How do you solve #10^(-12x)+6=100#?

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Answer 1 · 2016-07-16T18:41:38

#x = -0.1644#

#10^(-12x)+6=100#

#10^(-12x)=94#

As the variable is in the index, we will use logs.

#log10^(-12x)=log94#

#-12x xxlog10=log94 " "log 10 = 1#

#x =log94/(-12)#

#x = -0.1644#