Exercise Zone : Logaritma (2)

Berikut ini adalah kumpulan soal mengenai logaritma. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

Tingkat Kesulitan :

Dasar SBMPTN

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No. 11

Sederhanakanlah
\log10^{\frac12}+\log\sqrt{10^3}

Grup Whatsapp

$\begin{array}{rl}\log {10}^{\frac{1}{2}}+\log \sqrt{{10}^3}& ={\displaystyle \frac{1}{2}}+\log {10}^{\frac{3}{2}}\\ & ={\displaystyle \frac{1}{2}}+{\displaystyle \frac{3}{2}}\\ & ={\displaystyle \frac{4}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$

Jadi, \log10^{\frac12}+\log\sqrt{10^3}=2.

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No. 12

Sederhanakanlah
^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}

Grup Whatsapp

$\begin{array}{rl}{}^2\log \sqrt{2}:{}^2\log \sqrt{8}& {=}^2\log 2^{\frac{1}{2}}:{}^2\log \sqrt{2^3}\\ & ={\displaystyle \frac{1}{2}}:{}^2\log 2^{\frac{3}{2}}\\ & ={\displaystyle \frac{1}{2}}:{\displaystyle \frac{3}{2}}\\ & ={\displaystyle \frac{1}{2}}\times {\displaystyle \frac{2}{3}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{3}}}}}}\end{array}$

Jadi, ^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}=\dfrac13.

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No. 13

Sederhanakanlah
^4\negthinspace\log2+{^4\negthinspace\log32}

Grup Whatsapp

$\begin{array}{rl}{}^4\log 2+{}^4\log 32& ={}^{2^2}\log 2+{}^{2^2}\log 2^5\\ & ={\displaystyle \frac{1}{2}}+{\displaystyle \frac{5}{2}}\\ & ={\displaystyle \frac{6}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

5

Jadi, ^4\negthinspace\log2+{^4\negthinspace\log32}=3.

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No. 14

Jika x\gt0 dan y\gt0, maka \dfrac{3-3\log^2xy}{1-\log x^3y^2+2\log x\sqrt{y}}= ....

UN SMA 2018

$\begin{array}{rl}{\displaystyle \frac{3-3{\log }^{2}xy}{1-\log x^3{y}^2+2\log x\sqrt{y}}}& ={\displaystyle \frac{3(1-{\log }^{2}xy)}{1-(\log x^3{y}^2-\log {\left(x\sqrt{y}\right)}^2)}}\\ & ={\displaystyle \frac{3(1+\log xy)(1-\log xy)}{1-(\log x^3{y}^2-\log x^{2}y)}}\\ & ={\displaystyle \frac{3(1+\log xy)(1-\log xy)}{1-\log {\displaystyle \frac{x^3{y}^2}{x^{2}y}}}}\\ & ={\displaystyle \frac{3(1+\log xy)\cancel{(1-\log xy)}}{\cancel{1-\log xy}}}\\ & =3+3\log xy\end{array}$

Jadi,

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No. 15

Diketahui nilai dari ^{45}\negthinspace\log72=a dan ^{20}\negthinspace\log180=b, maka nilai dari ^3\negthinspace\log5 adalah

2. 
   
3. 
   

5. 
   

$\begin{array}{r}{}^{45}\log 72=a\\ {\displaystyle \frac{{}^3\log 72}{{}^3\log 45}}& =a\\ {\displaystyle \frac{{}^3\log (3^2\cdot 2^3)}{{}^3\log (3^2\cdot 5)}}& =a\\ {\displaystyle \frac{2+3{}^3\log 2}{2+{}^3\log 5}}& =a\\ 2+3{}^3\log 2& =2a+a{}^3\log 5\\ 3{}^3\log 2-a{}^3\log 5& =2(a-1)\end{array}$

$\begin{array}{rl}{}^{20}\log 180& =b\\ {\displaystyle \frac{{}^3\log 180}{{}^3\log 20}}& =b\\ {\displaystyle \frac{{}^3\log (3^2\cdot 2^2\cdot 5)}{{}^3\log (2^2\cdot 5)}}& =b\\ {\displaystyle \frac{2+2{}^3\log 2+{}^3\log 5}{2{}^3\log 2+{}^3\log 5}}& =b\\ 2+2{}^3\log 2+{}^3\log 5& =2b{}^3\log 2+b{}^3\log 5\\ 2(b-1){}^3\log 2+(b-1){}^3\log 5& =2\end{array}$

$\begin{array}{rlr}2(b-1){}^3\log 2+(b-1){}^3\log 5& =2& \times 3\\ 3{}^3\log 2-a{}^3\log 5& =2(a-1)& \times 2(b-1)\end{array}$

$\begin{array}{rl}6(b-1){}^3\log 2+3(b-1){}^3\log 5& =6\\ 6(b-1){}^3\log 2-2a(b-1){}^3\log 5& =4(a-1)(b-1)-\\ (3+2a)(b-1){}^3\log 5& =6-4(a-1)(b-1)\\ {}^3\log 5& ={\displaystyle \frac{6-4(a-1)(b-1)}{(3+2a)(b-1)}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{6-4(a-1)(b-1)}{(2a+3)(b-1)}}}}}}\end{array}$

Jadi, ^3\negthinspace\log5=\dfrac{6-4(a-1)(b-1)}{(2a+3)(b-1)}.
JAWAB: B

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No. 16

Hasil dari \left(\dfrac{^9\negthinspace\log4\cdot{^8\negthinspace\log3}+{^3\negthinspace\log9}}{^3\negthinspace\log54-{^3\negthinspace\log2}}\right)^2=

2. 
   
