Easy exponents question I have the GRE Friday... I got hung up on this easy exponents problem (I think it was these exponents, don't recall exactly)
$$\frac{6^{14}}{2^7 \times 3^5} = ? $$
The answer is $2^73^9$, but could anyone double check for me?
 A: It follows from:
$$
\frac{6^{14}}{2^7 3^5}=\frac{(2\times 3)^{14}}{2^7 3^5}=\frac{2^{14} 3^{14}}{2^73^5}=2^73^9
$$
EDIT: To really make that last step clear. Yes, you subtract the exponents. To make this easy to see:
$$
\frac{2^{14} 3^{14}}{2^73^5}=\big(2^{14}2^{-7}\big)\big(3^{14}3^{-5}\big)=2^{14-7}3^{14-5}=2^73^9
$$
A: Essentially yes. You need to use the rules $(xy)^a = x^a y^a$ and also $x^ax^b = x^{a+b}$ (the latter of which sometimes looks like $\frac{x^a}{x^b} = x^a x^{-b} = x^{a-b}$ for division). Here you first factor $6$ and then simplify. Explicitly:
$$
\frac{6^{14}}{2^7\cdot 3^5} = \frac{(2\cdot3)^{14}}{2^7\cdot 3^5} = \frac{2^{14}\cdot3^{14}}{2^7\cdot 3^5} = 2^{14-7}\cdot3^{14-5} = 2^7\cdot 3^{9}.
$$
Good luck on the GRE!
A: Alternatively, we can do
$$\frac{6^{14}}{2^73^5} \Longrightarrow \frac{6^{14}}{6^52^2} \Longrightarrow \frac{6^9}{2^2} \Longrightarrow 2^73^9$$
A: We have: $\dfrac{6^{14}}{2^{7}\times3^{5}}$
Now, $6$ is the product of $2$ and $3$, so we can rewrite this expression as:
$=\dfrac{(2\times3)^{14}}{2^{7}\times3^{5}}$
$=\dfrac{2^{14}\times3^{14}}{2^{7}\times3^{5}}$
When dividing numbers raised to powers we subtract the exponents:
$=2^{14-7}\times3^{14-5}$
$=2^{7}3^{9}$
