Question

use the properties of logarithms to write the following expression as one loganthm.\\(\frac{log s}{log r} + 8log_{r} s - 3log_{r} t\\)\
\\(\bigcirc\\) \\(\frac{log_{r} s^{9}}{log_{r} t^{3}}\\)\
\\(\bigcirc\\) \\(\log_{r}(s^{9})(t^{3})\\)\
\\(\bigcirc\\) \\(\log_{r} \frac{s^{9}}{t^{3}}\\)\
\\(\bigcirc\\) \\(\log_{r}(s^{9} + t^{3})\\)

Explanation:

Step1: Identify the change of base formula

The term \(\frac{\log s}{\log r}\) can be rewritten using the change of base formula for logarithms, which states that \(\log_b a=\frac{\log a}{\log b}\). So, \(\frac{\log s}{\log r}=\log_r s\).

Step2: Apply the power rule of logarithms

The power rule of logarithms states that \(n\log_b a = \log_b a^n\). For the term \(8\log_r s\), applying the power rule gives \(\log_r s^8\) (wait, no, actually \(8\log_r s=\log_r s^8\)? Wait, no, wait, in the original expression, after step 1, we have \(\log_r s+ 8\log_r s- 3\log_r t\). Wait, no, first, \(\frac{\log s}{\log r}=\log_r s\) (change of base formula: \(\log_r s=\frac{\log s}{\log r}\)). Then, \(8\log_r s=\log_r s^8\)? Wait, no, the power rule is \(n\log_b a=\log_b(a^n)\), so \(8\log_r s=\log_r(s^8)\)? Wait, but in the options, we have \(s^9\). Wait, maybe there is a typo in the original problem? Wait, no, maybe I misread. Wait, the first term is \(\frac{\log s}{\log r}\) which is \(\log_r s\), then the second term is \(8\log_r s\), so adding those two: \(\log_r s + 8\log_r s=9\log_r s\). Ah, right! I made a mistake earlier. So \(\log_r s+8\log_r s=(1 + 8)\log_r s = 9\log_r s\). Then, applying the power rule to \(9\log_r s\), we get \(\log_r s^9\) (since \(n\log_b a=\log_b a^n\), so \(9\log_r s=\log_r s^9\)).

Step3: Apply the quotient rule of logarithms

The quotient rule of logarithms states that \(\log_b a-\log_b c=\log_b\frac{a}{c}\). We have \(9\log_r s-3\log_r t\). First, apply the power rule to \(3\log_r t\) to get \(\log_r t^3\). Then, using the quotient rule on \(\log_r s^9-\log_r t^3\), we get \(\log_r\frac{s^9}{t^3}\).

Answer:

\(\boldsymbol{\log_{r}\frac{s^{9}}{t^{3}}}\) (which corresponds to the third option: \(\log_{r}\frac{s^{9}}{t^{3}}\))