# Simplifying an expression with algebraic indices

I don't know why but I just can't simplify this expression and its driving me mad!! Could someone please help me with the steps?

$\frac {2^n 9^{2n+1}}{6^{n-2}}$

I know the answer is 4 x $3^{3n + 4}$ I just can't get there.

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Notice that:

$$6^{n-2}=2^{n-2}3^{n-2}$$

and

$$9^{2n+1}=3^{4n+2}$$

Then use exponent rules.

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Thank you!!! Can do it now. Couldn't see that step by myself. Much appreciated. –  Dani Aug 4 '14 at 5:00
You're welcome!!! Glad you are able to solve the problem now. –  user71352 Aug 4 '14 at 5:00
Oh no I still can't get it!!! I've done: $\frac {2^{n} 3^{4n+2}}{2^{n} 3^{4n} 3^{n} 3^{-2}}$ $\frac {2^{n} 3^{4n} 3^{2} 2^{2} 3^{2}}{2^{n} 3^{n}}$ $\frac {2^{n}3^{4n}18^{2}}{6^{n}}$ $\frac {3^{4n} 18^{2}}{3}$ I can see this isn't going to lead anywhere useful! Can you please help set me in the right direction a bit more? Thank you –  Dani Aug 4 '14 at 5:12
Can you show me what you are getting? –  user71352 Aug 4 '14 at 5:15
Ok the denominator on that first one should say $2^{n} 2^{-2} 3^{n} 3^{-2}$ –  Dani Aug 4 '14 at 5:19

$$\frac {2^n 9^{2n+1}}{6^{n-2}}$$ $$\frac {2^n 3^{4n+2}}{2^{n-2}3^{n-2}}$$ $$2^{(n)-(n-2)}3^{(4n+2)-(n-2)}=4*3^{3n+4}$$

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