Let's solve each of the given problems step by step:

(a) $\log(0.0270)$

This is a logarithm with base 10 (common logarithm).

$\log(0.0270) \approx -1.5686$

(b) $\ln(47.89)$

This is the natural logarithm (logarithm with base $e$).

$\ln(47.89) \approx 3.868$

(c) $\log_7(126.4)$

For this, we will use the change of base formula: $\log_b(a) = \frac{\log(a)}{\log(b)}$ Here, $a = 126.4$ and $b = 7$.

$\log_7(126.4) = \frac{\log(126.4)}{\log(7)} \approx \frac{2.1014}{0.8451} \approx 2.487$

Final Results:

• (a) $\log(0.0270) \approx -1.5686$
• (b) $\ln(47.89) \approx 3.868$
• (c) $\log_7(126.4) \approx 2.487$

Would you like a detailed explanation of any of the steps?

Here are 5 related questions to explore further:

1. What is the change of base formula, and why is it used?
2. How does the logarithm of a number less than 1 result in a negative value?
3. What are the differences between common logarithms and natural logarithms?
4. How can logarithms be used to solve exponential equations?
5. What are some real-world applications of logarithmic calculations?

Tip: Always remember to check the base of the logarithm when solving problems, as it determines the method you use for calculations.