(7 to the power of -3)to the power of 7 times 7 to the power of 20 / 6 to the power of -2. Solve the expression. Answer with improper fraction

1 answer

Let's simplify the expression step by step:

We start with:

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

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

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

Now, we substitute this back into the expression:

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

Next, we simplify the numerator \(7^{-21} \cdot 7^{20}\):

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

\[ 7^{-21} \cdot 7^{20} = 7^{-21 + 20} = 7^{-1} \]

Now our expression looks like this:

\[ \frac{7^{-1}}{6^{-2}} \]

Next, we can rewrite \(6^{-2}\) in the denominator:

\[ \frac{7^{-1}}{6^{-2}} = 7^{-1} \cdot 6^2 = \frac{1}{7} \cdot 36 = \frac{36}{7} \]

Thus, the final answer is:

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