to EXPONENTIAL form
          {{{x^2 + 21x = 100}}}
 {{{x^2 + 21x - 100 = 0}}}
(x - 4)(x + 25) = 0 --- Factoring above TRINOMIAL
  x - 4 = 0     or       x + 25 = 0
        x = 4     or                x = - 25 (IGNORE)

The value, 4, for x, is > 0, but - 25 is NOT! Therefore, ONLY x = 4 is valid/ACCEPTABLE, while - 25 is an EXTRANEOUS solution.</font></font></font></b></pre>