Simplify Base-9 Expression: 9^300 × 1/9^-252 × 9^-549

Solve the following problem:

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=\text{?}$

Apply the laws of exponents for negative exponents, in the opposite direction:

$\frac{1}{a^n} = a^{-n}$

for the middle term in the multiplication in the problem:

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=9^{300}\cdot9^{-(-252)}\cdot9^{-549}=9^{300}\cdot9^{252}\cdot9^{-549}$

In the first stage we'll apply the aforementioned law of exponents, carefully given that the term in the denominator of the fraction has a negative exponent. Therefore we used parentheses. We then simplified the expression in the exponent,

Next we'll recall the law of exponents for multiplication between terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

and we'll apply this law to the last expression that we obtained:

$9^{300}\cdot9^{252}\cdot9^{-549}=9^{300+252+(-549)}=9^{300+252-549}=9^3$

Let's summarize the steps so far:

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=9^{300}\cdot9^{252}\cdot9^{-549} =9^{300+252-549}=9^3$

Note that there isn't such an answer among the answer choices, however we can always represent the expression that we obtained as a term with a negative exponent by taking the minus sign outside the parentheses in the exponent as follows:

$9^3=9^{-(-3)}$

We'll once again use the law of negative exponents:

$a^{-n}=\frac{1}{a^n}$

Let's apply it to the last expression that we obtained:

$9^3=9^{-(-3)}=\frac{1}{9^{-3}}$

Therefore :

$9^{300}\cdot\frac{1}{9^{-252}}\cdot9^{-549}=9^3 =9^{-(-3)}=\frac{1}{9^{-3}}$

The correct answer is answer A.