To solve the expression \((4^3 \cdot 7^{-2})^{-5} / (7^9 \cdot 4^{-13})\), we start by applying the properties of integer exponents.
First, we will simplify the numerator \((4^3 \cdot 7^{-2})^{-5}\):
- Apply the property \((a^m \cdot b^n)^p = a^{mp} \cdot b^{np}\): \[ (4^3 \cdot 7^{-2})^{-5} = 4^{3 \cdot (-5)} \cdot 7^{-2 \cdot (-5)} = 4^{-15} \cdot 7^{10} \]
Now let's rewrite the entire expression: \[ \frac{4^{-15} \cdot 7^{10}}{7^9 \cdot 4^{-13}} \]
Next, we simplify the denominator \(7^9 \cdot 4^{-13}\): \[ = 7^9 \cdot 4^{-13} \]
Now, we can rewrite the fraction: \[ \frac{4^{-15} \cdot 7^{10}}{7^9 \cdot 4^{-13}} = \frac{4^{-15}}{4^{-13}} \cdot \frac{7^{10}}{7^9} \]
Using the property \(\frac{a^m}{a^n} = a^{m-n}\):
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For the bases \(4\): \[ \frac{4^{-15}}{4^{-13}} = 4^{-15 - (-13)} = 4^{-15 + 13} = 4^{-2} \]
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For the bases \(7\): \[ \frac{7^{10}}{7^{9}} = 7^{10 - 9} = 7^{1} = 7 \]
Now we combine the results: \[ 4^{-2} \cdot 7 = \frac{7}{4^2} \]
Calculating \(4^2\): \[ 4^2 = 16 \]
Thus, we have: \[ \frac{7}{16} \]
The final answer, simplified and with no remaining exponents, is: \[ \boxed{\frac{7}{16}} \]