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To solve \( \log_6 \frac{3}{7} \) using the provided values, we can apply the property of logarithms:

\[
\log_b \frac{a}{c} = \log_b a - \log_b c
\]

In this case:

\[
\log_6 \frac{3}{7} = \log_6 3 - \log_6 7
\]

Given:
- \( \log_6 3 \approx 0.613 \)
- \( \log_6 7 \approx 1.086 \)

Substitute these values into the expression:

\[
\log_6 \frac{3}{7} \approx 0.613 - 1.086 = -0.473
\]

So, the approximate value of \( \log_6 \frac{3}{7} \) is **-0.473**.

Would you like further details, or have any questions?

Here are some related questions for practice:
1. How would you evaluate \( \log_6 \frac{7}{3} \) using the same properties?
2. What is the value of \( \log_6 21 \) given \( \log_6 3 \) and \( \log_6 7 \)?
3. How can you use the change of base formula to verify these calculations?
4. If \( \log_6 9 \) is needed, how would you use \( \log_6 3 \) to find it?
5. What is the approximate value of \( \log_6 49 \)?

**Tip:** Memorizing key logarithmic properties, like the quotient, product, and power rules, can make solving these problems much easier.

Use log₆ 3 ≈ 0.613 and log₆ 7 ≈ 1.086 to evaluate the logarithm log₆ (3/7).

Learn how to evaluate log₆ (3/7) using given values for log₆ 3 and log₆ 7. This problem demonstrates the use of logarithmic properties, specifically the quotient rule. Suitable for high school students studying logarithms in Grades 10-12.

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