# exponents

• May 27th 2007, 07:45 AM
Cals
exponents

n+1 2n-1
12 x 9
n 1-n
36 x 8
• May 27th 2007, 08:15 AM
Jhevon
Quote:

Originally Posted by Cals

n+1 2n-1
12 x 9
n 1-n
36 x 8

i'm not quite sure what is going on here. do you mean

$\displaystyle \frac {12^{n+1} \cdot 9^{2n - 1}}{36^{n} \cdot 8^{1-n}}$ ?

the dots mean multiply
• May 27th 2007, 09:13 AM
Soroban
Hello, Cals!

I must assume you know the basic Exponent Rules . . .

Quote:

$\displaystyle \frac{12^{n+1}}{36^n}\cdot\frac{9^{2n-1}}{8^{1-n}}$
Note that all the numbers are made up of 2's and 3's . . .

We have: .$\displaystyle \frac{(2^2\!\cdot\!3)^{n+1}\cdot(3^2)^{2n+1}}{(2^2 \!\cdot\!3^2)^n\cdot(2^3)^{1-n}} \;=\;\frac{(2^2)^{n+1}\cdot(3)^{n+1}\cdot(3)^{2(2n +1)} }{(2^2)^n\cdot(3^2)^n\cdot(2)^{3(1-n)}} \;=\;\frac{2^{2n+2}\cdot3^{n+1}\cdot3^{4n+2}} {2^{2n}\cdot3^{2n}\cdot2^{3-3n}}$

Multipfy (add exponents): .$\displaystyle \frac{2^{2n+2}\cdot3^{5n+3}}{2^{3-n}\cdot3^{2n}}$

Divide (subtract exponents): .$\displaystyle \boxed{2^{3n-1}\!\cdot\!3^{3n+3}}$