Lanjutan: Eksponen dan Logaritma

\begin{array}{|c|c|}\hline \LARGE\textrm{Eksponen}&\LARGE\textrm{Logaritma}\\\hline \LARGE a^{c}=b&\LARGE^{a}\log b=c\\\hline \multicolumn{2}{|c|}{\begin{matrix}\\ \bullet \quad a^{^{^{a}\log b}}=b\\\\ \bullet \quad ^{a}\log b\times d=\: ^{a}\log b+ \: ^{a}\log d\\\\ \bullet \quad ^{a}\log \displaystyle \frac{b}{d}=\: ^{a}\log b-\: ^{a}\log d\\\\ \bullet \quad ^{a}\log b=\displaystyle \frac{^{^{x}}\log b}{^{^{x}}\log a}\\\\ \bullet \quad ^{a}\log b=\displaystyle \frac{1}{^{^{b}}\log a}\\\\ \bullet \quad ^{a}\log b=m\Rightarrow \: ^{b}\log a=\displaystyle \frac{1}{m}\\\\ \bullet \quad ^{^{a^{n}}}\log b^{m}=\: \displaystyle \frac{m}{n}\times \: ^{a}\log b\\\\ \bullet \quad ^{a}\log b\times \: ^{b}\log c\times \: ^{c}\log d=\: ^{a}\log d\\\\ \bullet \quad ^{a}\log a=1\\\\ \bullet \quad ^{a}\log 1=0\\\\ \bullet \quad ^{a}\log a^{n}=n\\\\ \bullet \quad \log b=\: ^{10}\log b\\ \end{matrix}}\\\hline \end{array}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ \fbox{1}.&^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24=...\: .\\\\ &\textrm{Jawab}:\\ &\begin{aligned}^{2}\log 12+\: ^{2}\log 8-\: ^{2}\log 24&=\: ^{2}\log \left ( \displaystyle \frac{12\times 8}{24} \right )\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=4 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Diketahui}\: \: ^{3}\log 7=a,\: \: ^{5}\log 2=b,\: \: \textrm{dan}\: \: ^{2}\log 3=c\\ &\textrm{Nyatakanlah logaritma berikut dalam bentuk}\: \: a,\: b,\: \textrm{dan}\: \: c,\: \: \textrm{yaitu}:\\ &\textrm{a}.\quad ^{7}\log 3\\ &\textrm{b}.\quad ^{4}\log 5\\ &\textrm{c}.\quad ^{21}\log 5\\ &\textrm{d}.\quad ^{6}\log 7\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{7}\log 3&=\displaystyle \frac{1}{^{3}\log 7}\\ &=\displaystyle \frac{1}{a}\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}^{4}\log 5&=\displaystyle \frac{1}{^{5}\log 4}\\ &=\displaystyle \frac{1}{^{5}\log 2^{2}}\\ &=\displaystyle \frac{1}{2\: ^{5}\log 2}\\ &=\displaystyle \frac{1}{2b} \end{aligned}\\\hline \begin{aligned}&\\ ^{21}\log 5&=\displaystyle \frac{^{...}\log 5}{^{...}\log 21}\\ &=\displaystyle \frac{^{2}\log 5}{^{2}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 21}{^{3}\log 2}}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{\displaystyle \frac{^{3}\log 3\times 7}{^{3}\log 2}}\\ &=\displaystyle \frac{1}{^{5}\log 2\times \: ^{2}\log 3\times \left ( ^{3}\log 3+\: ^{3}\log 7 \right )}\\ &=\displaystyle \frac{1}{bc(1+a)} \end{aligned}&\begin{aligned}&\\ ^{6}\log 7&=\displaystyle \frac{^{3}\log 7}{^{3}\log 6}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2\times 3}\\ &=\displaystyle \frac{^{3}\log 7}{^{3}\log 2+\: ^{3}\log 3}\\ &=\displaystyle \frac{^{3}\log 7}{\displaystyle \frac{1}{^{2}\log 3}+\: ^{3}\log 3}\\ &=\displaystyle \frac{a}{\displaystyle \frac{1}{c}+1}\\ &=\displaystyle \frac{ac}{1+c}\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}.

