https://file.aiquickdraw.com/mathbot/file/1729899861894l2z8v2p9.jpg

To evaluate \( \log_{25} 125 \), we need to express 125 and/or 25 in terms of common bases, preferably powers of the same number.

### Step-by-Step Solution
1. **Rewrite the bases**: 
   - \( 25 = 5^2 \)
   - \( 125 = 5^3 \)

2. **Rewrite the logarithmic expression**:
   \[
   \log_{25} 125 = \log_{5^2} 5^3
   \]

3. **Use the change of base formula**:
   When dealing with a logarithm in the form \( \log_{a^m} b^n \), we can use the property:
   \[
   \log_{a^m} b^n = \frac{n}{m} \log_a b
   \]

   Applying this here:
   \[
   \log_{5^2} 5^3 = \frac{3}{2} \log_5 5
   \]

4. **Evaluate \( \log_5 5 \)**:
   Since \( \log_5 5 = 1 \), we get:
   \[
   \frac{3}{2} \cdot 1 = \frac{3}{2}
   \]

### Final Answer
\[
\log_{25} 125 = \frac{3}{2}
\]

Would you like a detailed breakdown of any step?

---

Here are some related questions to further explore this concept:
1. How would you evaluate \( \log_{125} 25 \)?
2. What is the value of \( \log_5 125 \)?
3. How do properties of logarithms simplify expressions like \( \log_{a^n} b^m \)?
4. Can you evaluate \( \log_{4} 16 \) using a similar method?
5. How would you approach \( \log_{9} 27 \) using common bases?

**Tip:** Familiarize yourself with properties like \( \log_{a^m} b^n = \frac{n}{m} \log_a b \); they help simplify complex logarithmic expressions quickly!

Learn how to evaluate the logarithmic expression \( \log_{25} 125 \) using properties of logarithms and exponents. Follow a detailed step-by-step solution involving change of base and logarithmic properties. Suitable for high school students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation