To evaluate $\log_9(6)$ using the Change of Base Formula, we can use either natural logarithms ($\ln$) or common logarithms ($\log$).

The Change of Base Formula states:

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

Where $b$ is the base of the logarithm we want to convert, $a$ is the argument, and $c$ is the new base, which can be any logarithmic base (e.g., $c = 10$ for common logarithms, or $c = e$ for natural logarithms).

For this case:

$\log_9(6) = \frac{\log(6)}{\log(9)} \quad \text{(using common logarithms)}$

Now, we compute using a calculator:

$\log(6) \approx 0.778151 \quad \text{and} \quad \log(9) \approx 0.954243$

Therefore:

$\log_9(6) = \frac{0.778151}{0.954243} \approx 0.814814$

Thus, the value of $\log_9(6)$ correct to six decimal places is:

$\log_9(6) \approx 0.814814$

Let me know if you'd like further clarification or if you have any more questions!

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Related Questions:

1. How does the Change of Base Formula work when using natural logarithms instead of common logarithms?
2. What is the value of $\log_2(6)$ using the Change of Base Formula?
3. Can the Change of Base Formula be applied to logarithms with any base, or only to certain ones?
4. How do you use the Change of Base Formula when the base is less than 1?
5. What is the difference between the common logarithm (log base 10) and the natural logarithm (log base e)?

Tip:

When using the Change of Base Formula