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 · 2016-07-10T13:18:07

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

Explanation:

${log}_{e} (x + 1) - 1 = {log}_{e} (x - 1) \to {log}_{e} (\frac{x + 1}{e}) = {log e}_{e} (x - 1) \to \frac{x + 1}{e} = (x - 1)$ $log_e(x+1) - 1 = log_e(x-1)->log_e((x+1)/e)=loge_e(x-1)->(x+1)/e=(x-1)$

then

$x = \frac{e + 1}{e - 1} > 1 \to$ $x =(e+1)/(e-1) > 1->$ feasible solution

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