Exercise Zone : Logaritma [3]

Berikut ini adalah kumpulan soal mengenai Logaritma. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Tentukan {\log25\times{^2\negthinspace\log10}\times{^5\negthinspace\log4}}

Grup Telegram

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}\log 25\times {}^2\log 10\times {}^5\log 4& =\log 25\times {}^5\log 4\times {}^2\log 10\\ & =\log 25\times {}^5\log 2^2\times {}^2\log 10\\ & =\log 25\times 2\times {}^5\log 2\times {}^2\log 10\\ & =2\times \log 25\times {}^5\log 10\\ & =2\times {\displaystyle \frac{\log 25}{\log 5}}\\ & =2\times {}^5\log 25\\ & =2\times 2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4}}}}\end{array}$

Jadi, {\log25\times{^2\negthinspace\log10}\times{^5\negthinspace\log4}=4}.

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No.

Jika {{^2\negthinspace\log3}=a}, dan {{^2\negthinspace\log5}=b}, nyatakan ^6\negthinspace\log50 dalam a dan b.

Grup Telegram

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{}^6\log 50& ={\displaystyle \frac{{}^2\log 50}{{}^2\log 6}}\\ & ={\displaystyle \frac{{}^2\log (5^2\cdot 2)}{{}^2\log (3\cdot 2)}}\\ & ={\displaystyle \frac{{}^2\log 5^2+{}^2\log 2}{{}^2\log 3+{}^2\log 2}}\\ & ={\displaystyle \frac{2{}^2\log 5+{}^2\log 2}{{}^2\log 3+{}^2\log 2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2b+1}{a+1}}}}}}\end{array}$

Jadi, ^6\negthinspace\log50=\dfrac{2b+1}{a+1}.

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No.

{{^2\negthinspace\log1}+{^3\negthinspace\log3}-{^5\negthinspace\log2}+{^5\negthinspace\log50}-{^8\negthinspace\log32}=}

Grup Telegram

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{}^2\log 1+{}^3\log 3-{}^5\log 2+{}^5\log 50-{}^8\log 32& ={}^2\log 1+{}^3\log 3+{}^5\log 50-{}^5\log 2-{}^{2^3}\log 2^5\\ & =0+1+{}^5\log {\displaystyle \frac{50}{2}}-{\displaystyle \frac{5}{3}}\\ & =1+{}^5\log 25-{\displaystyle \frac{5}{3}}\\ & =1+2-{\displaystyle \frac{5}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{4}{3}}}}}}\end{array}$

Jadi, {^2\negthinspace\log1}+{^3\negthinspace\log3}-{^5\negthinspace\log2}+{^5\negthinspace\log50}-{^8\negthinspace\log32}=\dfrac43.

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No.

{{^5\negthinspace\log35}+{^5\negthinspace\log55}-{^5\negthinspace\log77}=}

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ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{}^5\log 35+{}^5\log 55-{}^5\log 77& ={}^5\log {\displaystyle \frac{35\cdot 55}{77}}\\ & ={}^5\log 25\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$

Jadi, {^5\negthinspace\log35}+{^5\negthinspace\log55}-{^5\negthinspace\log77}=2.

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No.

\dfrac{a^2}{b^2}=4, maka \log\dfrac{a^3}{b^3}=

Grup Telegram

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{a^2}{b^2}}& =4\\ {\left({\displaystyle \frac{a}{b}}\right)}^2& =4\\ {\displaystyle \frac{a}{b}}& =2\end{array}$

$\begin{array}{rl}\log {\displaystyle \frac{a^3}{b^3}}& =\log {\left({\displaystyle \frac{a}{b}}\right)}^3\\ & =\log 2^3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \log 8}}}}\end{array}$

Jadi, \log\dfrac{a^3}{b^3}=\log8.

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No.

Sederhanakan {^3\negthinspace\log\dfrac13}

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ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{}^3\log {\displaystyle \frac{1}{3}}& ={}^3\log 3^{-1}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -1}}}}\end{array}$

Jadi, {^3\negthinspace\log\dfrac13}=-1.

