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Exercise Zone : Logaritma [3]

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

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StandarSNBTHOTS


No.

Tentukan log 25 × 2log 10 × 5log 4
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} \log25\times{^2\negthinspace\log10}\times{^5\negthinspace\log4}&=\log25\times{^5\negthinspace\log4}\times{^2\negthinspace\log10}\\ &=\log25\times{^5\negthinspace\log2^2}\times{^2\negthinspace\log10}\\ &=\log25\times2\times{^5\negthinspace\log2}\times{^2\negthinspace\log10}\\ &=2\times\log25\times{^5\negthinspace\log10}\\ &=2\times\dfrac{\log25}{\log5}\\ &=2\times{^5\negthinspace\log25}\\ &=2\times2\\ &=\boxed{\boxed{4}} \end{aligned}
Jadi, log 25 × 2log 10 × 5log 4 = 4.

No.

Jika 2log 3 = a, dan 2log 5 = b, nyatakan 6log 50 dalam a dan b.
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} ^6\negthinspace\log50&=\dfrac{^2\negthinspace\log50}{^2\negthinspace\log6}\\ &=\dfrac{^2\negthinspace\log\left(5^2\cdot2\right)}{^2\negthinspace\log(3\cdot2)}\\ &=\dfrac{{^2\negthinspace\log5^2}+{^2\negthinspace\log2}}{{^2\negthinspace\log3}+{^2\negthinspace\log2}}\\ &=\dfrac{2\ {^2\negthinspace\log5}+{^2\negthinspace\log2}}{{^2\negthinspace\log3}+{^2\negthinspace\log2}}\\ &=\boxed{\boxed{\dfrac{2b+1}{a+1}}} \end{aligned}
Jadi, ^6\negthinspace\log50=\dfrac{2b+1}{a+1}.

No.

2log 1 + 3log 3 − 5log 2 + 5log 50 − 8log 32 =
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} {^2\negthinspace\log1}+{^3\negthinspace\log3}-{^5\negthinspace\log2}+{^5\negthinspace\log50}-{^8\negthinspace\log32}&={^2\negthinspace\log1}+{^3\negthinspace\log3}+{^5\negthinspace\log50}-{^5\negthinspace\log2}-{^{2^3}\negthinspace\log{2^5}}\\ &=0+1+{^5\negthinspace\log\dfrac{50}2}-\dfrac53\\ &=1+{^5\negthinspace\log25}-\dfrac53\\ &=1+2-\dfrac53\\ &=\boxed{\boxed{\dfrac43}} \end{aligned}
Jadi, {^2\negthinspace\log1}+{^3\negthinspace\log3}-{^5\negthinspace\log2}+{^5\negthinspace\log50}-{^8\negthinspace\log32}=\dfrac43.

No.

5log 35 + 5log 55 − 5log 77 =
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} {^5\negthinspace\log35}+{^5\negthinspace\log55}-{^5\negthinspace\log77}&={^5\negthinspace\log\dfrac{35\cdot55}{77}}\\ &={^5\negthinspace\log25}\\ &=\boxed{\boxed{2}} \end{aligned}
Jadi, 5log 35 + 5log 55 − 5log 77 = 2.

No.

\dfrac{a^2}{b^2}=4, maka \log\dfrac{a^3}{b^3}=
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} \dfrac{a^2}{b^2}&=4\\ \left(\dfrac{a}b\right)^2&=4\\ \dfrac{a}b&=2 \end{aligned}
\begin{aligned} \log\dfrac{a^3}{b^3}&=\log\left(\dfrac{a}b\right)^3\\ &=\log2^3\\ &=\boxed{\boxed{\log8}} \end{aligned}
Jadi, \log\dfrac{a^3}{b^3}=\log8.

No.

Sederhanakan {^3\negthinspace\log\dfrac13}
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} {^3\negthinspace\log\dfrac13}&={^3\negthinspace\log3^{-1}}\\ &=\boxed{\boxed{-1}} \end{aligned}
Jadi, {^3\negthinspace\log\dfrac13}=-1.

No.

