Contoh Soal dan Pembahasan Logaritma 2

        Untuk mengerjakan soal-soal UK 1.3 ini, kita harus menguasai terlebih dahulu tentang sifat-sifat logaritma dengan baik dan benar. Logaritma tentu berkaitan erat dengan eksponen, logaritma dan eksponen saling berkebalikan.

         Pembahasan soal logaritma 2 ini kami sajikan sebagai bahan pertimbangan dalam menyelesaikan soal-soal logaritma yang ada pada buku kurikulum 2013 kelas X secara online. Jika ada kesalahan atau kekurangan dalam pembahasannya, mohon teman-teman untuk membantu mengkoreksinya dan memberikan masukan untuk memperbaikinya. Dengan mengerjakan kumpulan soal-soal logaritma uk 1.3 ini, harapannya siswa/siswi akan bisa lebih memperdalam konsep logaritma itu sendiri dengan baik dan benar.

         Soal-soal yang disajikan pada soal logaritma UK 1.3 kurikulum 2013 kelas x memang bervariasi dari yang paling mudah sampai yang paling sulit bahkan setingkat soal olimpiade. Semoga pembahasan pada artikel ini bisa membantu teman-teman dalam berlatih dan mengerjakan soal-soalnya.

         Berikut soal dan pembahasan soal logaritma UK 1.3 kurikulum 2013 kelas X.

Soal no. 1

Tuliskan dalam bentuk logaritma dari :

Gunakan definisi Logaritma : $a^c = b \Leftrightarrow {}^a \log b = c$
a). $5^3 = 125 \rightarrow {}^5 \log 125 = 3$
b). $10^2 = 100 \rightarrow {}^{10} \log 100 = 2$
c). $4^3 = 64 \rightarrow {}^{4} \log 64 = 3$
d). $6^1 = 6 \rightarrow {}^{6} \log 6 = 1$

Soal no. 2

Tuliskan dalam bentuk pangkat dari :

Gunakan definisi Logaritma : ${}^a \log b = c \Leftrightarrow a^c = b$
a). $\log 0,01 = -2 \rightarrow {}^{10} \log 0,01 = -2 \rightarrow 10^{-2} = 0,01$
b). ${}^{0,5} \log 0,0625 = 4 \rightarrow 0,5^4 = 0,0625$
c). ${}^{2} \log \sqrt[3]{2} = \frac{1}{3} \rightarrow 2^\frac{1}{3} = \sqrt[3]{2}$
d). ${}^{3} \log \frac{1}{9} = -2 \rightarrow 3^{-2} = \frac{1}{9}$

Soal no. 3

Hitunglah nilai setiap bentuk :

Gunakan Sifat-sifat logaritma :
${}^a \log b^n = n. {}^a \log b \, \, \,$ dan $\, \, {}^a \log a = 1$
a). $\log 10^4 = {}^{10} \log 10^4 = 4 \times {}^{10} \log 10 = 4 \times 1 = 4$
b). ${}^5 \log 125 = {}^{5} \log 5^3 = 3 \times {}^{5} \log 5 = 3 \times 1 = 3$
c). ${}^3 \log \frac{1}{27} = {}^{3} \log 3^{-3} = -3 \times {}^{3} \log 3 = -3 \times 1 = -3$
d). ${}^2 \log 0,25 = {}^2 \log \frac{1}{4} = {}^{2} \log 2^{-2} = -2 \times {}^{2} \log 2 = -2 \times 1 = -2$
e). ${}^4 \log 4^{10} = 10 \times {}^{4} \log 4 = 10 \times 1 = 10$
f). ${}^5 \log 1 = 0$

Soal no. 4

Diketahui $\log 2 = 0,3010 ; \, \log 3 = 0,4771 \,$ dan $\, \log 7 = 0,8451 \,$ tentukan :

Gunakan Sifat-sifat logaritma :
${}^a \log b^n = n. {}^a \log b ; \, {}^a \log b.c = {}^a \log b + {}^a \log c$
dan $\, {}^a \log \frac{b}{c} = {}^a \log b - {}^a \log c$
a). $\log 18$
$\begin{align} \log 18 & = \log 2.3^2 = \log 2 + \log 3^2 \\ & = \log 2 + 2.\log 3 = 0,3010 + 2.(0,4771) \\ & = 0,3010 + 0,9542 = 1,2552 \end{align}$
b). $\log 21$
$\begin{align} \log 21 & = \log 3.7 = \log 3 + \log 7 \\ & = 0,4771 + 0,8451 = 1,3222 \end{align}$
c). $\log 10,5$
$\begin{align} \log 10,5 & = \log \frac{105}{10} = \log \frac{21}{2} = \log 21 - \log 2 \\ & = 1,3222 - 0,3010 = 1,0212 \end{align}$
d). $\log \frac{1}{7}$
$\begin{align} \log \frac{1}{7} & = \log 7^{-1} = -1 \times \log 7 \\ & = -1 \times 0,8451 = -0,8451 \end{align}$

