4(2^(n-1))=16^n
4(2(n-1))=16n
And we can write 4 as 22 and 16 as 24.....and we have....
22(2(n-1))=(24)n ........and, by a property of exponents, we have
2(2 + n - 1) = 2(4n)
2( n + 1) = 2(4n)
And solving for the exponents, we have
n + 1 = 4n
1 = 3n
n = 1/3
Thiscanbewrittenas2^2\times2^{n-1}=(2^4)^nor2^{n+1}=2^{4n}Inotherwords$n+1=4n$or$n=13$
