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

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Answer 1 · 2018-05-25T08:56:32

$therefore x = \frac{e^2}{1 - e^2}$

Use the property of logarithms

$log(a)-log(b) = log(a/b)$

to write

$ln(x)-ln(x+1)=2 \iff ln(x/(x+1)) = 2$

Consider both sides as exponents of $e$ :

$e^{ln(x/(x+1))} = e^2$

By definition, $e^ln(z)=z$ , so

$x/(x+1) = e^2$

$therefore x = (x+1)e^2$

$therefore x = e^2x+e^2$

$therefore x - e^2x = e^2$

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

$therefore x = \frac{e^2}{1 - e^2}$

Answer 2 · 2018-05-25T08:57:39

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

Using logarithmic rules we have $ln(x/(x+1))=2=lne^2$

Then $x/(x+1)=e^2$

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

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