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No.

Jika {{^2\negthinspace\log5}=m} dan {{^2\negthinspace\log3}=n} nyatakan {^{50}\negthinspace\log90} dalam m dan n.

Grup Telegram

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{}^{50}\log 90& ={\displaystyle \frac{{}^2\log 90}{{}^2\log 50}}\\ & ={\displaystyle \frac{{}^2\log (5\cdot 3^2\cdot 2)}{{}^2\log (5^2\cdot 2)}}\\ & ={\displaystyle \frac{{}^2\log 5+2{}^2\log 3+{}^2\log 2}{2{}^2\log 5+{}^2\log 2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{m+2n+1}{2m+1}}}}}}\end{array}$

Jadi, {^{50}\negthinspace\log90}=\dfrac{m+2n+1}{2m+1}.

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No.

Nilai dari \dfrac{{^3\negthinspace\log36}\cdot{^6\negthinspace\log81}-{^4\negthinspace\log32}}{^{\frac19}\negthinspace\log27}=

1. 3
2. -9
3. \dfrac{11}2

4. -\dfrac{11}3
5. -11

Matematika Zone Id

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\displaystyle \frac{{}^3\log 36\cdot {}^6\log 81-{}^4\log 32}{{}^{\frac{1}{9}}\log 27}}& ={\displaystyle \frac{{}^3\log 6^2\cdot {}^6\log 81-{}^{2^2}\log 2^5}{{}^{\frac{1}{3^2}}\log 3^3}}\\ & ={\displaystyle \frac{2{}^3\log 6\cdot {}^6\log 81-{\displaystyle \frac{5}{2}}}{{}^{3^{-2}}\log 3^3}}\\ & ={\displaystyle \frac{2{}^3\log 81-{\displaystyle \frac{5}{2}}}{-{\displaystyle \frac{3}{2}}}}\\ & ={\displaystyle \frac{2(4)-{\displaystyle \frac{5}{2}}}{-{\displaystyle \frac{3}{2}}}}\\ & ={\displaystyle \frac{8-{\displaystyle \frac{5}{2}}}{-{\displaystyle \frac{3}{2}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{11}{2}}}{-{\displaystyle \frac{3}{2}}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -{\displaystyle \frac{11}{3}}}}}}\end{array}$

Jadi, \dfrac{{^3\negthinspace\log36}\cdot{^6\negthinspace\log81}-{^4\negthinspace\log32}}{^{\frac19}\negthinspace\log27}=-\dfrac{11}3.
JAWAB: D

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No.

Hasil dari {\left(^{c^{\frac72}}\negthinspace\log b^{14}\right)\cdot\left(^{a^{\frac53}}\negthinspace\log c^{10}\right)\cdot\left(^{b^{\frac32}}\negthinspace\log a^9\right)} adalah

1. 144
2. 155
3. 166

4. 177
5. 188

SKMP Juli 2020

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}({}^{c^{\frac{7}{2}}}\log b^{14})\cdot ({}^{a^{\frac{5}{3}}}\log c^{10})\cdot ({}^{b^{\frac{3}{2}}}\log a^9)& =({\displaystyle \frac{2}{7}}\cdot 14\cdot {}^c\log b)\cdot ({\displaystyle \frac{3}{5}}\cdot 10\cdot {}^a\log c)\cdot ({\displaystyle \frac{2}{3}}\cdot 9\cdot {}^b\log a)\\ & =(4\cdot {}^c\log b)\cdot (6\cdot {}^a\log c)\cdot (6\cdot {}^b\log a)\\ & =144\cdot {}^c\log b\cdot {}^a\log c\cdot {}^b\log a\\ & =144\cdot {}^a\log c\cdot {}^c\log b\cdot {}^b\log a\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 144}}}}\end{array}$/div>

Jadi, {\left(^{c^{\frac72}}\negthinspace\log b^{14}\right)\cdot\left(^{a^{\frac53}}\negthinspace\log c^{10}\right)\cdot\left(^{b^{\frac32}}\negthinspace\log a^9\right)}=144 .
JAWAB: A

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