Untuk cara penyelesaian seperti di atas, tidak harus demikian. Sebagai misal saya memberikan proses yang berbeda misal untuk No. 2 (c), tetapi tetap akan menghasilkan hasil yang sana, yaitu:

\begin{aligned}^{21}\log 5&=\displaystyle \frac{^{...}\log 5}{^{...}\log 21}\\ &=\displaystyle \frac{^{2}\log 5}{^{2}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{5}\log 2}}{^{2}\log 3\times \: ^{3}\log 21}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{2}\log 5}}{^{2}\log 3\times \: ^{3}\log (3\times 7)}\\ &=\displaystyle \frac{\displaystyle \frac{1}{^{2}\log 5}}{^{2}\log 3\times \: \left ( ^{3}\log 3+\: ^{3}\log 7 \right )}\\ &=\displaystyle \frac{\displaystyle \frac{1}{b}}{c(1+a)}\\ &=\displaystyle \frac{1}{bc(1+a)} \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui bahwa}\: \: \: ^{4}\log 5=a\\ &\textrm{a}.\quad \textrm{Carilah nilai}\: \: \: ^{4}\log 10\\ &\textrm{b}.\quad \textrm{Tunjukkan bahwa}\: \: \: ^{0,1}\log 1,25=\displaystyle \frac{2-2a}{2a+1}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}&\\ ^{4}\log 10&=\: ^{4}\log (2\times 5)\\ &=\: ^{4}\log 2+\: ^{4}\log 5\\ &=\: ^{2^{2}}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}.\: ^{2}\log 2+\: a\\ &=\: \displaystyle \frac{1}{2}+a\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ ^{0,1}\log 1,25&=\displaystyle \frac{^{4}\log 1,25}{^{4}\log 0,1}\\ &=\displaystyle \frac{^{4}\log \displaystyle \frac{125}{100}}{^{4}\log \displaystyle \frac{1}{10}}\\ &=\displaystyle \frac{^{4}\log 125-\: ^{4}\log 100}{^{4}\log 10^{-1}}\\ &=\displaystyle \frac{^{4}\log 5^{3}-\: ^{4}\log 10^{2}}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3.\: ^{4}\log 5-\: 2.\: ^{4}\log 10}{-\: ^{4}\log 10}\\ &=\displaystyle \frac{3a-2\left ( \displaystyle \frac{1}{2}+a \right )}{-\left ( \displaystyle \frac{1}{2}+a \right )}\\ &=\displaystyle \frac{a-1}{-a-\displaystyle \frac{1}{2}}\times \displaystyle \frac{-2}{-2}\\ &=\displaystyle \frac{2-2a}{2a+1}\qquad \blacksquare \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Jika}\: \: ^{2017}\log \displaystyle \frac{1}{x}=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2017}\: ,\: \textrm{maka hasil dari}\: \: \left ( 2x-3y \right )\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}^{2017}\log \displaystyle \frac{1}{x}&=\: ^{x}\log \displaystyle \frac{1}{y}=\: ^{y}\log \displaystyle \frac{1}{2017}\\ ^{2017}\log \displaystyle \frac{1}{x}&=\: ^{y}\log \displaystyle \frac{1}{2017}\\ ^{2017}\log \displaystyle x^{-1}&=\: ^{y}\log \displaystyle (2017)^{-1}\\ -\:\: ^{2017}\log x&=-\: \: ^{y}\log 2017\\ ^{2017}\log x&=\: ^{y}\log 2017,&\textnormal{dipenuhi saat}\: \: x=y=2017 \end{aligned}\\\\ &(2x-3y)=2x-3x=-x=-2017 \end{array}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Bila diberikan}\: \: ^{2}\log \left ( ^{8}\log x \right )=\: ^{8}\log \left ( ^{2}\log x \right ),\: \textrm{maka hasil dari}\: \: \left ( ^{2}\log x \right )^{2}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}^{2}\log \left ( ^{8}\log x \right )&=\: ^{8}\log \left ( ^{2}\log x \right )\\ ^{2}\log \left ( ^{8}\log x \right )&=\: ^{2^{3}}\log \left ( ^{2}\log x \right )\\ ^{2}\log \left ( ^{8}\log x \right )&=\: ^{2}\log \left ( ^{2}\log x \right )^{\frac{1}{3}}\\ ^{8}\log x&=\left ( ^{2}\log x \right )^{\frac{1}{3}}\\ \left (^{2^{3}}\log x \right )^{3}&=\: ^{2}\log x\\ \left ( \displaystyle \frac{1}{3}\: .^{2}\log x \right )^{3}&=\: ^{2}\log x\\ \displaystyle \frac{1}{27}\: .\left (^{2}\log x \right )^{3}&=\: ^{2}\log x\\ \left ( ^{2}\log x \right )^{2}.\left ( ^{2}\log x \right )&=27.\: ^{2}\log x\\ \left ( ^{2}\log x \right )^{2}&=27 \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Tentukanlah nilai}\: \: a+b\: \: \textrm{dimana}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{adalah bilangan riil positif}.