Given #a = log(4-x)# and #b = log(x+1/2)# solve #a^2+ab-2b^2=0# ?

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Answer 1 · 2017-06-15T08:38:52

#x = {0, 7/4, 1/2 (3 + 2 sqrt[6])}#

Calling #a = log(4-x)# and #b = log(x+1/2)# we have

#a^2+ab-2b^2=(a-b)(a+2b)=0# so the solutions are in

#a = b# and #a = -2b#

In the first case

#log(4-x)=log(x+1/2)# or

#4-x=x+1/2 rArr x = 7/4# which is feasible.

In the second option we have

#log(4-x)=-log(x+1/2)^2# or

#log((4-x)(x+1/2)^2)=0# or

#(4-x)(x+1/2)^2=1# or

#x(15/4 + 3 x - x^2) = 0# with feasible solutions

#x = 0# and #x = 1/2 (3 + 2 sqrt[6])#