How do you solve $Ln e^x = 5$ ?

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Answer 1 · 2016-08-16T19:36:48

Real solution: $x=5$

Complex solutions: $x=5+2kpi i " "$ for any $k in ZZ$

The only Real value of $x$ for which $ln e^x = 5$ is $5$ .

The function $x |-> e^x$ is one to one from $(-oo, oo)$ onto $(0, oo)$ .

The function $x |-> ln x$ is its inverse from $(0, oo)$ onto $(-oo, oo)$ .

So for any $x in (-oo, oo)$ we have $ln e^x = x$ .

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If we consider Complex solutions, note that $e^(2pii) = 1$

Hence we find solutions:

$x = 5 + 2kpii " "$ for any integer $k$

since:

$ln e^(5+2kpii) = ln (e^5 e^(2kpii)) = ln(e^5 * (e^(2pii))^k) = ln (e^5*1^k) = ln e^5 = 5$

How do you solve Ln e^x = 5Ln e^x = 5?