How do you solve #2^(5n-5)+2=75#?

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Answer 1 · 2016-09-12T23:32:54

#2^(5n - 5) = 73#

#log(2^(5n - 5)) = log73#

#(5n - 5)log2 = log73#

#5nlog2 - 5log2 = log73#

#5nlog2 = log73 + log32#

#n= log2336/log32#

The decimal approximation will be #n ~=2.24#.

Hopefully this helps!