3. 
   

5. 
   

$\begin{array}{rl}{\left({\displaystyle \frac{{}^9\log 4\cdot {}^8\log 3+{}^3\log 9}{{}^3\log 54-{}^3\log 2}}\right)}^2& ={\left({\displaystyle \frac{{}^{3^2}\log 2^2\cdot {}^{2^3}\log 3+2}{{}^3\log {\displaystyle \frac{54}{2}}}}\right)}^2\\ & ={\left({\displaystyle \frac{{\displaystyle \frac{1}{3}}\cdot {}^3\log 2\cdot {}^2\log 3+2}{{}^3\log 27}}\right)}^2\\ & ={\left({\displaystyle \frac{{\displaystyle \frac{1}{3}}+2}{3}}\right)}^2\\ & ={\left({\displaystyle \frac{{\displaystyle \frac{7}{3}}}{3}}\right)}^2\\ & ={\left({\displaystyle \frac{7}{9}}\right)}^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{49}{81}}}}}}\end{array}$

Jadi, \left(\dfrac{^9\negthinspace\log4\cdot{^8\negthinspace\log3}+{^3\negthinspace\log9}}{^3\negthinspace\log54-{^3\negthinspace\log2}}\right)^2=\dfrac{49}{81}.
JAWAB: A

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No. 17

Jika 3^a=5 dan 5^b=2 maka nilai dari {^{15}\negthinspace\log}40 adalah

1. \dfrac{2b+a}{1+a}
2. \dfrac{2ab+a}{2+a}
3. \dfrac{3ab+1}{1+a}

4. \dfrac{3ab+a}{1+a}
5. \dfrac{3b+1}{1+a}

$\begin{array}{rl}{3}^a& =5\\ {}^3\log 5& =a\end{array}$

$\begin{array}{rl}{5}^b& =2\\ {}^5\log 2& =b\end{array}$

$\begin{array}{rl}{}^3\log 5\cdot {}^5\log 2& =ab\\ {}^3\log 2& =ab\end{array}$

$\begin{array}{rl}{}^{15}\log 40& ={\displaystyle \frac{{}^3\log 40}{{}^3\log 15}}\\ & ={\displaystyle \frac{{}^3\log (2^3\cdot 5)}{{}^3\log (3\cdot 5)}}\\ & ={\displaystyle \frac{{}^3\log 2^3+{}^3\log 5}{{}^3\log 3+{}^3\log 5}}\\ & ={\displaystyle \frac{3{}^3\log 2+{}^3\log 5}{{}^3\log 3+{}^3\log 5}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{3ab+a}{1+a}}}}}}\end{array}$

Jadi, {^{15}\negthinspace\log}40=\dfrac{3ab+a}{1+a}.
JAWAB: D

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No. 18

Jika ^3\negmedspace\log x = 1{,}23, maka ^3\negmedspace\log 27x =

1. 3{,}690
2. -0{,}41
   
3. 3{,}23
   

4. -1{,}77
5. 4{,}23
   

Zenius.net

$\begin{array}{rl}{}^3\log 27x& ={}^3\log 27+{}^3\log x\\ & =3+1,23\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 4,23}}}}\end{array}$

Jadi, ^3\negmedspace\log 27x =4{,}23.
JAWAB: E

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No. 19

{^5\negmedspace\log 9+{^5\negmedspace\log 2}-{^5\negmedspace\log 450}=} ....

1. -2
2. -3
3. 2

4. 3
5. -1

Zenius.net

$\begin{array}{rl}{}^5\log 9+{}^5\log 2-{}^5\log 450& ={}^5\log {\displaystyle \frac{9\cdot 2}{450}}\\ & ={}^5\log {\displaystyle \frac{1}{25}}\\ & ={}^5\log {\displaystyle \frac{1}{5^2}}\\ & ={}^5\log 5^{-2}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -2}}}}\end{array}$

Jadi, {^5\negmedspace\log 9+{^5\negmedspace\log 2}-{^5\negmedspace\log 450}=-2}.
JAWAB: A

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No. 20

Hitunglah {{^2\negmedspace\log 48}-{^2\negmedspace\log 6}}

Grup Telegram

$\begin{array}{rl}{}^2\log 48-{}^2\log 6& ={}^2\log {\displaystyle \frac{48}{6}}\\ & ={}^2\log 8\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

Jadi, {{^2\negmedspace\log 48}-{^2\negmedspace\log 6}=3}.

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