SOLUTION: Hello Tutors! I'm preparing for NMAT. Can you please teach me how to solve problems like this: (3^n+2) ÷ (3^n+3 - 3^n+1)? Note: This symbol (^) is the exponent of the three #

Algebra ->  Absolute-value -> SOLUTION: Hello Tutors! I'm preparing for NMAT. Can you please teach me how to solve problems like this: (3^n+2) ÷ (3^n+3 - 3^n+1)? Note: This symbol (^) is the exponent of the three #       Log On


   

Question 905303: Hello Tutors!
I'm preparing for NMAT. Can you please teach me how to solve problems like this: (3^n+2) ÷ (3^n+3 - 3^n+1)?
Note: This symbol (^) is the exponent of the three # 3s. The (^n+2) is the exponent of the first # 3; (^n+3) is the exponent of the second # 3; and the (^n+1) is the exponents of the third # 3.
I would really appreciate your help.
Thank you. :-)
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i'll take a stab at it.

start with 3%5E%28n%2B2%29+%2F+%283%5E%28n%2B3%29+-+3%5E%28n%2B1%29%29

factor out the greatest common factor from the denominator to get:

3%5E%28n%2B2%29+%2F+%283%5E%28n%2B1%29+%2A+%283%5E2+-+1%29%29

the factoring is correct because, by the laws of exponent arithmetic:

3%5E%28n%2B1%29+%2A+3%5E2 = 3%5E%28n%2B1%2B2%29 = 3%5E%28n%2B3%29

and, by the same laws:

3%5E%28n%2B1%29+%2A+1 = 3%5E%28n%2B1%29

since 3%5E2+-+1 is equal to 8, your expression becomes:

3%5E%28n%2B2%29+%2F+%283%5E%28n%2B1%29+%2A+8%29

by the laws of exponent arithmetic:

3%5E%28n%2B2%29+%2F+3%5E%28n%2B1%29 = 3%5E%28%28n%2B2%29+-+%28n%2B1%29%29 = 3%5E%28n%2B2-n-1%29 = 3%5E%281%29 = 3

your expression becomes:

3%2F8

that's your solution.