Using exponent laws, find the value of the real number n for which 12^(2n+1) = 2^(3n+7) . 3^(3n-4)

Question

Using exponent laws, find the value of the real number n for which 12^(2n+1) = 2^(3n+7) . 3^(3n-4) 

Answer 1

Hello, Karlie

Since 12 = 2^2 times 3;

12^(2n+1) = 2^2(2n+1)*3^(2n+1) = 2^(4n +2)*3^(2n+1); according to the question;

12^(2n+1) = 2^(3n+7) . 3^(3n-4)

so; 2^(4n +2)*3^(2n+1) = 2^(3n+7)*3^(3n-4) then we have

2^(4n +2)=2^(3n+7) and 4n +2=3n+7...n =5

3^(2n+1) = 3^(3n-4) and 2n+1 = 3n-4...n =5

hope that helps