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

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

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Answer 1 · 2016-05-20T14:24:06

$x = 1 + \sqrt{1 + e}$ $x = 1 + sqrt[1 + e]$

Explanation:

From ${log}_{e} x + {log}_{e} (x - 2) = {log}_{e} (e)$ $log_e x + log_e(x-2) = log_e(e)$ we conclude
${log}_{e} (x (x - 2)) = {log}_{e} (e)$ $log_e(x(x-2))=log_e(e)$ and also
$x (x - 2) = e$ $x(x-2)=e$ . Solving for $x$ $x$ we obtain
$x = 1 + \sqrt{1 + e}$ $x = 1 + sqrt[1 + e]$

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

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