# Simplifying with negative exponents $(-11a^2)(-4a^{-7})$

$$(-11a^2)(-4a^{-7})$$

Can someone reformat, $a$ is second set of parenthesis is to the $-7$ power.

Change to reciprocal so we get $$\left(\frac{1}{-4a}\right)^7 * \frac{11a}{1} =$$ confused

Answer should be $$\frac{44}{a^5}$$

Unless,

$$(-11a^2)(-4a^{-7}) = 44a^{-5} = \frac{1}{44a^5}$$

but I guess, only $a$ was moved to the bottom because 44 is not negative.

Confused.

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Can someone edit that negative power so I know how to do it next time around? – Liger86 Oct 30 '11 at 0:42
$(ab)^c = a^cb^c \neq ab^c$ (as long as $a^c \neq a$, which is "mostly true") – kahen Oct 30 '11 at 0:48
The exponents only affect $a$, for example $2a^{-1}=\frac{2}{a}\neq(2a)^{-1}=\frac{1}{2a}$. – AMPerrine Oct 30 '11 at 0:52
The exponent -7 is only for $a$. – Hassan Muhammad Oct 30 '11 at 6:05

$(-11a^2)(-4a^{-7})$ means $(-11) \times a^2 \times (-4) \times a^{-7}$. (No real math there, just convention about how the notation works).
We can reorder the factors such that the two constants are multiplied together first and then use the rule $a^n a^k=a^{n+k}$ to get $$\Bigl((-11)\times(-4)\Bigr)\times a^{2-7} = 44\times a^{-5} = \frac{44}{a^5}$$