How do you solve #40 = 20(1.012)^x#?

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Answer 1 · 2018-07-03T20:46:49

#x=log2/\log1.012=58.10814961730409#

Given that

#40=20(1.012)^x#

#(1.012)^x=40/20#

#(1.012)^x=2#

Taking logarithms on both the sides as follows

#log(1.012)^x=\log2#

#x\log1.012=log2#

#x=\log2/log1.012#

#=58.10814961730409#