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To solve the given expression, we need to evaluate:

\[
4^{\log 25} \cdot 5^{\log 36} - 2^{\log 9}
\]

Let's break down each term step-by-step:

### Step-by-Step Evaluation:

1. **Evaluate \(4^{\log 25}\):**
   \[
   4 = 2^2 \quad \text{and} \quad 25 = 5^2
   \]
   So:
   \[
   4^{\log 25} = (2^2)^{\log (5^2)} = 2^{2 \cdot \log (5^2)} = 2^{2 \cdot 2 \log 5} = 2^{4 \log 5} = (2^{\log 5})^4
   \]
   Since \(2^{\log 5} = 5\), we have:
   \[
   (2^{\log 5})^4 = 5^4 = 625
   \]

2. **Evaluate \(5^{\log 36}\):**
   \[
   36 = 6^2 = (2 \cdot 3)^2 = 2^2 \cdot 3^2
   \]
   \[
   \log 36 = \log (2^2 \cdot 3^2) = 2 \log 2 + 2 \log 3 = 2 (\log 2 + \log 3)
   \]
   Therefore:
   \[
   5^{\log 36} = 5^{2 (\log 2 + \log 3)} = (5^{\log 2 + \log 3})^2 = (5^{\log 2} \cdot 5^{\log 3})^2 = (2 \cdot 3)^2 = 6^2 = 36
   \]

3. **Multiply the two results:**
   \[
   4^{\log 25} \cdot 5^{\log 36} = 625 \cdot 36
   \]
   \[
   625 \cdot 36 = 22500
   \]

4. **Evaluate \(2^{\log 9}\):**
   \[
   9 = 3^2 \implies \log 9 = \log (3^2) = 2 \log 3
   \]
   So:
   \[
   2^{\log 9} = 2^{2 \log 3} = (2^{\log 3})^2
   \]
   Since \(2^{\log 3} = 3\), we get:
   \[
   (2^{\log 3})^2 = 3^2 = 9
   \]

5. **Subtract the last term:**
   \[
   22500 - 9 = 22491
   \]

Given the choices, it seems the problem may have intended different properties of logarithms or base conversions, but assuming the straightforward interpretation above:

\[
\boxed{22491}
\]

However, none of the provided options match this result. The provided answer choices might indicate a different expected approach, such as base manipulation or simplifications assumed in a different way. Double-check the problem context for further clues!

Calculate the value of the expression 4^log25 * 5^log36 - 2^log9 and select the correct answer from the given options.

Learn how to evaluate the expression 4^log25 * 5^log36 - 2^log9 with a detailed step-by-step solution. This problem involves logarithmic and exponential functions, using properties like the product rule and change of base formula. Ideal for high school students in Grades 10-12.

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