Soal no. 5

Sederhanakanlah :

Gunakan Sifat-sifat logaritma :
${}^a \log b^n = n. {}^a \log b ; \, {}^a \log b.c = {}^a \log b + {}^a \log c$
dan $\, {}^a \log \frac{b}{c} = {}^a \log b - {}^a \log c$
a). $\frac{2}{3} \times {}^2 \log 64 - \frac{1}{2} \times {}^2 \log 16$
$\begin{align} \frac{2}{3} \times {}^2 \log 64 - \frac{1}{2} \times {}^2 \log 16 & = \frac{2}{3} \times {}^2 \log 2^6 - \frac{1}{2} \times {}^2 \log 2^4 \\ & = \frac{2}{3} \times 6 \times {}^2 \log 2 - \frac{1}{2} \times 4 \times {}^2 \log 2 \\ & = \frac{2}{3} \times 6 \times 1 - \frac{1}{2} \times 4 \times 1 \\ & = \frac{2}{3} \times 6 \times 1 - \frac{1}{2} \times 4 \times 1 \\ & = 4 - 2 = 2 \end{align}$
b). ${}^a \log 2x + 3({}^a \log x - {}^a \log y)$
$\begin{align} {}^a \log 2x + 3({}^a \log x - {}^a \log y) & = {}^a \log 2x + 3({}^a \log \frac{x}{y}) \\ & = {}^a \log 2x + {}^a \log \left( \frac{x}{y} \right)^3 \\ & = {}^a \log 2x + {}^a \log \frac{x^3}{y^3} \\ & = {}^a \log \left( 2x \times \frac{x^3}{y^3} \right) \\ & = {}^a \log \frac{2x^4}{y^3} \end{align}$
c). ${}^a \log \frac{a}{\sqrt{x}} - {}^a \log \sqrt{ax}$
$\begin{align} {}^a \log \frac{a}{\sqrt{x}} - {}^a \log \sqrt{ax} & = {}^a \log \frac{\frac{a}{\sqrt{x}}}{\sqrt{ax}} \\ & = {}^a \log \frac{a}{\sqrt{x}.\sqrt{a}.\sqrt{x}} \\ & = {}^a \log \frac{a}{x\sqrt{a}} \\ & = {}^a \log \frac{\sqrt{a}}{x} \\ & = {}^a \log \sqrt{a} - {}^a \log x \\ & = {}^a \log a^\frac{1}{2} - {}^a \log x \\ & = \frac{1}{2} \times {}^a \log a - {}^a \log x \\ & = \frac{1}{2} \times 1 - {}^a \log x \\ & = \frac{1}{2} - {}^a \log x \end{align}$
d). $\log \sqrt{a} + \log \sqrt{b} - \frac{1}{2} \log ab$
$\begin{align} \log \sqrt{a} + \log \sqrt{b} - \frac{1}{2} \log ab & = \log \sqrt{a} + \log \sqrt{b} - \log (ab)^\frac{1}{2} \\ & = \log \sqrt{a} + \log \sqrt{b} - \log \sqrt{ab} \\ & = \log \sqrt{a} . \sqrt{b} - \log \sqrt{ab} \\ & = \log \sqrt{ab} - \log \sqrt{ab} \\ & = 0 \end{align}$

Soal no. 6

Jika ${}^2 \log 3 = a \,$ dan $\, {}^3 \log 5 = b \,$ , nyatakan bentuk berikut dalam $a \,$ dan $b$ !

