SOLUTION: Solve fora in log(2a)- 3log2= 1/2log(a-3)

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Question 946643: Solve fora in log(2a)- 3log2= 1/2log(a-3)
Found 4 solutions by MathLover1, lwsshak3, ikleyn, MathTherapy:
Answer by MathLover1(20855) About Me  (Show Source):
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Solve for a in log%282a%29-3log%282%29=+%281%2F2%29log%28a-3%29
log%282%2Aa%29-3log%282%29=+%281%2F2%29log%28a-3%29
log%282a%29-log%282%5E3%29=+log%28%28a-3%29%5E%281%2F2%29%29
log%282a%29-log%288%29=+log%28sqrt%28%28a-3%29%29%29
log%28%282a%2F8%29%29=+log%28sqrt%28%28a-3%29%29%29
log%28%28a%2F4%29%29=+log%28sqrt%28%28a-3%29%29%29 ...if log same, then
a%2F4=+sqrt%28a-3%29...square both sides
%28a%2F4%29%5E2=+%28sqrt%28a-3%29%29%5E2
a%5E2%2F16=+%28a-3%29
a%5E2=+16%28a-3%29
a%5E2=+16a-48
a%5E2-16a%2B48=0...factor completely
a%5E2-4a+-12a%2B48=0...group
%28a%5E2-4a%29+-%2812a-48%29=0
a%28a-4%29+-12%28a-4%29=0
%28a-12%29%28a-4%29+=+0
solutions:
if %28a-12%29+=+0=> a=12
if %28a-4%29+=+0=> a=4

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for a in log(2a)- 3log2= 1/2log(a-3)
log(2a)- 3log2= 1/2log(a-3)
log(2a/(2^3))=log((a-3)^(1/2))
2a(2^3)=(a-3)^(1/2)
16a=√(a-3)
square both sides:
256a^2=a-3
256a^2-a+3=0
b^2-4ac<0
no real solution

Answer by ikleyn(53937) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve for a in log(2a)- 3log2= 1/2log(a-3).
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


        @lwsshar3 in his post gives the answer "no real solutions".
        This answer is incorrect.
        I came to bring a correct solution.


log(2a)- 3log2 = 1/2*log(a-3)
log(2a/(2^3)) = log((a-3)^(1/2))
2a/8 = (a-3)^(1/2)
a/4 = sqrt(a-3)
a = 4*sqrt(a-3)
square both sides:
a^2 = 16*(a-3)
a^2 = 16a - 48
a^2 - 16a + 48 = 0
Discriminant d = b^2-4ac = (-16)^2 - 4*16*48 = 64

a%5B1%2C2%5D = %2816+%2B-+sqrt%2864%29%29%2F2 = %2816+%2B-+8%29%2F2         (quadratic formula)

The solutions are a = 12 and a = 4.        ANSWER

Both solutions satisfy the original equation.

Solved correctly.



Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Solve fora in log(2a)- 3log2= 1/2log(a-3)
*****************************************
The SMALLEST log argument of the 3 log arguments, log (a - 3), signifies that a - 3 > 0, and so, a MUST BE > 3.
     log+%28%282a%29%29+-+3+log+%282%29+=+%281%2F2%29+log+%28%28a+-+3%29%29, 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+%28%282a%29%29%5E2+-+log+%28%28a+-+3%29%29+=+log+%28%282%5E6%29%29 ---- Applying a%2Alog+%28b%2C+%28c%29%29 = log+%28b%2C+%28c%5Ea%29%29
 log+%28%284a%5E2%29%29+-+log+%28%28a+-+3%29%29+=+log+%28%2864%29%29
               log+%28%284a%5E2%2F%28a+-+3%29%29%29+=+log+%28%2864%29%29 ----- Applying log+%28b%2C+%28a%29%29+-+log+%28b%2C+%28c%29%29 = log+%28b%2C+%28a%2Fc%29%29
                          4a%5E2%2F%28a+-+3%29+=+64 ---- If log+%28b%2C+%28c%29%29+=+log+%28b%2C+%28d%29%29, then c = d
                             4a%5E2+=+64%28a+-+3%29 ---- Cross-multiplying
                               a%5E2+=+16%28a+-+3%29 ---- Dividing each side by 4
                               a%5E2+=+16a+-+48
              a%5E2+-+16a+%2B+48+=+0
            (a - 12)(a - 4) = 0 --- Factorizing TRINOMIAL
                             highlight%28system%28a+=+12%2C+a+=+4%29%29

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