How do you solve #6^-t+1=22#?

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Answer 1 · 2016-12-28T19:08:54

#t= 1.700 (3s.f.)#

#6^(-t)+1=22#

subtract 1:

#6^(-t)=21#

convert to logarithmic form:

#a^m=n -> log_a(n)=m#

#6^(-t)=21 -> log_6(21)=-t#

solve using a calculator:

#log_6(21)= 1.69918...#

#-t = 1.69918...#

multiply by -1:

#t= 1.700 (3s.f.)#