To simplify the expression \((7^{-3}) \cdot 7^{20} / 6^{-2}\) using the properties of integer exponents, we can follow these steps:
-
Combine the exponents of the base 7: Using the property \(a^m \cdot a^n = a^{m+n}\), we can combine \(7^{-3}\) and \(7^{20}\):
\[ 7^{-3} \cdot 7^{20} = 7^{-3 + 20} = 7^{17} \]
-
Simplify the expression: Now, we rewrite the entire expression, substituting \(7^{17}\) for \((7^{-3}) \cdot 7^{20}\):
\[ \frac{7^{17}}{6^{-2}} \]
-
Convert the negative exponent in the denominator: Recall that \(\frac{1}{a^{-m}} = a^m\). Therefore, \(6^{-2}\) in the denominator can be written as:
\[ \frac{7^{17}}{6^{-2}} = 7^{17} \cdot 6^{2} = 7^{17} \cdot 6^{2} \]
-
Final expression: Now we have:
\[ 7^{17} \cdot 6^{2} \]
-
Calculating numerical values: To calculate this, we can compute the individual components:
- \(6^2 = 36\)
- \(7^{17}\) is a larger number, and we will leave it in its exponential form or we can calculate it if needed.
Thus our expression is:
\[ 7^{17} \cdot 36 \]
-
Conclusion: The final result can be expressed as:
\[ \frac{7^{17} \cdot 36}{1} \]
Therefore, the simplified expression with only positive exponents is:
\[ 7^{17} \cdot 36 \]
Resulting in the answer as follows:
\[ \boxed{\frac{36 \cdot 7^{17}}{1}} \quad \text{or simply } \quad 36 \cdot 7^{17} \text{ as an improper fraction.} \]
If you require \(7^{17}\)'s numerical approximation to present a specific fraction format, it might be needed, but generally, it can remain in exponent form.