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

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

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Answer 1 · 2016-06-26T19:55:01

Do some algebra and apply some log properties to get $x = - \frac{e + 1}{1 - e} \approx 2.164$ $x=-(e+1)/(1-e)~~2.164$

Explanation:

Begin by collecting the $x$ $x$ terms on one side of the equation and the constant terms (numbers) on the other:
$ln (x + 1) - ln (x - 1) = 1$ $ln(x+1)-ln(x-1)=1$

Apply the natural log property $ln a - ln b = ln (\frac{a}{b})$ $lna-lnb=ln(a/b)$ :
$ln (\frac{x + 1}{x - 1}) = 1$ $ln((x+1)/(x-1))=1$

Raise both sides to the power of $e$ $e$ :
$e^{ln (\frac{x + 1}{x - 1})} = e^{1}$ $e^ln((x+1)/(x-1))=e^1$
$\to \frac{x + 1}{x - 1} = e$ $->(x+1)/(x-1)=e$

This equation is solved in a few steps:
$x + 1 = e (x - 1)$ $x+1=e(x-1)$
$x + 1 = e x - e$ $x+1=ex-e$
$x - e x = - e - 1$ $x-ex=-e-1$
$x (1 - e) = - e - 1$ $x(1-e)=-e-1$
$x = - \frac{e + 1}{1 - e} \approx 2.164$ $x=-(e+1)/(1-e)~~2.164$

Note that $\frac{- e - 1}{1 - e} = \frac{- 1 (e + 1)}{1 - e} = - \frac{e + 1}{1 - e}$ $(-e-1)/(1-e)=(-1(e+1))/(1-e)=-(e+1)/(1-e)$ .

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

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