Question

what is \\(\frac{6^{-1}}{6^{-3}} \div \frac{1}{3^{-2}}\\) simplified? 4 \\(\frac{1}{4}\\) \\(-2\\) \\(-4\\)

Explanation:

Step1: Simplify the fraction with exponents

Recall the rule of exponents: $\frac{a^m}{a^n}=a^{m - n}$. So, $\frac{6^{-1}}{6^{-3}} = 6^{-1-(-3)} = 6^{2}$.

Step2: Divide by a fraction (multiply by reciprocal)

Dividing by $\frac{1}{3^{-2}}$ is the same as multiplying by $3^{-2}$. So now we have $6^{2}\times3^{-2}$.

Step3: Rewrite with same base or simplify

Note that $6 = 2\times3$, so $6^{2}=(2\times3)^{2}=2^{2}\times3^{2}$. Then the expression becomes $2^{2}\times3^{2}\times3^{-2}$.

Step4: Simplify the exponents of 3

Using the rule $a^m\times a^n = a^{m + n}$, for the 3 terms: $3^{2}\times3^{-2}=3^{2+(-2)} = 3^{0}=1$.

Step5: Simplify the remaining term

We are left with $2^{2}\times1 = 4$.

Answer:

4