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

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Answer 1 · 2018-08-01T07:34:33

$x = - 1, 3$ $x = -1, 3$

$log (x - 1) - log (x + 3) = log (\frac{1}{x})$ $log(x-1) - log(x+3) = log (1/x)$

$log x - log y = log (\frac{x}{y})$ $log x - log y = log (x/y)$

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

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

$x^2 - x = x + 3$

$x^2 - 2x - 3 = 0$

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

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

$x = -1, 3$

How do you evaluate \log ( x - 1) - \log ( x + 3) = \log \frac { 1} { x }log(x−1)−log(x+3)=log1x\log ( x - 1) - \log ( x + 3) = \log \frac { 1} { x }?