Question

log0.6 (0.36)

Answer 1

To find \( \log_{0.6}(0.36) \), we can use the change of base formula which states:

\[
\log_a(b) = \frac{\log_c(b)}{\log_c(a)}
\]

for any base \( c \). A common choice for \( c \) is 10 (common logarithm) or \( e \) (natural logarithm). Here, we can use base 10:

\[
\log_{0.6}(0.36) = \frac{\log_{10}(0.36)}{\log_{10}(0.6)}
\]

Now, we can calculate \( \log_{10}(0.36) \) and \( \log_{10}(0.6) \).

1. Calculate \( \log_{10}(0.36) \):
- \( 0.36 = 36 \times 10^{-2} \)
- \( \log_{10}(0.36) = \log_{10}(36) - 2 \)

Since \( 36 = 6^2 \):
- \( \log_{10}(36) = 2 \log_{10}(6) \)

The approximate value of \( \log_{10}(6) \) is about 0.7781 (you can use a calculator for a more precise value).
- Therefore, \( \log_{10}(36) \approx 2 \times 0.7781 = 1.5562 \)

So:

\[
\log_{10}(0.36) \approx 1.5562 - 2 = -0.4438
\]

2. Calculate \( \log_{10}(0.6) \):
- The approximate value of \( \log_{10}(0.6) \) is about -0.2218 (you can check with a calculator).

Now we can substitute these values back into the change of base formula:

\[
\log_{0.6}(0.36) \approx \frac{-0.4438}{-0.2218} \approx 2.0009
\]

Thus, the final result is:

\[
\log_{0.6}(0.36) \approx 2
\]