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Exercise Zone : Logaritma [2]

Written By Anonymous Thursday, July 16, 2020 Add Comment Edit
Berikut ini adalah kumpulan soal mengenai logaritma tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

No. 11

Sederhanakanlah
log⁡1012+log⁡103\log10^{\frac12}+\log\sqrt{10^3}
log⁡1012+log⁡103=12+log⁡1032=12+32=42=2\begin{aligned} \log10^{\frac12}+\log\sqrt{10^3}&=\dfrac12+\log10^{\frac32}\\[5pt] &=\dfrac12+\dfrac32\\[5pt] &=\dfrac42\\ &=\boxed{\boxed{2}} \end{aligned}
Jadi, log⁡1012+log⁡103=2\log10^{\frac12}+\log\sqrt{10^3}=2.

No. 12

Sederhanakanlah
2 ⁣log⁡2:2 ⁣log⁡8^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}
2 ⁣log⁡2:2 ⁣log⁡8=2 ⁣log⁡212:2 ⁣log⁡23=12:2 ⁣log⁡232=12:32=12×23=13\begin{aligned} ^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}&=^2\negthinspace\log2^{\frac12}:{^2\negthinspace\log\sqrt{2^3}}\\[5pt] &=\dfrac12:{^2\negthinspace\log2^{\frac32}}\\[5pt] &=\dfrac12:\dfrac32\\[5pt] &=\dfrac12\times\dfrac23\\ &=\boxed{\boxed{\dfrac13}} \end{aligned}
Jadi, 2 ⁣log⁡2:2 ⁣log⁡8=13^2\negthinspace\log\sqrt2:{^2\negthinspace\log\sqrt8}=\dfrac13.

No. 13

Sederhanakanlah
4 ⁣log⁡2+4 ⁣log⁡32^4\negthinspace\log2+{^4\negthinspace\log32}
4 ⁣log⁡2+4 ⁣log⁡32=22 ⁣log⁡2+22 ⁣log⁡25=12+52=62=3\begin{aligned} ^4\negthinspace\log2+{^4\negthinspace\log32}&={^{2^2}\negthinspace\log2}+{^{2^2}\negthinspace\log2^5}\\ &=\dfrac12+\dfrac52\\[5pt] &=\dfrac62\\ &=\boxed{\boxed{3}} \end{aligned}
Jadi, 4 ⁣log⁡2+4 ⁣log⁡32=3^4\negthinspace\log2+{^4\negthinspace\log32}=3.

No. 14

Jika x>0x\gt0 dan y>0y\gt0, maka 3−3log⁡2xy1−log⁡x3y2+2log⁡xy=\dfrac{3-3\log^2xy}{1-\log x^3y^2+2\log x\sqrt{y}}= ....
UN SMA 2018
3−3log⁡2xy1−log⁡x3y2+2log⁡xy=3(1−log⁡2xy)1−(log⁡x3y2−log⁡(xy)2)=3(1+log⁡xy)(1−log⁡xy)1−(log⁡x3y2−log⁡x2y)=3(1+log⁡xy)(1−log⁡xy)1−log⁡x3y2x2y=3(1+log⁡xy)(1−log⁡xy)1−log⁡xy=3+3log⁡xy\begin{aligned} \dfrac{3-3\log^2xy}{1-\log x^3y^2+2\log x\sqrt{y}}&=\dfrac{3\left(1-\log^2xy\right)}{1-\left(\log x^3y^2-\log\left(x\sqrt{y}\right)^2\right)}\\ &=\dfrac{3\left(1+\log xy\right)\left(1-\log xy\right)}{1-\left(\log x^3y^2-\log x^2y\right)}\\ &=\dfrac{3\left(1+\log xy\right)\left(1-\log xy\right)}{1-\log\dfrac{x^3y^2}{x^2y}}\\ &=\dfrac{3\left(1+\log xy\right)\cancel{\left(1-\log xy\right)}}{\cancel{1-\log xy}}\\ &=3+3\log xy \end{aligned}
Jadi,

