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 · 2016-05-08T05:08:34

$x = \frac{e + 1}{e - 1} = coth (\frac{1}{2}) = 2.164$ $x = (e+1)/(e-1)=coth (1/2)=2.164$ , nearly.

Explanation:

Rearranging,

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

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

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

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

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

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

$= coth (\frac{1}{2})$ $=coth (1/2)$

$\frac{e + 1}{e - 1} = 2.164$ $(e+1)/(e-1)=2.164$ , nearly.

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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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