To find the value of \(\log_6 8\) using the change of base formula with base 4, we can use the following change of base formula:
\[ \log_a b = \frac{\log_c b}{\log_c a} \]
For our case, we want to evaluate \(\log_6 8\) with base 4, so we can write:
\[ \log_6 8 = \frac{\log_4 8}{\log_4 6} \]
We also have previously computed \(\log_6 8\) using base \(e\), which has the value of \(1.161\). We can express this with natural logarithms as:
\[ \log_6 8 = \frac{\ln 8}{\ln 6} \]
Using the change of base formula, we can change the base from \(e\) to 4. Thus, we can relate these logarithms:
\[ \log_4 8 = \frac{\ln 8}{\ln 4} \quad \text{and} \quad \log_4 6 = \frac{\ln 6}{\ln 4} \]
Substituting into our formula, we get:
\[ \log_6 8 = \frac{\frac{\ln 8}{\ln 4}}{\frac{\ln 6}{\ln 4}} = \frac{\ln 8}{\ln 6} \]
Since we know \(\frac{\ln 8}{\ln 6}\) from above is \(1.161\):
Now we can find \(\log_4 8\) and \(\log_4 6\):
- \(\log_4 8 = \log_4 (2^3) = 3 \log_4 2\)
- To find \(\log_4 2\), we use the change of base again: \(\log_4 2 = \frac{\ln 2}{\ln 4} = \frac{1}{2}\) (because \(2^2 = 4\)).
- Therefore, \(\log_4 8 = 3 \cdot \frac{1}{2} = 1.5\).
Doing the same for \(\log_4 6\):
- \(\log_4 6 = \frac{\ln 6}{\ln 4}\).
- The value of \(\ln 6\) can be found using \(\ln 6 = \ln (2 \cdot 3) = \ln 2 + \ln 3\). We can approximate this as \(0.693 + 1.099 = 1.792\).
- Thus, \(\log_4 6 = \frac{1.792}{1.386} \approx 1.292\).
Now we can substitute these values into our ratio for \(\log_6 8\):
\[ \log_6 8 = \frac{\log_4 8}{\log_4 6} = \frac{1.5}{1.292} \approx 1.161 \]
Hence, the value of \(\log_6 8\) when evaluated using the change of base formula with base 4 aligns with the options offered; thus, the correct answer is:
1.161