No. 15

Diketahui nilai dari 45 ⁣log⁡72=a^{45}\negthinspace\log72=a dan 20 ⁣log⁡180=b^{20}\negthinspace\log180=b, maka nilai dari 3 ⁣log⁡5^3\negthinspace\log5 adalah



45 ⁣log⁡72=a3 ⁣log⁡723 ⁣log⁡45=a3 ⁣log⁡(32⋅23)3 ⁣log⁡(32⋅5)=a2+3 3 ⁣log⁡22+3 ⁣log⁡5=a2+3 3 ⁣log⁡2=2a+a 3 ⁣log⁡53 3 ⁣log⁡2−a 3 ⁣log⁡5=2(a−1)\begin{aligned} ^{45}\negthinspace\log72=a\\ \dfrac{^3\negthinspace\log72}{^3\negthinspace\log45}&=a\\ \dfrac{^3\negthinspace\log\left(3^2\cdot2^3\right)}{^3\negthinspace\log\left(3^2\cdot5\right)}&=a\\ \dfrac{2+3\ ^3\negthinspace\log2}{2+{^3\negthinspace\log5}}&=a\\ 2+3\ ^3\negthinspace\log2&=2a+a\ ^3\negthinspace\log5\\ 3\ ^3\negthinspace\log2-a\ ^3\negthinspace\log5&=2(a-1) \end{aligned}

20 ⁣log⁡180=b3 ⁣log⁡1803 ⁣log⁡20=b3 ⁣log⁡(32⋅22⋅5)3 ⁣log⁡(22⋅5)=b2+2 3 ⁣log⁡2+3 ⁣log⁡52 3 ⁣log⁡2+3 ⁣log⁡5=b2+2 3 ⁣log⁡2+3 ⁣log⁡5=2b 3 ⁣log⁡2+b 3 ⁣log⁡52(b−1) 3 ⁣log⁡2+(b−1) 3 ⁣log⁡5=2\begin{aligned} ^{20}\negthinspace\log180&=b\\ \dfrac{^3\negthinspace\log180}{^3\negthinspace\log20}&=b\\ \dfrac{^3\negthinspace\log\left(3^2\cdot2^2\cdot5\right)}{^3\negthinspace\log\left(2^2\cdot5\right)}&=b\\ \dfrac{2+2\ ^3\negthinspace\log2+{^3\negthinspace\log5}}{2\ ^3\negthinspace\log2+{^3\negthinspace\log5}}&=b\\ 2+2\ ^3\negthinspace\log2+{^3\negthinspace\log5}&=2b\ ^3\negthinspace\log2+b\ ^3\negthinspace\log5\\ 2(b-1)\ ^3\negthinspace\log2+(b-1)\ ^3\negthinspace\log5&=2 \end{aligned}

2(b−1) 3 ⁣log⁡2+(b−1) 3 ⁣log⁡5=2×33 3 ⁣log⁡2−a 3 ⁣log⁡5=2(a−1)×2(b−1)\begin{aligned} 2(b-1)\ ^3\negthinspace\log2+(b-1)\ ^3\negthinspace\log5&=2\qquad&\color{red}{\times3}\\ 3\ ^3\negthinspace\log2-a\ ^3\negthinspace\log5&=2(a-1)\qquad&\color{red}{\times2(b-1)} \end{aligned}

