# Proving with exponents

• Jan 9th 2011, 06:32 PM
thatsjustlovely
Proving with exponents
Proving identities of any sort has never been my strongpoint. No, this is not trig identities. I have attempted to figure out this problem multiple times with rearranging and such and I have yet to be able to prove it. The answers do not show any steps or anything, it just says: True. The question looks like:

[3^(2n) 27^(5n+1) 9^(6n)] / 81^(4n-7) = [ 9^(7n) 3^(8n+28)] / 27^(3n-1)

If anyone could explain to me their process of solving this I would be grateful.
• Jan 9th 2011, 06:42 PM
pickslides
I get

$\displaystyle \frac{3^{2n} 27^{5n+1} 9^{6n}}{ 81^{4n-7} }$

$\displaystyle \frac{3^{2n} (3^3)^{5n+1} (3^2)^{6n}}{ (3^4)^{4n-7} }$

$\displaystyle \frac{3^{2n} 3^{3(5n+1)} 3^{2\times 6n}}{ 3^{4(4n-7)} }$

$\displaystyle \frac{3^{2n} 3^{15n+3} 3^{12n}}{ 3^{16n-28} }$

$\displaystyle \frac{3^{2n+(15n+3)+(12n)}}{ 3^{16n-28} }$

$\displaystyle \frac{3^{29n+3}}{ 3^{16n-28)} }$

$\displaystyle 3^{(29n+3)-(16n-28)}$

Finish it.