How do you solve $ln x - ln (x - 1) = 2$ $ln x - ln (x-1) = 2$ ?

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Answer 1 · 2018-07-09T15:22:26

$x = \frac{- e^{2}}{1 - e^{2}} \approx 1.156517643$ $color(blue)(x=(-e^2)/(1-e^2)~~1.156517643)$

$ln a - ln b = ln (\frac{a}{b})$ $lna-lnb=ln(a/b)$

$ln x - ln (x - 1) = 2$ $lnx-ln(x-1)=2$

$ln (\frac{x}{x - 1}) = 2$ $ln(x/(x-1))=2$

$e^{ln (\frac{x}{x - 1})} = e^{2}$ $e^(ln(x/(x-1)))=e^2$

$\frac{x}{x - 1} = e^{2}$ $x/(x-1)=e^2$

$x - e^{2} x = - e^{2}$ $x-e^2x=-e^2$

$x (1 - e^{2}) = - e^{2}$ $x(1-e^2)=-e^2$

$x = \frac{- e^{2}}{1 - e^{2}} \approx 1.156517643$ $x=(-e^2)/(1-e^2)~~1.156517643$

How do you solve lnx−ln(x−1)=2ln x - ln (x-1) = 2?