Use the change-of-base formula to approximate the logarithm accurate to the nearest ten-thousandth.
log√2(6.5)

To approximate \( \log_{\sqrt{2}}(6.5) \) using the change-of-base formula, we can use the following formula:

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

where \( b = \sqrt{2} \) and \( a = 6.5 \), and \( c \) is typically 10 or \( e \) for convenience (we will use \( e \) for natural logarithms). 

Using natural logarithms (\( \ln \)):

\[
\log_{\sqrt{2}}(6.5) = \frac{\ln(6.5)}{\ln(\sqrt{2})}
\]

First, let's find the natural logarithms of 6.5 and \(\sqrt{2}\):

\[
\ln(6.5) \approx 1.8718
\]

\[
\ln(\sqrt{2}) = \ln(2^{1/2}) = \frac{1}{2} \ln(2) \approx \frac{1}{2} \times 0.6931 \approx 0.3466
\]

Now, we can calculate the logarithm:

\[
\log_{\sqrt{2}}(6.5) = \frac{1.8718}{0.3466} \approx 5.4001
\]

Thus, the value of \( \log_{\sqrt{2}}(6.5) \) accurate to the nearest ten-thousandth is approximately \( 5.4001 \).

Do you want details or have any questions?

Here are five numbered questions for you to ask:
1. How do you find natural logarithms of numbers?
2. Can you explain why we use the change-of-base formula?
3. What are other methods to calculate logarithms besides using the change-of-base formula?
4. How do logarithmic properties apply to solving equations?
5. Can you demonstrate how to use logarithms in a real-world problem?

Tip: When working with logarithms, remember that the change-of-base formula is a powerful tool for simplifying calculations, especially when the base is not one of the common logarithm bases (10 or \( e \)).

Learn how to approximate the logarithm \( \log_{\sqrt{2}}(6.5) \) using the change-of-base formula. Understand the steps involved in calculating logarithms with specific examples and solutions. Suitable for Grades 10-12.

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