To evaluate $\log_{0.3}(12.1)$, you can use the change of base formula:

$\log_b(a) = \frac{\log(a)}{\log(b)}$

Here, $b = 0.3$ and $a = 12.1$. You can use either common logarithms (base 10) or natural logarithms (base $e$).

Let’s apply the change of base formula using common logarithms (base 10):

$\log_{0.3}(12.1) = \frac{\log(12.1)}{\log(0.3)}$

Now, using a calculator:

• $\log(12.1) \approx 1.0839$
• $\log(0.3) \approx -0.5229$

Thus:

$\log_{0.3}(12.1) = \frac{1.0839}{-0.5229} \approx -2.073$

So, the value of $\log_{0.3}(12.1)$ is approximately $-2.073$.

Would you like more details on how the change of base formula works? Here are 5 related questions to explore:

1. How do natural logarithms compare to common logarithms?
2. What happens to the logarithmic value if the base is greater than 1?
3. How does the change of base formula relate to exponential functions?
4. Can logarithms be applied to complex numbers?
5. How do logarithmic scales work in real-world applications, such as sound intensity or earthquake magnitude?

Tip: Logarithms with bases between 0 and 1, like $\log_{0.3}$, yield negative values for numbers greater than 1.