How do you solve $ln x = -5$ ?

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Answer 1 · 2015-12-19T12:31:12

$x = 1/e^5$

From the definition of a logarithm, we have the property $e^ln(x) = x$ . From this:

$ln(x) = -5$

$=> e^(ln(x)) = e^(-5)$

$=> x = 1/e^5$

Answer 2 · 2015-12-19T12:36:29

$x = e^(-5) ~= 0.6738xx10^(-2)$

If
$color(white)("XXX")ln(x)=-5$
then
$color(white)("XXX")e^(ln(x)) =e^(-5)$
which can be evaluated using a calculator as $0.006738$

Why?

$ln(x)$ means the same thing as $log_e (x)$
$b^(log_b(a)) = a$
$color(white)("XXX")$ because $log_b(a)$ means
$color(white)("XXX")$ the value, $c$ , needed as an exponent to make $b^c = a$

How do you solve ln x = -5ln x = -5?