To simplify the expression \((4^3 \cdot 7^{-2})^{-5} / (7^9 \cdot 4^{-13})\) using the properties of integer exponents, we'll follow these steps:
- Apply the Power of a Product Rule: \((a \cdot b)^n = a^n \cdot b^n\)
- Apply the Power of a Power Rule: \((a^m)^n = a^{m \cdot n}\)
- Convert Negative Exponents: \(a^{-n} = \frac{1}{a^n}\)
Let's start by simplifying \((4^3 \cdot 7^{-2})^{-5}\): \[ (4^3 \cdot 7^{-2})^{-5} = (4^3)^{-5} \cdot (7^{-2})^{-5} \]
Using the Power of a Power Rule: \[ (4^3)^{-5} = 4^{-15}, \quad (7^{-2})^{-5} = 7^{10} \] So, \[ (4^3 \cdot 7^{-2})^{-5} = 4^{-15} \cdot 7^{10} \]
Now combining this with the denominator \(7^9 \cdot 4^{-13}\): \[ \frac{4^{-15} \cdot 7^{10}}{7^9 \cdot 4^{-13}} \]
Now, split this into separate fractions: \[ = \frac{4^{-15}}{4^{-13}} \cdot \frac{7^{10}}{7^9} \]
Using the Quotient of Powers Rule, \(a^m/a^n = a^{m-n}\): \[ \frac{4^{-15}}{4^{-13}} = 4^{-15 - (-13)} = 4^{-2} = \frac{1}{4^2} = \frac{1}{16} \] \[ \frac{7^{10}}{7^9} = 7^{10 - 9} = 7^{1} = 7 \]
Now we can combine these parts: \[ \frac{1}{16} \cdot 7 = \frac{7}{16} \]
Thus, the simplified expression with only positive exponents is: \[ \frac{7}{16} \]
Final answer: \[ \frac{7}{16} \]