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

2 Answers
Dec 19, 2015

#x = 1/e^5#

Explanation:

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#

Dec 19, 2015

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

Explanation:

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?

  1. #ln(x)# means the same thing as #log_e (x)#
  2. #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#