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

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Answer 1 · 2016-06-13T11:25:03

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

Using that $ln (a) - ln (b) = ln (\frac{a}{b})$ $ln(a)-ln(b) = ln(a/b)$ :

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

$\Rightarrow ln (x + 1) - ln (x - 1) = 1$ $=>ln(x+1)-ln(x-1)=1$

$\Rightarrow ln (\frac{x + 1}{x - 1}) = 1$ $=>ln((x+1)/(x-1))=1$

$\Rightarrow e^{ln (\frac{x + 1}{x - 1})} = e^{1}$ $=>e^ln((x+1)/(x-1))=e^1$

$\Rightarrow \frac{x + 1}{x - 1} = e$ $=>(x+1)/(x-1)=e$

$\Rightarrow x + 1 = e x - e$ $=>x+1=ex-e$

$\Rightarrow e x - x = e + 1$ $=>ex-x = e+1$

$\Rightarrow (e - 1) x = e + 1$ $=>(e-1)x = e+1$

$:.x = (e+1)/(e-1)$

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