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

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Answer 1 · 2018-04-04T22:05:18

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

First note that

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

So for your problem we can write

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

Now take the exponential of both sides

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

(Another way to think of this is if $ln a = b$ $lna=b$ , then $a = e^{b}$ $a=e^b$ .)

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

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

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