High School

Which expression is equivalent to [tex]\log_5 82[/tex]?

A. [tex]82 \log 5[/tex]

B. [tex]\frac{\log 82}{\log 5}[/tex]

C. [tex]5 \log 82[/tex]

D. [tex]\frac{\log 5}{\log 82}[/tex]

Answer :

To determine which expression is equivalent to [tex]\(\log_5 82\)[/tex], we need to use the change of base formula. The change of base formula allows us to convert a logarithm from one base to another. It states that:

[tex]\[
\log_b a = \frac{\log_k a}{\log_k b}
\][/tex]

where [tex]\(k\)[/tex] is any positive number. A common choice for [tex]\(k\)[/tex] is 10 (common logarithm) or [tex]\(e\)[/tex] (natural logarithm), but it can be any valid base.

Using this formula, we can convert [tex]\(\log_5 82\)[/tex] to a base that is more commonly used, like the base 10 logarithm ([tex]\(\log\)[/tex]):

[tex]\[
\log_5 82 = \frac{\log 82}{\log 5}
\][/tex]

Looking at the given options, we find that option B corresponds to this expression:

- A. [tex]\(82 \log 5\)[/tex]: This is not correct because it implies multiplying the number by [tex]\(\log 5\)[/tex], not changing the base.
- B. [tex]\(\frac{\log 82}{\log 5}\)[/tex]: This is correct as it matches the change of base formula for [tex]\(\log_5 82\)[/tex].
- C. [tex]\(5 \log 82\)[/tex]: This is not correct because it shows multiplication by 5, not changing the base.
- D. [tex]\(\frac{\log 5}{\log 82}\)[/tex]: This is not correct; while it uses division, it's the inverse of what we need.

Therefore, the correct equivalent expression is option B: [tex]\(\frac{\log 82}{\log 5}\)[/tex].