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

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Answer 1 · 2016-03-19T12:44:48

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

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

$ln (\frac{x}{x + 2}) = ln e$ $ln (x/(x+2))=ln e" " "$ because $ln e = 1$ $ln e=1$

$\frac{x}{x + 2} = e$ $x/(x+2)=e$

$x = e (x + 2)$ $x=e(x+2)$

$x = e x + 2 e$ $x=ex+2e$

$x - e x = 2 e$ $x-ex=2e$

$x (1 - e) = 2 e$ $x(1-e)=2e$

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

$(xcancel((1-e)))/cancel((1-e))=(2e)/(1-e)$

$x=(2e)/(1-e)$

God bless...I hope the explanation is useful.

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