Question
page 14 of 14 Changing Ljavascript:void(0)ogarithm Bases Quick Check 1 of 51 of 5 Items Question Which of the following correctly changes log25200 to an equivalent expression?(1 point) Responses log25log200 Start Fraction logarithm 25 over logarithm 200 End Fraction log200log525 Start Fraction logarithm 200 over logarithm subscript 5 baseline 25 End Fraction log200log25 Start Fraction logarithm 200 over logarithm 25 End Fraction log200ln25
Answers
GPT-4o mini
To change the base of a logarithm, you can use the formula:
\[
\log_b(a) = \frac{\log_k(a)}{\log_k(b)}
\]
where \(k\) can be any positive number.
In this case, we want to rewrite \(\log_{25}(200)\). To do this, we'll use the base 200 as follows:
\[
\log_{25}(200) = \frac{\log_{200}(200)}{\log_{200}(25)}
\]
Since \(\log_{200}(200) = 1\), the expression simplifies to:
\[
\log_{25}(200) = \frac{1}{\log_{200}(25)}
\]
However, if we rewrite it using the equivalent basis to express it directly in terms of logs, we arrive at:
\[
\log_{25}(200) = \frac{\log(200)}{\log(25)}
\]
So, among the provided options, the correct equivalent expression for \(\log_{25}(200)\) is:
**\(\frac{\log(200)}{\log(25)}\)**
Based on your provided options, that corresponds to:
**log200 / log25**
Thus, the answer is:
**log200log25 Start Fraction logarithm 200 over logarithm 25 End Fraction**
\[
\log_b(a) = \frac{\log_k(a)}{\log_k(b)}
\]
where \(k\) can be any positive number.
In this case, we want to rewrite \(\log_{25}(200)\). To do this, we'll use the base 200 as follows:
\[
\log_{25}(200) = \frac{\log_{200}(200)}{\log_{200}(25)}
\]
Since \(\log_{200}(200) = 1\), the expression simplifies to:
\[
\log_{25}(200) = \frac{1}{\log_{200}(25)}
\]
However, if we rewrite it using the equivalent basis to express it directly in terms of logs, we arrive at:
\[
\log_{25}(200) = \frac{\log(200)}{\log(25)}
\]
So, among the provided options, the correct equivalent expression for \(\log_{25}(200)\) is:
**\(\frac{\log(200)}{\log(25)}\)**
Based on your provided options, that corresponds to:
**log200 / log25**
Thus, the answer is:
**log200log25 Start Fraction logarithm 200 over logarithm 25 End Fraction**