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To solve for \( \log_{12} 36 \) using the given approximations \( \log_{12} 3 \approx 0.4421 \) and \( \log_{12} 7 \approx 0.7831 \), we can express 36 as a product of the numbers whose logarithms we know:

\[
36 = 3^2 \times 4 = 3^2 \times (7 \times \frac{3}{7}) = 3^2 \times 7.
\]

Using the logarithmic properties:

\[
\log_{12} 36 = \log_{12} (3^2 \times 7).
\]

Now apply the product and power rule of logarithms:

\[
\log_{12} (3^2 \times 7) = \log_{12} 3^2 + \log_{12} 7.
\]

This simplifies to:

\[
\log_{12} 36 = 2 \log_{12} 3 + \log_{12} 7.
\]

Substituting the approximated values:

\[
\log_{12} 36 \approx 2(0.4421) + 0.7831 = 0.8842 + 0.7831 = 1.6673.
\]

Thus, \( \log_{12} 36 \approx 1.6673 \).

Would you like any additional details or clarifications?

Here are 5 related questions to expand on this concept:
1. How does the change of base formula work for logarithms?
2. Can you apply logarithmic rules to divide numbers?
3. What are the main properties of logarithms that simplify expressions?
4. How do you approximate logarithms for numbers not directly given?
5. How does the relationship between exponents and logarithms assist in solving equations?

**Tip:** Always remember to use the logarithmic properties—product, quotient, and power rules—when breaking down complex expressions.

Use log_12 3 ≈ 0.4421 and log_12 7 ≈ 0.7831 to approximate log_12 36.

Learn how to approximate log_12 36 using the given values log_12 3 ≈ 0.4421 and log_12 7 ≈ 0.7831. This solution applies logarithmic properties such as the product rule and the power rule, suitable for students in Grades 9-12.

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