\\ &^{7}\log \left ( 1+a^{2} \right )-\: ^{7}\log 25=\: ^{7}\log \left ( 2ab-15 \right )-\: ^{7}\log \left ( 25+b^{2} \right )\\\\ &\textrm{Jawab}:\\ &\begin{aligned}^{7}\log \displaystyle \frac{\left ( 1+a^{2} \right )}{25}&=\: ^{7}\log \displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ \textrm{diambil}&\: \textrm{persamaannya, maka}\\ \displaystyle \frac{\left ( 1+a^{2} \right )}{25}&=\displaystyle \frac{\left ( 2ab-15 \right )}{\left ( 25+b^{2} \right )}\\ \displaystyle \left ( 1+a^{2} \right )\left ( 25+b^{2} \right )&=25\left ( 2ab-15 \right )\\ &\begin{cases} \left ( 1+a^{2} \right ) & \text{ faktor dari} \: \: 25,\: \: a> 0,\: a\in \mathbb{R} \\ &\textrm{atau}\\ \left ( 25+b^{2} \right ) & \text{ faktor dari }\: \: 25,\: \: b> 0,\: b\in \mathbb{R}\: \: \: \textrm{juga} \end{cases}\\ \end{aligned}\\\\ &\begin{array}{|c|l|c|l|c|c|c|}\hline \textrm{No}&a&\left ( 1+a^{2} \right )&b&\left ( 25+b^{2} \right )&\textrm{Keterangan}&a+b\\\hline 1&2&5&10&125&\textrm{Memenuhi}&12\\\hline 2&7&50&\cdots &\cdots &\textrm{tidak memenuhi}&\cdots \\\hline 3&\cdots &\cdots &5&50&\textrm{tidak memenuhi}&\cdots \\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{7}.&\textrm{Jika}\: \: 60^{a}=3\: \: \textrm{dan}\: \: 60^{b}=5\: ,\: \textrm{maka hasil dari}\: \: 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}\\\\ &\textrm{Jawab}:\\ &\\ &\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ &\begin{cases} 60^{a}=3 & \Rightarrow \: ^{60}\log 3=a \\ 60^{b}=5 & \Rightarrow \: ^{60}\log 5=b \end{cases}\\ & \end{aligned}}\\\hline (1-a-b)&2(1-b)&\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )\\\hline \begin{aligned}1-a-b&=1-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log 60-\: ^{60}\log 3-\: ^{60}\log 5\\ &=\: ^{60}\log \displaystyle \frac{60}{3\times 5}\\ &=\: ^{60}\log 4\\ &=\: ^{60}\log 2^{2} \end{aligned}&\begin{aligned}2(1-b)&=2\left ( 1-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log 60-\: ^{60}\log 5 \right )\\ &=2\left ( ^{60}\log \displaystyle \frac{60}{5} \right )\\ &=2\left ( ^{60}\log 12 \right )\\ &=\: ^{60}\log 12^{2} \end{aligned}&\begin{aligned}\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )&=\displaystyle \frac{\: ^{60}\log 2^{2}}{\: ^{60}\log 12^{2}}\\ &=\displaystyle \frac{2\: ^{60}\log 2}{2\: ^{60}\log 12}\\ &=\displaystyle \frac{\: ^{60}\log 2}{\: ^{60}\log 12}\\ &=\: ^{12}\log 2 \end{aligned}\\\hline \multicolumn{3}{|c|}{\begin{aligned}&\\ 12^{^{\left ( \displaystyle \frac{1-x-y}{2(1-b)} \right )}}&=12^{^{\: ^{12}\log 2}}=2\\ & \end{aligned}}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ \fbox{8}.&\textrm{Diberikan bilangan riil positif}\: \: x,\: y,\: \textrm{dan}\: z\: \: \textrm{yang memenuhi persamaan}\\ &2\: ^{x}\log (2y)=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0.\\ &\textrm{Jika nilai}\: \: xy^{5}z\: \: \textrm{dapat dinyatakan dengan}\: \: \displaystyle \frac{1}{2^{\displaystyle \frac{p}{q}}}\: \: \textrm{dengan}\: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{bilangan asli yang saling prima},\\ &\textrm{maka nilai dari}\: \: p+q=....\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}2\: ^{x}\log (2y)&=2\: ^{2x}\log (4z)=2\: ^{4x}\log (8yz)\neq 0\\ \textrm{maka}&\\ ^{x}\log (2y)&=\: ^{2x}\log (4y)\quad\Rightarrow \quad \log (2y)\times \log (2x)=\log x\times \log (4y)............\textcircled{1}\\ ^{x}\log (2y)&=\: ^{4x}\log (8yz)\quad\Rightarrow \quad \log (2y)\times \log (4x)=\log x\times \log (8yz)............\textcircled{2}\\ ^{2x}\log (4y)&=\: ^{4x}\log (8yz)\quad\Rightarrow \quad \log (4y)\times \log (4x)=\log (2x)\times \log (8yz)............ \textcircled{3}\\ & \end{aligned} \end{array}.

