How do you find a value of n that makes #4^(3n)*4^(n-2)=4^18# a true statement?

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Answer 1 · 2016-11-04T01:46:00

#n=5#

We have:

#4^(3n)xx4^(n-2)=4^18#

Let's first combine the left side:

#4^(3n+(n-2))=4^18#

We can now take the #log_4# of both sides:

#log_(4)4^(3n+(n-2))=log_(4)4^18#

#3n+(n-2)=18#

#4n-2=18#

#4n=20#

#n=5#