Question #f8e12

1 Answer
sente
Dec 4, 2016

#x=100e^-13.57~~1.278xx10^-4#

Explanation:

We will use the following properties of logarithms and exponents:

  • #ln(xy) = ln(x)+ln(y)#
  • #e^ln(x) = x#
  • #e^-x = 1/e^x#
  • #e^(x+y) = e^xe^y#

With those:

#ln(0.01x) = -13.57#

#=> ln(0.01)+ln(x) = -13.57#

#=> ln(x) = -13.57-ln(0.01)#

#=> e^ln(x) = e^(-13.57-ln(0.01))#

#:. x = e^(-(13.57+ln(0.01)))#

#=1/e^(13.57+ln(0.01))#

#=1/(e^13.57e^ln(0.01))#

#=1/(0.01e^13.57)#

#=100e^-13.57~~1.278xx10^-4#

Related questions
See all questions in Natural Logs
Impact of this question
1110 views around the world
You can reuse this answer
Creative Commons License