To find the logarithm [tex]\(\log_{1/6} 8\)[/tex] using the change-of-base theorem, we can rewrite the expression in terms of natural logarithms.
The change-of-base theorem states that:
[tex]\[
\log_b a = \frac{\log_k a}{\log_k b}
\][/tex]
where [tex]\( b \)[/tex] is the base, [tex]\( a \)[/tex] is the value, and [tex]\( k \)[/tex] is the new base for the logarithm (commonly [tex]\( k \)[/tex] is chosen to be either 10 for common logarithms or [tex]\( e \)[/tex] for natural logarithms).
In this problem, we need to find [tex]\(\log_{1/6} 8\)[/tex]. Using the natural logarithm (base [tex]\( e \)[/tex]), the expression can be rewritten as:
[tex]\[
\log_{1/6} 8 = \frac{\ln 8}{\ln (1/6)}
\][/tex]
So, the logarithm [tex]\(\log_{1/6} 8\)[/tex] in terms of natural logarithms is:
[tex]\[
\log_{1/6} 8 = \frac{\ln 8}{\ln (1/6)}
\][/tex]
This is the expression provided in terms of natural logarithms without evaluating or simplifying.
