Question

rewrite using a single positive exponent. 9^{-6} cdot 9^{2}

Explanation:

Step1: Use exponent - product rule

According to the rule $a^m\cdot a^n=a^{m + n}$, for $a = 9$, $m=-6$ and $n = 2$.
$9^{-6}\cdot9^{2}=9^{-6 + 2}$

Step2: Simplify the exponent

Calculate $-6 + 2=-4$. So $9^{-6+2}=9^{-4}$.

Step3: Convert to positive exponent

Use the rule $a^{-n}=\frac{1}{a^{n}}$, then $9^{-4}=\frac{1}{9^{4}}$. Another way is to rewrite $9^{-6}\cdot9^{2}$ as $\frac{1}{9^{6}}\cdot9^{2}=\frac{9^{2}}{9^{6}}$. And by the rule $\frac{a^{m}}{a^{n}}=a^{m - n}$, we have $\frac{9^{2}}{9^{6}}=9^{2-6}=9^{-4}=\frac{1}{9^{4}}$. We can also rewrite it as $9^{-6}\cdot9^{2}=9^{2-6}=9^{-4}=\frac{1}{9^{4}}$. If we want a single positive - exponent form in terms of $9$, we use the rule $a^{-n}=\frac{1}{a^{n}}$ to get $9^{-6}\cdot9^{2}=\frac{1}{9^{4}}$. But if we want to keep the base $9$ and have a positive exponent directly from the product rule $a^m\cdot a^n=a^{m + n}$, we can rewrite it as $9^{2+( - 6)}=9^{-4}=\frac{1}{9^{4}}$. A better form with a single positive exponent for the base $9$ is $9^{2-6}=9^{-4}=\frac{1}{9^{4}}$. The correct way using the product rule of exponents first:
$9^{-6}\cdot9^{2}=9^{-6 + 2}=9^{-4}=\frac{1}{9^{4}}$. If we rewrite it in terms of a single positive exponent for the base $9$ using the rule $a^m\cdot a^n=a^{m + n}$ and then convert the negative exponent to positive, we have:
$9^{-6}\cdot9^{2}=9^{-6+2}=9^{-4}=\frac{1}{9^{4}}$. The single positive - exponent form is $\frac{1}{9^{4}}$.

Answer:

$\frac{1}{9^{4}}$