Question

(7^-3)^7 x 7^20/6^-2 provide answer as an improper fraction

Answer 1

Let's simplify the expression step by step.

The expression is

\[
(7^{-3})^7 \times \frac{7^{20}}{6^{-2}}.
\]

First, simplify \((7^{-3})^7\):

\[
(7^{-3})^7 = 7^{-3 \times 7} = 7^{-21}.
\]

Now substitute this back into the expression:

\[
7^{-21} \times \frac{7^{20}}{6^{-2}}.
\]

Next, simplify \(\frac{7^{20}}{6^{-2}}\):

\[
\frac{7^{20}}{6^{-2}} = 7^{20} \times 6^2 = 7^{20} \times 36.
\]

Now substituting this back into the main expression gives:

\[
7^{-21} \times 7^{20} \times 36.
\]

Using the property of exponents that \(a^m \times a^n = a^{m+n}\):

\[
7^{-21} \times 7^{20} = 7^{-21 + 20} = 7^{-1} = \frac{1}{7}.
\]

So the entire expression now becomes:

\[
\frac{1}{7} \times 36 = \frac{36}{7}.
\]

Thus, the final answer as an improper fraction is:

\[
\boxed{\frac{36}{7}}.
\]