Jika 2log 5 = m dan 2log 3 = n nyatakan 50log 90 dalam m dan n.
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} {^{50}\negthinspace\log90}&=\dfrac{^2\negthinspace\log90}{^2\negthinspace\log50}\\ &=\dfrac{^2\negthinspace\log\left(5\cdot3^2\cdot2\right)}{^2\negthinspace\log\left(5^2\cdot2\right)}\\ &=\dfrac{{^2\negthinspace\log5}+2\ {^2\negthinspace\log3}+{^2\negthinspace\log2}}{2\ {^2\negthinspace\log5}+{^2\negthinspace\log2}}\\ &=\boxed{\boxed{\dfrac{m+2n+1}{2m+1}}} \end{aligned}
Jadi, {^{50}\negthinspace\log90}=\dfrac{m+2n+1}{2m+1}.

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
  1. -\dfrac{11}3
  2. −11
Matematika Zone Id
ALTERNATIF PENYELESAIAN
\begin{aligned} \dfrac{{^3\negthinspace\log36}\cdot{^6\negthinspace\log81}-{^4\negthinspace\log32}}{^{\frac19}\negthinspace\log27}&=\dfrac{{^3\negthinspace\log6^2}\cdot{^6\negthinspace\log81}-{^{2^2}\negthinspace\log2^5}}{^{\frac1{3^2}}\negthinspace\log3^3}\\ &=\dfrac{2\ {^3\negthinspace\log6}\cdot{^6\negthinspace\log81}-\dfrac52}{^{3^{-2}}\negthinspace\log3^3}\\ &=\dfrac{2\ {^3\negthinspace\log81}-\dfrac52}{-\dfrac32}\\ &=\dfrac{2(4)-\dfrac52}{-\dfrac32}\\ &=\dfrac{8-\dfrac52}{-\dfrac32}\\ &=\dfrac{\dfrac{11}2}{-\dfrac32}\\ &=\boxed{\boxed{-\dfrac{11}3}} \end{aligned}
Jadi, \dfrac{{^3\negthinspace\log36}\cdot{^6\negthinspace\log81}-{^4\negthinspace\log32}}{^{\frac19}\negthinspace\log27}=-\dfrac{11}3.
JAWAB: D

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
  1. 177
  2. 188
SKMP Juli 2020
ALTERNATIF PENYELESAIAN
\begin{aligned} \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)&=\left(\dfrac27\cdot14\cdot{^c\negthinspace\log b}\right)\cdot\left(\dfrac35\cdot10\cdot{^a\negthinspace\log c}\right)\cdot\left(\dfrac23\cdot9\cdot{^b\negthinspace\log a}\right)\\ &=\left(4\cdot{^c\negthinspace\log b}\right)\cdot\left(6\cdot{^a\negthinspace\log c}\right)\cdot\left(6\cdot{^b\negthinspace\log a}\right)\\ &=144\cdot{^c\negthinspace\log b}\cdot{^a\negthinspace\log c}\cdot{^b\negthinspace\log a}\\ &=144\cdot{^a\negthinspace\log c}\cdot{^c\negthinspace\log b}\cdot{^b\negthinspace\log a}\\ &=\boxed{\boxed{144}} \end{aligned}
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

No.

Hitunglah nilai dari:
  1. 2log 16
  2. 2log 64 − 3log 9 − 5log 25
  3. 2log 5 ⋅ 5log 6 ⋅ 6log 8
Latihan Pra UKK Semester Genap Kelas X
ALTERNATIF PENYELESAIAN

  1. \begin{aligned} {^2\negthinspace\log16} &={^2\negthinspace\log2^4} \\ &=\boxed{\boxed{4}} \end{aligned}

  2. \begin{aligned} {^2\negthinspace\log64}-{^3\negthinspace\log9}-{^5\negthinspace\log25}&={^2\negthinspace\log2^6}-{^3\negthinspace\log3^2}-{^5\negthinspace\log5^2}\\ &=6-2-2\\ &=\boxed{\boxed{2}} \end{aligned}

  3. \begin{aligned} {^2\negthinspace\log5}\cdot{^5\negthinspace\log6}\cdot{^6\negthinspace\log8}&={^2\negthinspace\log8}\\ &={^2\negthinspace\log}2^3\\ &=\boxed{\boxed{3}} \end{aligned}
Jadi,
  1. 2log 16 = 4
  2. 2log 64 − 3log 9 − 5log 25 = 2
  3. 2log 5 ⋅ 5log 6 ⋅ 6log 8 = 3


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