# Thread: Index Laws and more - HELP! URGENT

1. ## 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

2. Originally Posted by mibamars
Simplify
x^-3/4 * x^9/8
$\displaystyle x^{-3/4+9/8} = x^{-6/8+9/8} = x^{3/8}$
Sqroot of x^3 divided sqroot of x
$\displaystyle \frac{\sqrt{x^3}}{\sqrt{x}} = \frac{x^{3/2}}{x^{1/2}} = x^{3/2 - 1/2}= x$
(p + 3) (p + 3)^-2/5
$\displaystyle (p+3)^{1-2/5} = (p+3)^{5/5-2/5} = (p+3)^{3/5}$

3. Originally Posted by mibamars
Simplify

2^n * 9^2n+1 / 6^n-2
$\displaystyle \frac{2^n \times 9^{2n+1}}{6^{n-2}}=\frac{2^n \times (9^2)^n \times 9^1}{6^n \times 6^{-2}}=\frac{2^n \times 81^n}{6^n} \times \frac{9}{6^{-2}}= \left(2 \times 81 \times \frac{1}{6} \right)^n \times 9 \times 36=324\times27^n$ If someone knows an easier way please tell me, i haven't done these in a while

4. Hello, mibamars!

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

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

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

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

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