# Index Laws and more - HELP! URGENT

• Jul 4th 2007, 05:44 AM
mibamars
Index Laws and more - HELP! URGENT
Simplify

2^n * 9^2n+1 / 6^n-2

x^-3/4 * x^9/8

Sqroot of x^3 divided sqroot of x

(p + 3) (p + 3)^-2/5
• Jul 4th 2007, 06:34 AM
ThePerfectHacker
Quote:

Originally Posted by mibamars
Simplify
x^-3/4 * x^9/8

$x^{-3/4+9/8} = x^{-6/8+9/8} = x^{3/8}$
Quote:

Sqroot of x^3 divided sqroot of x
$\frac{\sqrt{x^3}}{\sqrt{x}} = \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2}= x$
Quote:

(p + 3) (p + 3)^-2/5
$(p+3)^{1-2/5} = (p+3)^{5/5-2/5} = (p+3)^{3/5}$
• Jul 4th 2007, 06:45 AM
DivideBy0
Quote:

Originally Posted by mibamars
Simplify

2^n * 9^2n+1 / 6^n-2

$\frac{2^n \times 9^{2n+1}}{6^{n-2}}=\frac{2^n \times (9^2)^n \times 9^1}{6^n \times 6^{-2}}=\frac{2^n \times 81^n}{6^n} \times \frac{9}{6^{-2}}= \left(2 \times 81 \times \frac{1}{6} \right)^n \times 9 \times 36=324\times27^n$ If someone knows an easier way please tell me, i haven't done these in a while
• Jul 4th 2007, 07:30 AM
Soroban
Hello, mibamars!

Quote:

Simplify: . $\frac{2^n\cdot9^{2n+1}}{6^{n-2}}$

We have: . $\frac{2^n\left(3^2\right)^{2n+1}}{(2\cdot3)^{n-2}}$

. . . . . . $= \;\frac{2^n\cdot3^{4n+2}}{2^{n-2}\cdot3^{n-2}}$

. . . . . . $= \;2^{n-(n-2)}\cdot3^{(4n+2)-(n-2)}$

. . . . . . $= \;2^2\cdot3^{3n+4}$