Gunakan Sifat-sifat logaritma : ${}^a \log b = \frac{{}^p \log b}{{}^p \log a}$
a). ${}^2 \log 15$
$\begin{align} {}^2 \log 15 & = \frac{\log 15 }{\log 2 } = \frac{{}^3 \log 15 }{{}^3 \log 2 } \\ & = \frac{{}^3 \log (3.5) }{{}^3 \log 2 } \\ & = \frac{{}^3 \log 3 + {}^3 \log 5 }{{}^3 \log 2 } \\ & = \frac{1 + b }{\frac{1}{a} } = a(1+b) = a + ab \end{align}$
b). ${}^4 \log 75$
$\begin{align} {}^4 \log 75 & = \frac{\log 75 }{\log 4 } = \frac{{}^3 \log 75 }{{}^3 \log 2^2 } \\ & = \frac{{}^3 \log (3.5^2) }{{}^3 \log 2^2 } = \frac{{}^3 \log 3 + {}^3 \log 5^2 }{2. {}^3 \log 2 } \\ & = \frac{{}^3 \log 3 + 2.{}^3 \log 5 }{2.{}^3 \log 2 } \\ & = \frac{1 + 2b }{2.\frac{1}{a} } = \frac{(1+2b)a}{2} = \frac{a+2ab}{2} \end{align}$
c). ${}^{25} \log 36$
$\begin{align} {}^{25} \log 36 & = \frac{\log 36 }{\log 25 } = \frac{{}^3 \log 6^2 }{{}^3 \log 5^2 } = \frac{2. {}^3 \log 6 }{2.{}^3 \log 5} \\ & = \frac{{}^3 \log (3.2) }{{}^3 \log 5 } = \frac{{}^3 \log 3 + {}^3 \log 2 }{ {}^3 \log 5 } \\ & = \frac{1 + \frac{1}{a} }{b } = \frac{1 + \frac{1}{a} }{b } . \frac{a}{a} = \frac{a + 1 }{ab } \end{align}$
d). ${}^{2} \log 5$
$\begin{align} {}^{2} \log 5 & = \frac{\log 5 }{\log 2 } = \frac{{}^3 \log 5 }{{}^3 \log 2 } = \frac{b }{\frac{1}{a} } =ab \end{align}$
e). ${}^{30} \log 150$
$ \begin{align} {}^{30} \log 150 & = \frac{\log 150 }{\log 30 } = \frac{{}^3 \log 150 }{{}^3 \log 30 } \\ & = \frac{{}^3 \log (3.2.5^2) }{{}^3 \log (3.2.5) } = \frac{{}^3 \log 3 + {}^3 \log 2 + {}^3 \log 5^2 }{{}^3 \log 3 + {}^3 \log 2 + {}^3 \log 5} \\ & = \frac{{}^3 \log 3 + {}^3 \log 2 + 2 . {}^3 \log 5 }{{}^3 \log 3 + {}^3 \log 2 + {}^3 \log 5} \\ & = \frac{1 + \frac{1}{a}

$\begin{align} \log \sqrt{a} + \log \sqrt{b} - \frac{1}{2} \log ab & = \log \sqrt{a} + \log \sqrt{b} - \log (ab)^\frac{1}{2} \\ & = \log \sqrt{a} + \log \sqrt{b} - \log \sqrt{ab} \\ & = \log \sqrt{a} . \sqrt{b} - \log \sqrt{ab} \\ & = \log \sqrt{ab} - \log \sqrt{ab} \\ & = 0 \end{align}$
Soal no. 6

Jika ${}^2 \log 3 = a \,$ dan $\, {}^3 \log 5 = b \,$ , nyatakan bentuk berikut dalam $a \,$ dan $b$ !

Gunakan Sifat-sifat logaritma : ${}^a \log b = \frac{{}^p \log b}{{}^p \log a}$
a). ${}^2 \log 15$
$\begin{align} {}^2 \log 15 & = \frac{\log 15 }{\log 2 } = \frac{{}^3 \log 15 }{{}^3 \log 2 } \\ & = \frac{{}^3 \log (3.5) }{{}^3 \log 2 } \\ & = \frac{{}^3 \log 3 + {}^3 \log 5 }{{}^3 \log 2 } \\ & = \frac{1 + b }{\frac{1}{a} } = a(1+b) = a + ab \end{align}$
b). ${}^4 \log 75$
$\begin{align} {}^4 \log 75 & = \frac{\log 75 }{\log 4 } = \frac{{}^3 \log 75 }{{}^3 \log 2^2 } \\ & = \frac{{}^3 \log (3.5^2) }{{}^3 \log 2^2 } = \frac{{}^3 \log 3 + {}^3 \log 5^2 }{2. {}^3 \log 2 } \\ & = \frac{{}^3 \log 3 + 2.{}^3 \log 5 }{2.{}^3 \log 2 } \\ & = \frac{1 + 2b }{2.\frac{1}{a} } = \frac{(1+2b)a}{2} = \frac{a+2ab}{2} \end{align}$
c). ${}^{25} \log 36$
$\begin{align} {}^{25} \log 36 & = \frac{\log 36 }{\log 25 } = \frac{{}^3 \log 6^2 }{{}^3 \log 5^2 } = \frac{2. {}^3 \log 6 }{2.{}^3 \log 5} \\ & = \frac{{}^3 \log (3.2) }{{}^3 \log 5 } = \frac{{}^3 \log 3 + {}^3 \log 2 }{ {}^3 \log 5 } \\ & = \frac{1 + \frac{1}{a} }{b } = \frac{1 + \frac{1}{a} }{b } . \frac{a}{a} = \frac{a + 1 }{ab } \end{align}$
d). ${}^{2} \log 5$
$\begin{align} {}^{2} \log 5 & = \frac{\log 5 }{\log 2 } = \frac{{}^3 \log 5 }{{}^3 \log 2 } = \frac{b }{\frac{1}{a} } =ab \end{align}$
e). ${}^{30} \log 150$
$\begin{align} {}^{30} \log 150 & = \frac{\log 150 }{\log 30 } = \frac{{}^3 \log 150 }{{}^3 \log 30 } \\ & = \frac{{}^3 \log (3.2.5^2) }{{}^3 \log (3.2.5) } = \frac{{}^3 \log 3 + {}^3 \log 2 + {}^3 \log 5^2 }{{}^3 \log 3 + {}^3 \log 2 + {}^3 \log 5} \\ & = \frac{{}^3 \log 3 + {}^3 \log 2 + 2 . {}^3 \log 5 }{{}^3 \log 3 + {}^3 \log 2 + {}^3 \log 5} \\ & = \frac{1 + \frac{1}{a} + 2 . b }{1 + \frac{1}{a} + b} = \frac{1 + \frac{1}{a} + 2 b }{1 + \frac{1}{a} + b} . \frac{a}{a} = \frac{a + 1 + 2ab }{a + 1 + ab} \end{align}$
f). ${}^{100} \log 50$
$\begin{align} {}^{100} \log 50 & = \frac{\log 50 }{\log 100 } = \frac{{}^3 \log (2 . 5^2) }{{}^3 \log (2^2 . 5^2)} \\ & = \frac{{}^3 \log 2 + {}^3 \log 5^2 }{ {}^3 \log 2^2 + {}^3 \log 5^2 } = \frac{{}^3 \log 2 + 2. {}^3 \log 5 }{ 2.{}^3 \log 2 + 2.{}^3 \log 5 } \\ & = \frac{\frac{1}{a} + 2. b }{ 2.\frac{1}{a} + 2.b } = \frac{\frac{1}{a} + 2b }{ 2.\frac{1}{a} + 2b } . \frac{a}{a} = \frac{1 + 2ab}{2 + 2ab} \end{align}$