\begin{aligned}\textrm{Perhatikan persamaan}\: \: \textcircled{2}&,\: \textrm{yaitu}:\\ \log (2y)\times \log (4x)&=\log x\times \log (8yz)\\ \log (2y)\times \left (\log (2x)+\log 2 \right )&=\log x\times \log (8yz)\\ \log (2y)\times \log (2x)+\log (2y)\times \log 2&=\log x\times \log (8yz)\\ \log x\times \log (4y)+\log (2y)\times \log 2&=\log x\times \log (8yz)&\textnormal{persamaan}\: \: \textcircled{1}\: \: \textrm{disubstitusikan}\\ \log (2y)&=\displaystyle \frac{\log x\times \log (8yz)-\log x\times \log (4y)}{\log 2}\\ \log (2y)&=\displaystyle \frac{\log x\times \left ( \log \displaystyle \frac{8yz}{4y} \right )}{\log 2}\\ \log (2y)&=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\: ..................\textcircled{4} \end{aligned}.

\begin{aligned}\textrm{Perhatikan juga persamaan}\: \: \textcircled{3}&,\: \textrm{yaitu}:\\ \log (4y)\times \log (4x)&=\log (2x)\times \log (8yz)\\ \left (\log (2y)+\log 2 \right )\times \log (4x)&=\log (2x)\times \log (8yz)\\ \log (2y)\times \log (4x)+\log 2\times \log (4x)&=\log (2x)\times \log (8yz)\\ \log x\times \log (8yz)+\log 2\times \log (4x)&=\log (2x)\times \log (8yz)&\textnormal{persamaan}\: \: \textcircled{2}\: \: \textrm{disubstitusikan}\\ \log 2\times \log (4x)&=\log (2x)\times \log (8yz)-\log x\times \log (8yz)\\ \log 2\times \log (4x)&=\log (8yz)\times \left ( \log (2x)-\log x \right )\\ \log 2\times \log (4x)&=\log (8yz)\times \left ( \log \displaystyle \frac{2x}{x} \right )\\ \log 2\times \log (4x)&=\log (8yz)\times \log 2\\ \log 4x&=\log (8yz)\\ 4x&=8yz\\ \displaystyle \frac{x}{z}&=2y\: .....................................\textcircled{5} \end{aligned}.

