Rumus-rumus di atas untuk membantu dan memahami soal-soal berikut ini :
$\boxed{1.}$ $\log 2+\log 18-\log 6+\log 5-\log 3=\ldots$
A. 90
Jawaban:
$\begin{aligned}
&\log 2+\log 18-\log 6+\log 5-\log 3\\
&=\log\left( \dfrac{2\times18\times5}{6\times3} \right)\\
&=\log 2\times5\Leftrightarrow {}^{10}\log 10\\
&=1
\end{aligned}$
$\boxed{2.}$ ${ }^5 \log 150-{ }^5 \log 24+{ }^5 \log 4=\ldots$.
Jawaban:
$\begin{aligned}
&{ }^5 \log 150-{ }^5 \log 24+{ }^5 \log 4\\
&={ }^5 \log\left( 150\times \dfrac{1}{24}\times4\right)\\
&={ }^5 \log25\\
&={ }^5 \log5^{2}\\
&=2\cdot{ }^5 \log5\\
&=2
\end{aligned}$
$\boxed{3.}$ $\log 30-\dfrac{1}{{ }^{48} \log 10}+\dfrac{1}{{ }^{16} \log 10}=\ldots$
A. 0
Jawaban:
$\begin{aligned}
\log 30-\dfrac{1}{{ }^{48} \log 10}+\dfrac{1}{{ }^{16} \log 10}&=\log 30-{{ }^{10} \log 48}+{{ }^{10} \log 16}=\ldots\\
&=\log\left( 30\times\frac{1}{48}\times16 \right)\\
&=\log(10) \\
&=1\end{aligned}$
$\boxed{4.}$ $\quad$ Nilai dari $\dfrac{{ }^2 \log \sqrt{5}+2 \cdot{ }^4 \log 5}{{ }^2 \log 3 \cdot{ }^3 \log 5}=\ldots$
A. 3
Jawaban:
$\begin{aligned}
\dfrac{{ }^2 \log \sqrt{5}+2 \cdot{ }^4 \log 5}{{ }^2 \log 3 \cdot{ }^3 \log 5}&=\dfrac{{ }^2 \log {5}^{\frac{1}{2}}+2 \cdot{ }^{2^2} \log 5}{{ }^2 \log 5}\\
&=\dfrac{{ }\frac{1}{2}.^2 \log {5}+2\cdot\frac{1}{2}\cdot{ }^{2} \log 5}{{ }^2 \log 5}\\
&=\frac{{}^2\log5\left( \frac{1}{2}+1 \right)}{{}^2\log5}\\
&=\frac{3}{2}
\end{aligned}$
$\boxed{1.}$ $\log 2+\log 18-\log 6+\log 5-\log 3=\ldots$
A. 90
B. 10
C. 1
D. 0
E. -1
D. 0
E. -1
Jawaban:
$\begin{aligned}
&\log 2+\log 18-\log 6+\log 5-\log 3\\
&=\log\left( \dfrac{2\times18\times5}{6\times3} \right)\\
&=\log 2\times5\Leftrightarrow {}^{10}\log 10\\
&=1
\end{aligned}$
$\boxed{2.}$ ${ }^5 \log 150-{ }^5 \log 24+{ }^5 \log 4=\ldots$.
