Question 946643
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Solve fora in log(2a)- 3log2= 1/2log(a-3)
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The SMALLEST log argument of the 3 log arguments, log (a - 3), signifies that a - 3 > 0, and so, a MUST BE > 3.
     {{{log ((2a)) - 3 log (2) = (1/2) log ((a - 3))}}}, with a being > 3.
  2 log (2a) - 6 log (2) = log (a - 3) ----- Multiplying by LCD, 2
2 log (2a) - log (a - 3) = 6 log (2) --- Subtracting log (a - 3) and adding 6 log (2), to both sides
  {{{log ((2a))^2 - log ((a - 3)) = log ((2^6))}}} ---- Applying {{{a*log (b, (c))}}} = {{{log (b, (c^a))}}}
 {{{log ((4a^2)) - log ((a - 3)) = log ((64))}}}
               {{{log ((4a^2/(a - 3))) = log ((64))}}} ----- Applying {{{log (b, (a)) - log (b, (c))}}} = {{{log (b, (a/c))}}}
                          {{{4a^2/(a - 3) = 64}}} ---- If {{{log (b, (c)) = log (b, (d))}}}, then c = d
                             {{{4a^2 = 64(a - 3)}}} ---- Cross-multiplying
                               {{{a^2 = 16(a - 3)}}} ---- Dividing each side by 4
                               {{{a^2 = 16a - 48}}}
              {{{a^2 - 16a + 48 = 0}}}
            (a - 12)(a - 4) = 0 --- Factorizing TRINOMIAL
                             {{{highlight(system(a = 12, a = 4))}}}

a = 12 > 3, and a = 4 > 3. Therefore, 12 and 4 are VALID/ACCEPTABLE values of a.</font></font></font></b></pre>