Question #b1a17

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Answer 1 · 2017-10-07T11:28:14

#17.0673# years

Assuming that you are familiar with logarithms...

#2^x = 100000 #

#=2^x=10^5 #

#=log_2(x) =log_10x(5) #

#=log_2(x) = 5 #

#=x=5/(log_2)#

#=16.0673+1# (for the first year)

#=17.0673# years