A. 1
B. 2
C. 3
D. 4
E. 5
D. 4
E. 5
$\begin{aligned}
&{ }^5 \log 150-{ }^5 \log 24+{ }^5 \log 4\\
&={ }^5 \log\left( 150\times \dfrac{1}{24}\times4\right)\\
&={ }^5 \log25\\
&={ }^5 \log5^{2}\\
&=2\cdot{ }^5 \log5\\
&=2
\end{aligned}$
$\boxed{3.}$ $\log 30-\dfrac{1}{{ }^{48} \log 10}+\dfrac{1}{{ }^{16} \log 10}=\ldots$
A. 0
B. 1
C. 10
D. 18
E. 60
D. 18
E. 60
Jawaban:
$\begin{aligned}
\log 30-\dfrac{1}{{ }^{48} \log 10}+\dfrac{1}{{ }^{16} \log 10}&=\log 30-{{ }^{10} \log 48}+{{ }^{10} \log 16}=\ldots\\
&=\log\left( 30\times\frac{1}{48}\times16 \right)\\
&=\log(10) \\
&=1\end{aligned}$
$\boxed{4.}$ $\quad$ Nilai dari $\dfrac{{ }^2 \log \sqrt{5}+2 \cdot{ }^4 \log 5}{{ }^2 \log 3 \cdot{ }^3 \log 5}=\ldots$
A. 3
B. 2
C. $\frac{3}{2}$
D. $\frac{2}{3}$
E. $\frac{1}{2}$
D. $\frac{2}{3}$
E. $\frac{1}{2}$
Jawaban:
$\begin{aligned}
\dfrac{{ }^2 \log \sqrt{5}+2 \cdot{ }^4 \log 5}{{ }^2 \log 3 \cdot{ }^3 \log 5}&=\dfrac{{ }^2 \log {5}^{\frac{1}{2}}+2 \cdot{ }^{2^2} \log 5}{{ }^2 \log 5}\\
&=\dfrac{{ }\frac{1}{2}.^2 \log {5}+2\cdot\frac{1}{2}\cdot{ }^{2} \log 5}{{ }^2 \log 5}\\
&=\frac{{}^2\log5\left( \frac{1}{2}+1 \right)}{{}^2\log5}\\
&=\frac{3}{2}
\end{aligned}$
$\boxed{5.}$ Ditentukan
${ }^P \log 81-2{ }^P \log 27+{ }^P \log 27+{ }^P \log 243=6$ Nilai $p=\ldots$.
A. $\sqrt{3}$
Jawaban:
$\begin{aligned}
{ }^P \log 81-2{ }^P \log 27+{ }^P \log 27+{ }^P \log 243&=6\\
{ }^P \log 3^4-2\cdot{ }^P \log 3^3+{ }^P \log 3^3+{ }^P \log 3^5&=6\\
4\cdot{ }^P \log 3-2\cdot3\cdot{ }^P \log 3+3\cdot{ }^P \log 3+5\cdot{ }^P \log 3&=6\\
\left( 4-6+3+5 \right){}^P\log 3&=6\\
\left( 6 \right){}^P\log 3&=6\\
{}^P\log 3&=\frac{6}{6}\\
{}^P\log 3&=1 \Leftrightarrow {}^a\log a=1\\
\text{Jadi: nilai } P&=3
\end{aligned}$
Jawaban:
$\begin{aligned}
&{ }^6 \log 12 \sqrt{2}+{ }^6 \log \sqrt{3}+{ }^6 \log 18-{ }^6 \log 2-{ }^6 \log 3\\
&={}^6 \log\left( \frac{12 \sqrt{2}\times\sqrt{3}\times18}{2\times3} \right)\\
&={}^6 \log\left( {12 \sqrt{2}\times\sqrt{3}\times3} \right)\\
&={}^6 \log\left( {36 \sqrt{6}} \right)\\
&={}^{6^2} \log\left( {36 \sqrt{6}} \right)^2\\
&={}^{6^2} \log {36^2 \cdot(\sqrt{6})} ^2\\
&={}^{6^2} \log {(6^2)^2}\cdot({6} )\\
&={}^{6^2} \log {6^5}\\
&=\frac{5}{2}\cdot {}^6\log6 \Leftrightarrow \frac{5}{2}\cdot 1\\
&=\frac{5}{2} \text{atau:} \Leftrightarrow 2\frac{1}{2}\Leftrightarrow 2,5\end{aligned}$
Soal 22.
${ }^P \log 81-2{ }^P \log 27+{ }^P \log 27+{ }^P \log 243=6$ Nilai $p=\ldots$.
A. $\sqrt{3}$
B. 2
C. 3
D. $3 \sqrt{3}$
E. 9
D. $3 \sqrt{3}$
E. 9
$\begin{aligned}
{ }^P \log 81-2{ }^P \log 27+{ }^P \log 27+{ }^P \log 243&=6\\
{ }^P \log 3^4-2\cdot{ }^P \log 3^3+{ }^P \log 3^3+{ }^P \log 3^5&=6\\
4\cdot{ }^P \log 3-2\cdot3\cdot{ }^P \log 3+3\cdot{ }^P \log 3+5\cdot{ }^P \log 3&=6\\
\left( 4-6+3+5 \right){}^P\log 3&=6\\
\left( 6 \right){}^P\log 3&=6\\
{}^P\log 3&=\frac{6}{6}\\
{}^P\log 3&=1 \Leftrightarrow {}^a\log a=1\\
\text{Jadi: nilai } P&=3
\end{aligned}$
Soal 21.
