\begin{tabular}{lll}\hline Example 2: H.W \ \begin{tabular}{lll} a $$ \log _{3}(27)+1 $$ & b $$ \log _{4}(16)+3 $$ & c $$ 3 \log _{5}(2)-2 $$ \ d $$ 2+3 \log _{10}(x) $$ & e $$ 4 \log _{3}(2)-2 \log _{3}(6)+2 $$ & f $$ \frac{1}{2}+3 \log _{10}\left(x^{2} ight) $$ \ \hline\end{tabular}\end{tabular}

Question

\begin{tabular}{lll}\hline Example 2: H.W \ \begin{tabular}{lll} a $$ \log _{3}(27)+1 $$ & b $$ \log _{4}(16)+3 $$ & c $$ 3 \log _{5}(2)-2 $$ \ d $$ 2+3 \log _{10}(x) $$ & e $$ 4 \log _{3}(2)-2 \log _{3}(6)+2 $$ & f $$ \frac{1}{2}+3 \log _{10}\left(x^{2}ight) $$ \ \hline\end{tabular}\end{tabular}

UpStudy AI · Accepted Answer

a) We have
$$
\log_{3}(27)+1.
$$
Since $$27=3^3$$, it follows that 
$$
\log_{3}(27)=3.
$$
Thus, 
$$
3+1=4.
$$

---

b) We have
$$
\log_{4}(16)+3.
$$
Since $$16=4^2$$, we get 
$$
\log_{4}(16)=2.
$$
Thus,
$$
2+3=5.
$$

---

c) We have
$$
3\log_{5}(2)-2.
$$
First, rewrite using the power rule:
$$
3\log_{5}(2)=\log_{5}(2^3)=\log_{5}(8).
$$
The expression becomes
$$
\log_{5}(8)-2.
$$
Express the constant $$2$$ as a logarithm:
$$
2=\log_{5}(5^2)=\log_{5}(25).
$$
Thus,
$$
\log_{5}(8)-\log_{5}(25)=\log_{5}\left(\frac{8}{25}\right).
$$

---

d) We have
$$
2+3\log_{10}(x).
$$
Express the constant $$2$$ as a logarithm:
$$
2=\log_{10}(10^2)=\log_{10}(100).
$$
Also, apply the power rule:
$$
3\log_{10}(x)=\log_{10}(x^3).
$$
Thus,
$$
\log_{10}(100)+\log_{10}(x^3)=\log_{10}(100x^3).
$$

---

e) We have
$$
4\log_{3}(2)-2\log_{3}(6)+2.
$$
Apply the power rule to the logarithmic terms:
$$
4\log_{3}(2)=\log_{3}(2^4)=\log_{3}(16)
$$
and 
$$
2\log_{3}(6)=\log_{3}(6^2)=\log_{3}(36).
$$
Thus, the expression becomes
$$
\log_{3}(16)-\log_{3}(36)+2=\log_{3}\left(\frac{16}{36}\right)+2=\log_{3}\left(\frac{4}{9}\right)+2.
$$
Express the constant $$2$$ as a logarithm:
$$
2=\log_{3}(3^2)=\log_{3}(9).
$$
Combine the logarithms:
$$
\log_{3}\left(\frac{4}{9}\right)+\log_{3}(9)=\log_{3}\left(\frac{4}{9}\cdot9\right)=\log_{3}(4).
$$

---

f) We have
$$
\frac{1}{2}+3\log_{10}\left(x^{2}\right).
$$
Apply the power rule to the logarithmic term:
$$
3\log_{10}\left(x^{2}\right)=\log_{10}\left((x^2)^3\right)=\log_{10}(x^6).
$$
Express the constant $$\frac{1}{2}$$ as a logarithm:
$$
\frac{1}{2}=\log_{10}(10^{1/2})=\log_{10}\left(\sqrt{10}\right).
$$
Thus, combining the logarithms:
$$
\log_{10}\left(\sqrt{10}\right)+\log_{10}(x^6)=\log_{10}\left(\sqrt{10}\,x^6\right).
$$
a) 4 b) 5 c) $$\log_{5}\left(\frac{8}{25}\right)$$ d) $$\log_{10}(100x^3)$$ e) $$\log_{3}(4)$$ f) $$\log_{10}\left(\sqrt{10}\,x^6\right)$$

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