log(450/32) + log(25/128) - log(225/64) - log(25/32)

We can solve the expression:

\[
\log\left(\frac{450}{32}\right) + \log\left(\frac{25}{128}\right) - \log\left(\frac{225}{64}\right) - \log\left(\frac{25}{32}\right)
\]

### Step 1: Apply the logarithmic addition and subtraction rules
We know that:

\[
\log(a) + \log(b) = \log(a \cdot b) \quad \text{and} \quad \log(a) - \log(b) = \log\left(\frac{a}{b}\right)
\]

So we can combine the logs in pairs:

\[
\log\left(\frac{450}{32} \cdot \frac{25}{128}\right) - \log\left(\frac{225}{64} \cdot \frac{25}{32}\right)
\]

### Step 2: Multiply the terms inside the logs

First, calculate:

\[
\frac{450}{32} \cdot \frac{25}{128} = \frac{450 \times 25}{32 \times 128} = \frac{11250}{4096}
\]

and

\[
\frac{225}{64} \cdot \frac{25}{32} = \frac{225 \times 25}{64 \times 32} = \frac{5625}{2048}
\]

### Step 3: Apply subtraction of logs

Now subtract:

\[
\log\left(\frac{11250}{4096}\right) - \log\left(\frac{5625}{2048}\right)
\]

Using the subtraction rule again:

\[
\log\left(\frac{\frac{11250}{4096}}{\frac{5625}{2048}}\right) = \log\left(\frac{11250 \times 2048}{4096 \times 5625}\right) = \log\left(\frac{23040000}{23040000}\right)
\]

\[
\log(1) = 0
\]

### Final Answer:
The result of the expression is:

\[
\boxed{0}
\]

Would you like more details or have any questions?

Here are 5 related questions:
1. How do the logarithmic rules simplify complex expressions?
2. What are the properties of logarithms for different bases?
3. How can logarithms be used to solve exponential equations?
4. What is the change of base formula for logarithms?
5. How are logarithms applied in real-world situations?

**Tip:** Always look for opportunities to apply logarithmic rules (like addition or subtraction) to simplify expressions early.

Learn how to simplify the logarithmic expression log(450/32) + log(25/128) - log(225/64) - log(25/32) by applying the logarithmic addition and subtraction rules. This problem covers essential logarithmic concepts and rules, suitable for students in Grades 10-12.

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