Hasil dari ${ }^6 \log 12 \sqrt{2}+{ }^6 \log \sqrt{3}+{ }^6 \log 18-{
}^6 \log 2-{ }^6 \log 3=$
A. 2
D. 5
E. 6
A. 2
B. 2,5
C. 3D. 5
E. 6
Jawaban:
$\begin{aligned}
&{ }^6 \log 12 \sqrt{2}+{ }^6 \log \sqrt{3}+{ }^6 \log 18-{ }^6 \log 2-{ }^6 \log 3\\
&={}^6 \log\left( \frac{12 \sqrt{2}\times\sqrt{3}\times18}{2\times3} \right)\\
&={}^6 \log\left( {12 \sqrt{2}\times\sqrt{3}\times3} \right)\\
&={}^6 \log\left( {36 \sqrt{6}} \right)\\
&={}^{6^2} \log\left( {36 \sqrt{6}} \right)^2\\
&={}^{6^2} \log {36^2 \cdot(\sqrt{6})} ^2\\
&={}^{6^2} \log {(6^2)^2}\cdot({6} )\\
&={}^{6^2} \log {6^5}\\
&=\frac{5}{2}\cdot {}^6\log6 \Leftrightarrow \frac{5}{2}\cdot 1\\
&=\frac{5}{2} \text{atau:} \Leftrightarrow 2\frac{1}{2}\Leftrightarrow 2,5\end{aligned}$
Soal 22.
Diketahui:
$
\begin{aligned}
& x={ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10 \\
& y={ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10
\end{aligned}
$
Nilai $\dfrac{y}{x}=\ldots$.
A. $\frac{1}{2}$
Soal 26.
$
\begin{aligned}
& x={ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10 \\
& y={ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10
\end{aligned}
$
Nilai $\dfrac{y}{x}=\ldots$.
A. $\frac{1}{2}$
B. $\frac{5}{8}$
C. 1
D. $\frac{8}{5}$
E. 2
D. $\frac{8}{5}$
E. 2
Jawaban:
$\begin{aligned}
\text{Diketahui: }\\
x&={ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10 \\
y&={ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10\\
\text{maka : }\\
\frac{y}{x}&=\dfrac{{ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10}{{ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10}\\
\\
\text{untuk : } \\
y&={ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10\\
&={}^5\log\left( \dfrac{ 7 \times 5 \times 10}{14} \right)\Leftrightarrow {}^5\log(5\times5)\Leftrightarrow {}^5\log5^2\\
&=2\cdot1\\
&=2\\
\\
\text{untuk : } \\
x&={ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10\\
&={ }^3 \log\left( \frac{5\times12}{2\times10} \right)\\
&={ }^3 \log3\\
&=1\\
\text{Jadi :}\dfrac{y}{x}&=\dfrac{2}{1}\\
&=2
\end{aligned}$
Soal 23.
Jawaban :
$\begin{aligned}
&4^{{}^2 \log 5}+8^{{}^2 \log {\frac{1}{2}}}-{ }^{256} \log 2\\
&=(2^2)^{{}^2 \log 5}+(2^3)^{{}^2 \log {\frac{1}{2}}}-{ }^{2^8} \log 2\\
&=(2)^{2\cdot{}^2 \log 5}+(2)^{3\cdot{}^2 \log {\frac{1}{2}}}-{ }^{2^8} \log 2\\
&=2^{{}^2 \log 5^2}+2^{{}^2 \log {(\frac{1}{2}})^3}-{ }^{2^8} \log 2\\
&=5^2+(\frac{1}{2})^3-\frac{1}{8}\\
&=25+\frac{1}{8}-\frac{1}{8}\\
&=25
\end{aligned}$
Soal 25.
