Question 946643
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Solve for a in log(2a)- 3log2= 1/2log(a-3).
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        @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[1,2]}}} = {{{(16 +- sqrt(64))/2}}} = {{{(16 +- 8)/2}}}          (quadratic formula)


The solutions are a = 12 and a = 4.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>ANSWER</U>


Both solutions satisfy the original equation.


Solved correctly.