To find [tex]\(\log_{\sqrt{3}} 7\)[/tex], 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]\(c \neq 1\)[/tex]):
[tex]\[
\log_b(a) = \frac{\log_c(a)}{\log_c(b)}
\][/tex]
In this case, we want to find [tex]\(\log_{\sqrt{3}} 7\)[/tex]. We can choose any base [tex]\(c\)[/tex] for our logarithms, but a common choice is the natural logarithm (base [tex]\(e\)[/tex]), denoted as [tex]\(\ln\)[/tex].
Applying the change-of-base theorem, we have:
[tex]\[
\log_{\sqrt{3}} 7 = \frac{\ln 7}{\ln \sqrt{3}}
\][/tex]
Now, let's calculate this step-by-step:
1. Compute the natural logarithm of [tex]\(7\)[/tex]:
[tex]\[
\ln 7 \approx 1.9459101490553132
\][/tex]
2. Compute the natural logarithm of [tex]\(\sqrt{3}\)[/tex]:
[tex]\[
\ln \sqrt{3} = \ln(3^{1/2}) = \frac{1}{2} \ln 3
\][/tex]
We know that:
[tex]\[
\ln 3 \approx 1.0986122886681098
\][/tex]
Thus,
[tex]\[
\ln \sqrt{3} = \frac{1}{2} \times 1.0986122886681098 \approx 0.5493061443340549
\][/tex]
3. Divide the natural logarithm of [tex]\(7\)[/tex] by the natural logarithm of [tex]\(\sqrt{3}\)[/tex]:
[tex]\[
\log_{\sqrt{3}} 7 = \frac{1.9459101490553132}{0.5493061443340549} \approx 3.5424874983228447
\][/tex]
Therefore, the value of [tex]\(\log_{\sqrt{3}} 7\)[/tex] is approximately [tex]\(3.5424874983228447\)[/tex].
