help with solving for n in exponents?

Question

4^n+3 * 16^n = 8^3n
and
3 * 9^2n = (3^n+1)^3

Please show your steps in solving for 1 or both and give the right answer. Thanks.

Answer 1

2^(2n+6)*2^4n=2^9n
2n+6+4n=9n
n=2

Answer 2

1. 4^(n+3) = 2^(2n+6), 16^n = 2^(4n) and 8^(3n) = 2^(9n)
so we have (2^(2n+6)) * 2^(4n) = 2^(9n)
with the rule that n^p * n^r = n^(p+r) we know that 2n+6+4n = 9n so n = 2
2. We know (x^y)^z = x^(yz) so we have 3^1 * (3^2)^(2n) = 3^(3n+3) this gives us 1+4n=3n+3 so n=2

Answer 3

I assume you mean

4^(n+3) * 16^n = 8^3n get everything in powers of 2
2^(2(n+3))*2^(4(n))=2^(3*3n)
2^(2n+4+4n)=2^(9n)
6n+4=9n
3n=4
n=4/3

3 * 9^(2n) = (3^(n+1))^3 get everything in powers of 3
3*3^(2*2n)=3^((n+1)*3)
3^(4n+1)=3^(3n+3)
4n+1=3n+3
n=2

Answer 4

(2^(2n+6))(2^5n) = 2^9n
2^(7n+6) = 2^9n

3^(4n+1) = 3^(3n+3)