Jawaban:
$\begin{aligned}
&\dfrac{2 \cdot{ }^3 \log \dfrac{1}{9}+{ }^4 \log 2}{{ }^3 \log 2 \cdot{ }^{25} \log 3 \cdot{ }^2 \log 5}\\
\\
&=\dfrac{2 \cdot{ }^3 \log 3^{-2}+{ }^{2^2} \log 2}{{ }^3 \log 2 \cdot{ }^{5^2} \log 3 \cdot{ }^2 \log 5}\\
\\
&=\dfrac{2 \cdot(-2)\cdot{ }^3 \log 3+\frac{1}{2}\cdot{ }^{2} \log 2}{{ }^3 \log 2 \cdot\frac{1}{2 }\cdot^{5}\log 3 \cdot{ }^2 \log 5}\\
\\
&=\dfrac{-4+\frac{1}{2}}{\frac{1}{2}}\\
&=(-4+\frac{1}{2})\cdot2\\
&=-7
\end{aligned}$
$\begin{aligned}
\text{Diketahui: }\\
x&={ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10 \\
y&={ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10\\
\text{maka : }\\
\frac{y}{x}&=\dfrac{{ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10}{{ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10}\\
\\
\text{untuk : } \\
y&={ }^5 \log 7+{ }^5 \log 5-{ }^5 \log 14+{ }^5 \log 10\\
&={}^5\log\left( \dfrac{ 7 \times 5 \times 10}{14} \right)\Leftrightarrow {}^5\log(5\times5)\Leftrightarrow {}^5\log5^2\\
&=2\cdot1\\
&=2\\
\\
\text{untuk : } \\
x&={ }^3 \log 5+{ }^3 \log 12-{ }^3 \log 2-{ }^3 \log 10\\
&={ }^3 \log\left( \frac{5\times12}{2\times10} \right)\\
&={ }^3 \log3\\
&=1\\
\text{Jadi :}\dfrac{y}{x}&=\dfrac{2}{1}\\
&=2
\end{aligned}$
Soal 23.
Diketahui $p={ }^3 \log \dfrac{1}{125} \cdot{ }^8 \log 9 \cdot{ }^5 \log 64$. Nilai p yang memenuhi adalah ....
A. 27
B. 12
C. 6
D. -12
E. -27
A. 27
B. 12
C. 6
D. -12
E. -27
Jawaban :
$\begin{aligned}
&p={ }^3 \log \dfrac{1}{125} \cdot{ }^8 \log 9 \cdot{ }^5 \log 64\\
&p={ }^3 \log \dfrac{1}{5^3} \cdot{ }^{2^3} \log 3^2 \cdot{ }^5 \log 2^6\\
&p={ }^3 \log {5^{-3}} \cdot\frac{2}{3}\cdot {}^2\log 3 \cdot6\cdot{ }^5 \log 2\\
&p=-3\cdot({ }^3 \log {5}) \cdot\frac{2}{3}\cdot ({}^2\log 3) \cdot6\cdot({ }^5 \log 2)\\
&p=(-3\times\frac{2}{3}\times6)\cdot({ }^3 \log {5} \cdot{ }^5 \log 2\cdot {}^2\log 3 )\\
&p=(-12)\cdot({ }^3 \log 3 )\Leftrightarrow p=-12\cdot1\\
&p=-12
\end{aligned}$
Soal 24.
$\begin{aligned}
&p={ }^3 \log \dfrac{1}{125} \cdot{ }^8 \log 9 \cdot{ }^5 \log 64\\
&p={ }^3 \log \dfrac{1}{5^3} \cdot{ }^{2^3} \log 3^2 \cdot{ }^5 \log 2^6\\
&p={ }^3 \log {5^{-3}} \cdot\frac{2}{3}\cdot {}^2\log 3 \cdot6\cdot{ }^5 \log 2\\
&p=-3\cdot({ }^3 \log {5}) \cdot\frac{2}{3}\cdot ({}^2\log 3) \cdot6\cdot({ }^5 \log 2)\\
&p=(-3\times\frac{2}{3}\times6)\cdot({ }^3 \log {5} \cdot{ }^5 \log 2\cdot {}^2\log 3 )\\
&p=(-12)\cdot({ }^3 \log 3 )\Leftrightarrow p=-12\cdot1\\
&p=-12
\end{aligned}$
Soal 24.