6(b−1) 3 ⁣log⁡2+3(b−1) 3 ⁣log⁡5=66(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=6−4(a−1)(b−1)(3+2a)(b−1)=6−4(a−1)(b−1)(2a+3)(b−1)\begin{aligned} 6(b-1)\ ^3\negthinspace\log2+3(b-1)\ ^3\negthinspace\log5&=6\\ 6(b-1)\ ^3\negthinspace\log2-2a(b-1)\ ^3\negthinspace\log5&=4(a-1)(b-1)\qquad-\\\hline (3+2a)(b-1)\ ^3\negthinspace\log5&=6-4(a-1)(b-1)\\ ^3\negthinspace\log5&=\dfrac{6-4(a-1)(b-1)}{(3+2a)(b-1)}\\ &=\boxed{\boxed{\dfrac{6-4(a-1)(b-1)}{(2a+3)(b-1)}}} \end{aligned}
Jadi, 3 ⁣log⁡5=6−4(a−1)(b−1)(2a+3)(b−1)^3\negthinspace\log5=\dfrac{6-4(a-1)(b-1)}{(2a+3)(b-1)}.
JAWAB: B

No. 16

Hasil dari (9 ⁣log⁡4⋅8 ⁣log⁡3+3 ⁣log⁡93 ⁣log⁡54−3 ⁣log⁡2)2=\left(\dfrac{^9\negthinspace\log4\cdot{^8\negthinspace\log3}+{^3\negthinspace\log9}}{^3\negthinspace\log54-{^3\negthinspace\log2}}\right)^2=



(9 ⁣log⁡4⋅8 ⁣log⁡3+3 ⁣log⁡93 ⁣log⁡54−3 ⁣log⁡2)2=(32 ⁣log⁡22⋅23 ⁣log⁡3+23 ⁣log⁡542)2=(13⋅3 ⁣log⁡2⋅2 ⁣log⁡3+23 ⁣log⁡27)2=(13+23)2=(733)2=(79)2=4981\begin{aligned} \left(\dfrac{^9\negthinspace\log4\cdot{^8\negthinspace\log3}+{^3\negthinspace\log9}}{^3\negthinspace\log54-{^3\negthinspace\log2}}\right)^2&=\left(\dfrac{^{3^2}\negthinspace\log2^2\cdot{^{2^3}\negthinspace\log3}+2}{^3\negthinspace\log\dfrac{54}2}\right)^2\\ &=\left(\dfrac{\dfrac13\cdot{^3\negthinspace\log2}\cdot{^2\negthinspace\log3}+2}{^3\negthinspace\log27}\right)^2\\ &=\left(\dfrac{\dfrac13+2}3\right)^2\\ &=\left(\dfrac{\dfrac73}3\right)^2\\ &=\left(\dfrac79\right)^2\\ &=\boxed{\boxed{\dfrac{49}{81}}} \end{aligned}
Jadi, (9 ⁣log⁡4⋅8 ⁣log⁡3+3 ⁣log⁡93 ⁣log⁡54−3 ⁣log⁡2)2=4981\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

No. 17

Jika 3a=53^a=5 dan 5b=25^b=2 maka nilai dari 15 ⁣log⁡40{^{15}\negthinspace\log}40 adalah
  1. 2b+a1+a\dfrac{2b+a}{1+a}
  2. 2ab+a2+a\dfrac{2ab+a}{2+a}
  3. 3ab+11+a\dfrac{3ab+1}{1+a}
  1. 3ab+a1+a\dfrac{3ab+a}{1+a}
  2. 3b+11+a\dfrac{3b+1}{1+a}
3a=53 ⁣log⁡5=a\begin{aligned} 3^a&=5\\ ^3\negthinspace\log5&=a \end{aligned}

5b=25 ⁣log⁡2=b\begin{aligned} 5^b&=2\\ ^5\negthinspace\log2&=b \end{aligned}

3 ⁣log⁡5⋅5 ⁣log⁡2=ab3 ⁣log⁡2=ab\begin{aligned} ^3\negthinspace\log5\cdot{^5\negthinspace\log}2&=ab\\ ^3\negthinspace\log2&=ab \end{aligned}

Related:


