To condense logarithmic expressions, we use logarithm properties like the product, quotient, and power rules. Here are the steps to match each expression from Side 1 with its condensed version from Side 2:
1. Expression A: [tex]\(\log_2 8 + \log_2 3\)[/tex]
- Using the product rule: [tex]\(\log_b x + \log_b y = \log_b (xy)\)[/tex]
- [tex]\(\log_2 8 + \log_2 3 = \log_2 (8 \cdot 3) = \log_2 24\)[/tex]
- Thus, Expression A ([tex]\(\log_2 8 + \log_2 3\)[/tex]) matches with Side 2 ([tex]\(\log_2 24\)[/tex]).
2. Expression B: This one has a log base of 2 but does not have a match
- Since there is no mathematical expression to simplify here, it directly matches with a log base of 2 without a proper pair.
- Thus, Expression B ([tex]\(This one has a log base of 2 but does not have a match\)[/tex]) matches with Side 2 ([tex]\(\log_2 11\)[/tex]).
3. Expression C: [tex]\(\log_3 25 - \log_3 2\)[/tex]
- Using the quotient rule: [tex]\(\log_b x - \log_b y = \log_b \left(\frac{x}{y}\right)\)[/tex]
- [tex]\(\log_3 25 - \log_3 2 = \log_3 \left(\frac{25}{2}\right)\)[/tex]
- Thus, Expression C ([tex]\(\log_3 25 - \log_3 2\)[/tex]) matches with Side 2 ([tex]\(\log_3 \frac{25}{2}\)[/tex]).
4. Expression D: This one has a log base of 3 but does not have a match
- Similar to Expression B, it directly relates to a log base of 3 without a proper pair.
- Thus, Expression D ([tex]\(This one has a log base of 3 but does not have a match\)[/tex]) matches with Side 2 ([tex]\(\log_3 23\)[/tex]).
5. Expression E: [tex]\(2 \log_4 x + 3 \log_4 y\)[/tex]
- Using the power rule: [tex]\(a \log_b x = \log_b (x^a)\)[/tex] and then applying product rule.
- [tex]\(2 \log_4 x + 3 \log_4 y = \log_4 (x^2) + \log_4 (y^3) = \log_4 (x^2 y^3)\)[/tex]
- Thus, Expression E ([tex]\(2 \log_4 x + 3 \log_4 y\)[/tex]) matches with Side 2 ([tex]\(\log_4 x^2 y^3\)[/tex]).
6. Expression F: This one has a log base of 4 but does not have a match
- Similar to previous unmatched expressions, it directly relates to a log base of 4 without a proper pair.
- Thus, Expression F ([tex]\(This one has a log base of 4 but does not have a match\)[/tex]) matches with Side 2 ([tex]\(\log_4 6 xy\)[/tex]).
Based on this analysis, here is the final matching:
[tex]\[
\begin{tabular}{|l|l|}
\hline Side 1 & Side 2 \\
\hline A.) \(\log _2 8+\log _2 3\) & K.) \(\log _2 24\) \\
\hline \begin{tabular}{l}
B.) This one has a log base of 2 but does \\
not have a match \\
\end{tabular} & H.) \(\log _2 11\) \\
\hline C.) \(\log _3 25-\log _3 2\) & J.) \(\log _3 \frac{25}{2}\) \\
\hline \begin{tabular}{l}
D.) This one has a log base of 3 but does \\
not have a match \\
\end{tabular} & L.) \(\log _3 23\) \\
\hline E.) \(2 \log _4 x+3 \log _4 y\) & G.) \(\log _4 x^2 y^3\) \\
\hline \begin{tabular}{l}
F.) This one has a log base of 4 but does \\
not have a match \\
\end{tabular} & I.) \(\log _4 6 x y\) \\
\hline
\end{tabular}
\][/tex]
