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Answer 1 · 2017-05-27T15:12:01

#n = 1#

I'm going to assume the question is #5^(n + 1) = sqrt(25^(3n - 1))#

Square both sides to get:

#(5^(n + 1))^2 =(sqrt(25^(3n - 1)))^2#

Use #(a^n)^m = a^(nm)#

#5^(2n + 2) = 25^(3n - 1)#

#5^(2n + 2) = (5^2)^(3n - 1)#

#5^(2n + 2) = 5^(6n - 2)#

#2n + 2 = 6n - 2#

#4 = 4n#

#1= n#

Hopefully this helps!

Answer 2 · 2017-05-27T15:56:09

#color(brown)(n=1#

#5^(n+1)=(sqrt25)^(3n-1)#

#:.5^(n+1)=5^(3n-1)#

base numbers are the same then:

#:.n+1=3n-1#

#:.n-3n=-1-1#

#:.-2n=-2#

multiply both sides by #-1#

#:.2n=2#

#:.color(brown)(n=1#