Question 258141
Remember when you add logs this is the same a multiplying big information. When you subtract logs this is the same as dividing big information. Using rules of logs we get
(i) {{{log(2,x(x+2)/(4-x))=2}}}
now, rewriting using exponents, we get
(ii) {{{2^2 = x(x+2)/(4-x)}}}
cross multiply to get
(ii) {{{4(4-x) = x^2 + 2x}}}
setting = 0, we get
(iv) {{{x^2 + 2x + 4x - 16 = 0}}}
simplify to get
(v) {{{x^2 + 6x - 16 = 0}}}
factor to get
(vi) {{{(x+8)(x-2) = 0}}}
solving for x, we get
x = -8
or
x = 2.
Since x = -8 is not a solution, we are left with x = 2.