How do you evaluate #\log ( x - 1) - \log ( x + 3) = \log \frac { 1} { x }#?

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Answer 1 · 2018-08-01T07:34:33

#x = -1, 3#

#log(x-1) - log(x+3) = log (1/x)#

#log x - log y = log (x/y)#

#:. log((x-1) / (x+3)) = log (1/x)#

#(x-1) / (x+3) = 1/x#, Removing log on both sides.

#x^2 - x = x + 3#

#x^2 - 2x - 3 = 0#

#x^2 + x - 3x - 3 = 0#

#x(x+1)-3(x+1) = 0#

#x = -1, 3#