15 ⁣log⁡40=3 ⁣log⁡403 ⁣log⁡15=3 ⁣log⁡(23⋅5)3 ⁣log⁡(3⋅5)=3 ⁣log⁡23+3 ⁣log⁡53 ⁣log⁡3+3 ⁣log⁡5=3 3 ⁣log⁡2+3 ⁣log⁡53 ⁣log⁡3+3 ⁣log⁡5=3ab+a1+a\begin{aligned} {^{15}\negthinspace\log}40&=\dfrac{^3\negthinspace\log40}{^3\negthinspace\log15}\\[8pt] &=\dfrac{^3\negthinspace\log\left(2^3\cdot5\right)}{^3\negthinspace\log(3\cdot5)}\\[8pt] &=\dfrac{^3\negthinspace\log2^3+{^3\negthinspace\log}5}{^3\negthinspace\log3+{^3\negthinspace\log}5}\\[8pt] &=\dfrac{3\ {^3\negthinspace\log}2+{^3\negthinspace\log}5}{^3\negthinspace\log3+{^3\negthinspace\log}5}\\ &=\boxed{\boxed{\dfrac{3ab+a}{1+a}}} \end{aligned}
Jadi, 15 ⁣log⁡40=3ab+a1+a{^{15}\negthinspace\log}40=\dfrac{3ab+a}{1+a}.
JAWAB: D

No. 18

Jika 3 ⁣log⁡x=1,23^3\negmedspace\log x = 1{,}23, maka 3 ⁣log⁡27x=^3\negmedspace\log 27x =
  1. 3,6903{,}690
  2. −0,41-0{,}41
  3. 3,233{,}23
  1. −1,77-1{,}77
  2. 4,234{,}23
Zenius.net
3 ⁣log⁡27x=3 ⁣log⁡27+3 ⁣log⁡x=3+1,23=4,23\begin{aligned} ^3\negmedspace\log 27x&={^3\negmedspace\log}27+{^3\negmedspace\log}x\\ &=3+1{,}23\\ &=\boxed{\boxed{4{,}23}} \end{aligned}
Jadi, 3 ⁣log⁡27x=4,23^3\negmedspace\log 27x =4{,}23.
JAWAB: E

No. 19

5 ⁣log⁡9+5 ⁣log⁡2−5 ⁣log⁡450={^5\negmedspace\log 9+{^5\negmedspace\log 2}-{^5\negmedspace\log 450}=} ....
  1. −2-2
  2. −3-3
  3. 22
  1. 33
  2. −1-1
Zenius.net
5 ⁣log⁡9+5 ⁣log⁡2−5 ⁣log⁡450=5 ⁣log⁡9⋅2450=5 ⁣log⁡125=5 ⁣log⁡152=5 ⁣log⁡5−2=−2\begin{aligned} ^5\negmedspace\log 9+{^5\negmedspace\log 2}-{^5\negmedspace\log 450}&={^5\negmedspace\log\dfrac{9\cdot2}{450}}\\[8pt] &={^5\negmedspace\log\dfrac1{25}}\\[8pt] &={^5\negmedspace\log\dfrac1{5^2}}\\[8pt] &={^5\negmedspace\log5^{-2}}\\ &=\boxed{\boxed{-2}} \end{aligned}
Jadi,

No. 20

Hitunglah:
  1. 2 ⁣log⁡4{^2\negmedspace\log 4}
  2. 2 ⁣log⁡8{^2\negmedspace\log 8}
  3. 2 ⁣log⁡16{^2\negmedspace\log 16}
  4. 3 ⁣log⁡27{^3\negmedspace\log 27}
  5. log⁡1000{\log 1000}
zenius.net
  1. 2 ⁣log⁡4=2{^2\negmedspace\log 4}=2
  2. 2 ⁣log⁡8=3{^2\negmedspace\log 8}=3
  3. 2 ⁣log⁡16=4{^2\negmedspace\log 16}=4
  4. 3 ⁣log⁡27=3{^3\negmedspace\log 27}=3
  5. log⁡1000=3{\log 1000}=3
Jadi,

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