<s>[INST] How would you explain to a 9th grader the negative exponent rule? Let us assume that the students haven't been exposed to these two rules: $a^{x+y} = a^{x}a^{y}$ and $\frac{a^x}{a^y} = a^{x-y}$. They have just been introduced to the generalization: $a^{-x} = \frac{1}{a^x}$ from the pattern method: $2^2 = 4, 2^1 = 2, 2^0 = 1, 2^{-1} = \frac{1}{2}$ etc. However, some students confuse $2^{-3}$ to be $(-2)(-2)(-2)$ since they are familiar with $2^{3} = 2 \cdot 2 \cdot 2$. This is a low-income urban school and most kids in this algebra class struggle with math dealing with exponents, fractions and decimals. What would be the best approach to reach all 32 students? 
 [/INST] *

*Help for confusing $$ 2^{-3}\neq (-2)(-2)(-2)$$
Advise them to watch out for the exact position of the - sign,
and tell them since it isn't next to  2, it simply does not belong to the 2, instead it belongs to the 3.
Example would be 
$$2^{-3} = \frac{1}{2^3} = \frac{1}{2\times2\times2} = \frac{1}{8}$$
as already proposed by @DoktoroReichhard

*Learning the rule:
$$a^{x+y} = a^{x}a^{y}$$
First example (with x = 2, y = 3) would be
$$ 10^{2+3} = 10^{5} = 100 ~000= 100 \cdot 1000  = (10\cdot10) ~ \cdot ~ (10\cdot10\cdot10) =   10^{2}\cdot10^{3} $$
Now substitute 10 with a and you get
$$a^{2+3}= a^{5} = aaaaa =aa\cdot aaa =a^{2}\cdot a^{3}=a^{2+3} $$
Then use 2 instead of 10 and calculate each single step on the board together with class

*Learning the rule:
$$a^{x-y} = \frac{a^{x}}{a^{y}}$$
First example (with x = 2, y = 3) would be 
$$ 10^{2-3} = 10^{-1} =\frac{1}{10}= \frac {100}{1000}  = \frac {10^{2}}{10^{3}} $$
Now substitute 10 with a and you get
$$a^{2-3}=a^{-1} = \frac{1}{a}= \frac {aa}{aaa}  = \frac {a^{2}}{a^{3}}  $$
Then use 2 instead of 10 and calculate each single step on the board together with class



In the end feel free to point out (for the ambitious students) that these rules will also work with crazy numbers like 
 $$ 2^{1.45 - 2.45} =2^{-1} =\frac{1}{2} $$
It's great that you are so passionate about teaching your class.
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