Nilai $4^{2 \log 5}+8^{2 \log {\frac{1}{2}}}-{ }^{256} \log 2=\ldots$
A. 23
B. 24
C. 25
D. 26
E. 27
A. 23
B. 24
C. 25
D. 26
E. 27
Jawaban :
$\begin{aligned}
&4^{{}^2 \log 5}+8^{{}^2 \log {\frac{1}{2}}}-{ }^{256} \log 2\\
&=(2^2)^{{}^2 \log 5}+(2^3)^{{}^2 \log {\frac{1}{2}}}-{ }^{2^8} \log 2\\
&=(2)^{2\cdot{}^2 \log 5}+(2)^{3\cdot{}^2 \log {\frac{1}{2}}}-{ }^{2^8} \log 2\\
&=2^{{}^2 \log 5^2}+2^{{}^2 \log {(\frac{1}{2}})^3}-{ }^{2^8} \log 2\\
&=5^2+(\frac{1}{2})^3-\frac{1}{8}\\
&=25+\frac{1}{8}-\frac{1}{8}\\
&=25
\end{aligned}$
Nilai dari $\dfrac{2 \cdot{ }^3 \log \dfrac{1}{9}+{ }^4 \log 2}{{ }^3 \log 2 \cdot{ }^{25} \log 3 \cdot{ }^2 \log 5}=\ldots$
A. -16
B. -9
C. -7
D. 7
E. 9
A. -16
B. -9
C. -7
D. 7
E. 9
$\begin{aligned}
&\dfrac{2 \cdot{ }^3 \log \dfrac{1}{9}+{ }^4 \log 2}{{ }^3 \log 2 \cdot{ }^{25} \log 3 \cdot{ }^2 \log 5}\\
\\
&=\dfrac{2 \cdot{ }^3 \log 3^{-2}+{ }^{2^2} \log 2}{{ }^3 \log 2 \cdot{ }^{5^2} \log 3 \cdot{ }^2 \log 5}\\
\\
&=\dfrac{2 \cdot(-2)\cdot{ }^3 \log 3+\frac{1}{2}\cdot{ }^{2} \log 2}{{ }^3 \log 2 \cdot\frac{1}{2 }\cdot^{5}\log 3 \cdot{ }^2 \log 5}\\
\\
&=\dfrac{-4+\frac{1}{2}}{\frac{1}{2}}\\
&=(-4+\frac{1}{2})\cdot2\\
&=-7
\end{aligned}$
Soal 26.
Nilai dari
$\dfrac{{}^{0,5} \log 72+{ }^2 \log 144}{{ }^3 \log \sqrt{5}-{
}^3 \log 3 \sqrt{5}}=\ldots$.
A. -2
Jawaban:
$\begin{aligned}
&\dfrac{{}^{0,5} \log 72+{ }^2 \log 144}{{ }^3 \log \sqrt{5}-{ }^3 \log 3 \sqrt{5}}\\
\\
&=\dfrac{{}^{\frac{1}{2}} \log (8\cdot9)+{ }^2 \log (12)^2}{{ }^3 \log(\frac{\sqrt{5}}{3\sqrt{5}})}\\
&=\dfrac{{}^{2} \log (2^3\cdot3^2)^{-1}+{ }^2 \log (2^2\cdot3)^2}{{ }^3 \log(\frac{1}{3})}\\
&=\dfrac{{}^{2} \log (2^{-3}\cdot3^{-2}\cdot2^4\cdot3^2)}{-1}\\
&=\dfrac{{}^{2} \log (2\cdot3^0)}{-1}\Leftrightarrow \dfrac{1}{-1}\\
&=-1\end{aligned}$
A. -2
B. -1
C. 0
D. 1
E. 2
D. 1
E. 2
Jawaban:
$\begin{aligned}
&\dfrac{{}^{0,5} \log 72+{ }^2 \log 144}{{ }^3 \log \sqrt{5}-{ }^3 \log 3 \sqrt{5}}\\
\\
&=\dfrac{{}^{\frac{1}{2}} \log (8\cdot9)+{ }^2 \log (12)^2}{{ }^3 \log(\frac{\sqrt{5}}{3\sqrt{5}})}\\
&=\dfrac{{}^{2} \log (2^3\cdot3^2)^{-1}+{ }^2 \log (2^2\cdot3)^2}{{ }^3 \log(\frac{1}{3})}\\
&=\dfrac{{}^{2} \log (2^{-3}\cdot3^{-2}\cdot2^4\cdot3^2)}{-1}\\
&=\dfrac{{}^{2} \log (2\cdot3^0)}{-1}\Leftrightarrow \dfrac{1}{-1}\\
&=-1\end{aligned}$
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