To find [tex]\(\log_3{12}\)[/tex] in terms of natural logarithms, we can use the change-of-base theorem. The change-of-base theorem states that for any positive numbers [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] where [tex]\(a \neq 1\)[/tex] and [tex]\(b \neq 1\)[/tex],
[tex]\[
\log_b{a} = \frac{\log_c{a}}{\log_c{b}}
\][/tex]
In this case, we want to express [tex]\(\log_3{12}\)[/tex] using natural logarithms, which use the base [tex]\(e\)[/tex] (denoted as [tex]\(\ln\)[/tex]).
According to the change-of-base theorem, we have:
[tex]\[
\log_3{12} = \frac{\ln{12}}{\ln{3}}
\][/tex]
We then compute each of the natural logarithms separately:
1. Compute [tex]\(\ln{12}\)[/tex]
2. Compute [tex]\(\ln{3}\)[/tex]
Finally, divide the natural logarithm of 12 by the natural logarithm of 3:
[tex]\[
\log_3{12} = \frac{\ln{12}}{\ln{3}} \approx \frac{2.4849066497880004}{1.0986122886681098} \approx 2.2618595071429146
\][/tex]
Thus, in terms of natural logarithms,
[tex]\[
\log_3{12} \approx 2.26
\][/tex]