\begin{aligned}\textrm{dari persamaan}\: \: \textcircled{4}&\: \: \textrm{dan}\: \: \textcircled{5}\\ \log (2y)&=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ \log \left ( \displaystyle \frac{x}{z} \right )&=\displaystyle \frac{\log x\times \log (2z)}{\log 2}\\ \log 2\left ( \log x-\log z \right )&=\log x\times \log (2z)\\ \log 2\times \log x-\log 2\times \log z&=\log x\times \left ( \log 2+\log z \right )\\ \log 2\times \log x-\log 2\times \log z&=\log x\times \log 2+\log x\times \log z\\ -\log 2\times \log z&=\log x\times \log z\\ \log 2^{-1}&=\log x\\ \displaystyle \frac{1}{2}&=x\: ..................................................\textcircled{6} \end{aligned}.

\begin{array}{|c|c|c|c|}\hline \textrm{persamaan}\: \: \textcircled{6}&\textrm{persamaan}\: \: \textcircled{5}&\textrm{persamaan}\: \: \textcircled{2}&\textrm{Menentukan nilai}\: \: z\\\hline \begin{aligned}&\\ x=\displaystyle \frac{1}{2}&\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned} \log 2y\times \log (4x)&=\log x\times \log (8yz)\\ \log 2y\times \log (4(2yz))&=\log x\times \log (8yz)\\ \log 2y\times \log (8yz)&=\log x\times \log (8yz)\\ \log (2y)&=\log x\\ 2y&=x\\ y&=\displaystyle \frac{1}{2}x\\ &=\displaystyle \frac{1}{2}\times \frac{1}{2}\\ &=\displaystyle \frac{1}{4}\: ..................\textcircled{7}\\ & \end{aligned}&\begin{aligned}&\\ x&=2yz\\ \displaystyle \frac{1}{2}&=2\left ( \displaystyle \frac{1}{4} \right )z\\ 1&=z\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \multicolumn{4}{|c|}{\begin{aligned}\textrm{maka nilai untuk}&\: \: xy^{5}z\: \: \textrm{adalah}\\ xy^{5}z&=\left ( \displaystyle \frac{1}{2} \right ).\left ( \displaystyle \frac{1}{4} \right )^{5}.1\\ &=\displaystyle \frac{1}{2\times 4^{5}}\\ &=\displaystyle \frac{1}{2\times \left ( 2^{2} \right )^{5}}=\displaystyle \frac{1}{2^{1+10}}\\ &=\displaystyle \frac{1}{2^{11}}=\displaystyle \frac{1}{2^{^{\frac{11}{1}}}}=\displaystyle \frac{1}{2^{^{\frac{p}{q}}}}\quad \begin{cases} p & =11 \\ q & =1 \end{cases}\quad \textrm{dan jelas bahwa} \: \: p\: \: \textrm{dan}\: \: q\: \: \textrm{saling prima}\\ \textrm{Jadi},&\\ p+q&=11+1=12\\ & \end{aligned}}\\\hline \end{array}.

Sumber Referensi

  1. Idris, M., Rusdi, I. 2015. Langkah Awal Meraih Medali Emas Olimpiade Matematika SMA. Bandung: YRAMA WIDYA.
  2. Kurnianingsih, S., Kuntarti, dan Sulistiyono. 2007. Matematika SMA dan MA untuk Kelas X Semester 1 Standar Isi 2006. Jakarta: ESIS.
  3. Sembiring, S. 2002. Olimpiade Matematika untuk SMU. Bandung: YRAMA WIDYA.
  4. Sembiring, S., Suparmin, S. 2015. Pena Emas OSN Matematika SMA. Bandung: YRAMA WIDYA.

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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