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Let's break down the expression using properties of logarithms and solve it step by step:

The given expression is:

\[
5\log_{25}(5) - 4\log_{25}(5) + \log_{25}\left(\frac{1}{625}\right)
\]

### Step 1: Combine like terms
The first two terms both involve \(\log_{25}(5)\). So, we can simplify them as follows:

\[
(5 - 4)\log_{25}(5) = \log_{25}(5)
\]

### Step 2: Evaluate \(\log_{25}(5)\)
Recall that \(25 = 5^2\), so:

\[
\log_{25}(5) = \frac{1}{2}
\]

Thus, the expression becomes:

\[
\frac{1}{2} + \log_{25}\left(\frac{1}{625}\right)
\]

### Step 3: Simplify \(\log_{25}\left(\frac{1}{625}\right)\)
First, express \(625\) as a power of \(25\). Since \(625 = 25^2\), we can rewrite the logarithm as:

\[
\log_{25}\left(\frac{1}{625}\right) = \log_{25}(25^{-2})
\]

By the logarithmic power rule, \(\log_b(x^n) = n \log_b(x)\), this becomes:

\[
\log_{25}(25^{-2}) = -2 \log_{25}(25)
\]

Since \(\log_{25}(25) = 1\), the expression simplifies to:

\[
-2
\]

### Step 4: Combine everything
Now, substitute back into the expression:

\[
\frac{1}{2} - 2 = \frac{1}{2} - \frac{4}{2} = -\frac{3}{2}
\]

Thus, the value of the expression is:

\[
\boxed{-\frac{3}{2}}
\]

Would you like further details on any of the steps? 

Here are 5 related questions for further exploration:
1. How do the logarithmic properties apply when changing bases?
2. Can you explain why \(\log_{25}(25)\) equals 1?
3. What is the logarithmic rule for division and powers?
4. How would you solve if the base was different, like \(\log_5\)?
5. How does this simplify when logarithmic arguments are fractions?

**Tip**: Always try breaking down the expression using basic logarithmic identities like the product, quotient, and power rules before simplifying further.

Use properties of logarithms to evaluate the expression without using a calculator: 5log_{25}(5) - 4log_{25}(5) + log_{25}(1/625) = ?

Solve the expression 5log_25(5) - 4log_25(5) + log_25(1/625) step-by-step using the properties of logarithms. This solution explains the logarithmic power rule, simplifies fractional exponents, and provides intermediate steps suitable for high school students.

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