Question

Apply the Properties of Integer Exponents to generate equivalent expressions to (7^−3)^7 ⋅ 7^20/6^−2 with only positive exponents, then solve the expression. Provide the answer as an improper fraction.(1 point) (7^−3)^7 ⋅ 7^20/6^−2=

Answer 1

To solve the expression \((7^{-3})^7 \cdot 7^{20} / 6^{-2}\) using the properties of integer exponents, we will first simplify each part.

1. **Simplify \((7^{-3})^7\)**:
Using the power of a power property, \((a^m)^n = a^{m \cdot n}\):
\[
(7^{-3})^7 = 7^{-3 \cdot 7} = 7^{-21}
\]

2. **Combine the terms**:
Now we substitute back into the expression:
\[
7^{-21} \cdot 7^{20}
\]
Using the product of powers property, \(a^m \cdot a^n = a^{m+n}\):
\[
7^{-21 + 20} = 7^{-1}
\]

3. **Now substitute into the entire expression**:
We have:
\[
\frac{7^{-1}}{6^{-2}}
\]
We can rewrite \(6^{-2}\) in the denominator using the rule \(\frac{1}{a^{-n}} = a^n\):
\[
\frac{7^{-1}}{6^{-2}} = 7^{-1} \cdot 6^{2}
\]

4. **Rewrite \(7^{-1}\)**:
\[
= \frac{6^2}{7^1} = \frac{6^2}{7}
\]

5. **Calculate \(6^2\)**:
\[
= \frac{36}{7}
\]

Thus, the final simplified expression in improper fraction form is:
\[
\boxed{\frac{36}{7}}
\]