Question

log9 (1/729)

Answer 1

To calculate \( \log_9 \left(\frac{1}{729}\right) \), we can first express \( \frac{1}{729} \) in terms of powers of 9.

We know that:
\[
729 = 9^3
\]
Thus:
\[
\frac{1}{729} = \frac{1}{9^3} = 9^{-3}
\]

Now we can rewrite the logarithm:
\[
\log_9 \left(\frac{1}{729}\right) = \log_9 \left(9^{-3}\right)
\]

Using the property of logarithms \( \log_b(a^n) = n \cdot \log_b(a) \), we have:
\[
\log_9 \left(9^{-3}\right) = -3 \cdot \log_9(9)
\]

Since \( \log_9(9) = 1 \), we find:
\[
-3 \cdot \log_9(9) = -3 \cdot 1 = -3
\]

Therefore:
\[
\log_9 \left(\frac{1}{729}\right) = -3
\]

The final answer is:
\[
\boxed{-3}
\]