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#