Soal no. 7

Jika $b = a^4, \, a \,$ dan $b \,$ bilangan real positif, $a \neq 1, b \neq 1 , \,$ tentukan nilai ${}^a \log b - {}^b \log a !$

Gunakan sifat : ${{}^a}^m \log {b}^n = \frac{n}{m} {}^a \log b$
Substitusi bentuk $b = a^4$
$\begin{align} {}^a \log b - {}^b \log a & = {}^a \log a^4 - {{}^a}^4 \log a^1 \\ & = 4 . {}^a \log a - \frac{1}{4} . {}^a \log a \\ & = 4 . 1 - \frac{1}{4} . 1 = 4 - \frac{1}{4} = 3\frac{3}{4} \end{align}$
Jadi, nilai ${}^a \log b - {}^b \log a = 3\frac{3}{4} . \heartsuit$

Soal no. 8

Jika ${}^a \log b = 4 , \, {}^c \log b = 4 \,$ dan $a, b, c \,$ bilangan positif, $a , c \neq 1, \,$ tentukan nilai $\left[ {}^a \log (bc)^4 \right]^\frac{1}{2} !$

Gunakan sifat : ${}^a \log {b} = \frac{{}^p \log b}{{}^p \log a}$
$\begin{align} \left[ {}^a \log (bc)^4 \right]^\frac{1}{2} & = \left[ 4.{}^a \log (bc) \right]^\frac{1}{2} \\ & = \left[ 4.\frac{{}^b \log ab}{{}^b \log a} \right]^\frac{1}{2} \\ & = \left[ 4.\frac{{}^b \log a + {}^b \log b}{{}^b \log a} \right]^\frac{1}{2} \\ & = \left[ 4.\frac{\frac{1}{4} + 1}{\frac{1}{4}} \right]^\frac{1}{2} \\ & = \left[ 4.\frac{\frac{5}{4} }{\frac{1}{4}} \right]^\frac{1}{2} \\ & = \left[ 4.5 \right]^\frac{1}{2} \\ & = 4^\frac{1}{2}.5^\frac{1}{2} = 2 \sqrt{5} \end{align}$
Jadi, nilai $\left[ {}^a \log (bc)^4 \right]^\frac{1}{2} = 2 \sqrt{5} . \heartsuit$

Soal no. 9

Buktikan $\log 1 = 0 \,$ dan $\log 10 = 1$ !

Definisi logaritma : ${}^a \log b = c \Leftrightarrow a^c = b$
Kita membuktikan berdasarkan definisi logaritma di atas :
$\begin{align} \log 1 = 0 \rightarrow {}^{10} \log 1 & = 0 \\ 10^0 & = 1 \\ \log 10 = 1 \rightarrow {}^{10} \log 10 & = 1 \\ 10^1 & = 10 \end{align}$
Jadi, berdasarkan definisi logaritma, terbukti yang diinginkan. $\heartsuit$