Question #f8e12
1 Answer
Dec 4, 2016
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:
#=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#