Question 377694
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Hi
log 2 (x-1) - log 2 (x+3) = log 2 (1/x)
log 2 (x-1) - log 2 (x+3) - log 2 (1/x)=0
*[tex \large\ \ log_bx - log_by = log_b(x/y) ]
{{{log_2((x-1)/(x+3)(1/x))}}}=0
*[tex \LARGE \ \ \ \ \ \ \ \ \ \ log_b(x) \ = \ y \ \ \Rightarrow\ \ b^y = x]
2^0 = (x-1)/(x+3)(1/x)
{{{1 = (x-1)/(x+3)(1/x)}}}
1 = x(x-1)/(x+3)
(x+3) = x^2 -x
x^2 -2x - 3 = 0
(x-3)(x+1)= 0
(x-3)=0    x = 3
(x+1)= 0   x = -1 reject as a solution
Any solution that makes an argument (or base) zero or negative must be rejected.
log 2 ((-1)-1) - log 2 ((-1)+3) = log 2 (1/(-1))