Lanjutan 2 (Fungsi Eksponen dan Logaritma)

\begin{array}{ll}\\ 7.&\textrm{Hitunglah} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \frac{ \sqrt{1+\sqrt{1+\sqrt{1+...}}}}{\sqrt{1-\sqrt{1-\sqrt{1-...}}}}&b.&\displaystyle \frac{ \sqrt{2+\sqrt{2+\sqrt{2+...}}}}{\sqrt{2-\sqrt{2-\sqrt{2-...}}}}\\ &&&&&&&\\ &c.&\displaystyle \frac{\sqrt{31-\sqrt{31-\sqrt{31-...}}}}{\sqrt{1+\sqrt{1+\sqrt{1+...}}}}&d.&\displaystyle \frac{\sqrt{31+\sqrt{31+\sqrt{31+...}}}}{\sqrt{3+\sqrt{3+\sqrt{3+...}}}}\\ \end{array}..

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textbf{Pembahasan}}\\\hline \begin{aligned}x&=\sqrt{1+\sqrt{1+\sqrt{1+...}}}\\ x^{2}&=1+\underset{x}{\underbrace{\sqrt{x+\sqrt{x+...}}}}\\ x^{2}&=1+x,\quad \textrm{gunakan rumus \textbf{abc}}\\ x^{2}&-x-1=0\begin{cases} & a=1 \\ & b=-1 \\ & c=-1 \end{cases}\\ x_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \end{aligned}&\begin{aligned}x&=\sqrt{1-\sqrt{1-\sqrt{1-...}}}\\ x^{2}&=1-\underset{x}{\underbrace{\sqrt{x-\sqrt{x-...}}}}\\ x^{2}&=1-x,\quad \textrm{gunakan rumus \textbf{abc}}\\ x^{2}&+x-1=0\begin{cases} & a=1 \\ & b=1 \\ & c=-1 \end{cases}\\ x_{1,2}&=\displaystyle \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \end{aligned}\\\hline \multicolumn{2}{|c|}{\textrm{kita ambil akar yang positif saja}}\\\hline x=\displaystyle \frac{1+\sqrt{5}}{2}&x=\displaystyle \frac{-1+\sqrt{5}}{2}\\\hline \multicolumn{2}{|c|}{\begin{aligned}\displaystyle \frac{ \sqrt{1+\sqrt{1+\sqrt{1+...}}}}{\sqrt{1-\sqrt{1-\sqrt{1-...}}}}&=\displaystyle \frac{\displaystyle \frac{1+\sqrt{5}}{2}}{\displaystyle \frac{-1+\sqrt{5}}{2}}\\ &=\displaystyle \frac{\sqrt{5}+1}{\sqrt{5}-1}\\ &=\displaystyle \frac{\sqrt{5}+1}{\sqrt{5}-1}\times \displaystyle \frac{\sqrt{5}+1}{\sqrt{5}+1}\\ &=\displaystyle \frac{\sqrt{25}+2\sqrt{5}+1}{\sqrt{25}-1}\\ &=\displaystyle \frac{5+2\sqrt{5}+1}{5-1}\\ &=\displaystyle \frac{6+2\sqrt{5}}{4}\\ &=\displaystyle \frac{1}{2}\left ( 3+\sqrt{5} \right ) \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{lp{15.0cm}}\\ 8.&\textrm{Perhatikanlah bentuk akar berikut yang dikemukankan oleh matematikawan India \textbf{Srinivasa Ramanujan} yang terkenal dengan istilah \textbf{Ramanujan's infinite radicals}, yaitu} \end{array}\\ \begin{array}{lll}\\ .\quad\quad&.&\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+...}}}}}}\\ &&\\ &&\textrm{Tentukanlah nilai akar tersebut}? \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\textrm{Untuk menemukan solusi soal di atas, saat} \quad x>0,\: \: \textrm{yaitu}:\\ &\begin{aligned}x^{2}&=x^{2}\\ x^{2}&=1+\left ( x^{2}-1 \right )\\ &=1+\left ( x-1 \right )\left ( x+1 \right )\\ &=1+\left ( x-1 \right )\sqrt{\left ( x+1 \right )^{2}}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1 \right )^{2}-1}\\ &=1+\left ( x-1 \right )\sqrt{1+\left ( x+1-1 \right )\left ( x+1+1 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\left ( x+2 \right )}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{\left ( x+2 \right )^{2}}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2 \right )^{2}-1}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+2-1 \right )\left ( x+2+1 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\left ( x+3 \right )}}\\ &=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{\left ( x+3 \right )^{2}}}}\\ x^{2}&=1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}\\ x&=\sqrt{1+\left ( x-1 \right )\sqrt{1+x\sqrt{1+\left ( x+1 \right )\sqrt{...}}}} \end{aligned} \end{aligned}.

.\quad \: \: \textrm{Jadi},\quad x=3.

\begin{array}{ll}\\ 9.&\textrm{Nyatakan dalam bentuk pangkat} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \sqrt{6}&b.&\displaystyle 6\sqrt{6}&c.&\displaystyle \sqrt{\sqrt{6\sqrt[3]{6}}}&d.&\displaystyle \sqrt{6\sqrt{6\sqrt{6\sqrt{6}}}}\\ &&&&&&&\\ &e.&\displaystyle \sqrt{\sqrt{\sqrt{\sqrt{6}}}}&f.&\displaystyle \sqrt[4]{\sqrt{\sqrt[3]{\sqrt{6}}}}&g.&\displaystyle \sqrt[3]{\sqrt{a^{2}+2ab+b^{2}}}&h.&\displaystyle \sqrt[5]{x\sqrt{x\sqrt{x\sqrt{x}}}}\left ( \sqrt{x\sqrt[3]{x}}+\sqrt[3]{x\sqrt{x}} \right )\\ &&&&&&&\\ &i.&\displaystyle \left ( \sqrt[3]{16\sqrt[3]{8\sqrt{4}}} \right )^{\frac{1}{4}}&j.&\displaystyle \sqrt[3]{\sqrt[5]{1024a^{15}b^{30}}}&k.&\displaystyle \sqrt[a]{16^{\frac{1}{2}a(a-1)}x^{2a^{2}-a}}&l.&\displaystyle \frac{\sqrt[4]{\left ( a^{4}b^{\frac{2}{3}} \right )^{2}}}{\sqrt[3]{\left ( a^{\frac{1}{2}}b^{3} \right )^{\frac{1}{2}}}}\times \displaystyle \frac{\sqrt[4]{\left ( a^{2\frac{1}{3}}b \right )^{2}}}{\sqrt[3]{a^{\frac{1}{4}}\sqrt[3]{b^{\frac{2}{3}}}}}\\ \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\sqrt{6}=6^{\frac{1}{2}}&b.&\begin{aligned}\displaystyle 6\sqrt{6}&=\displaystyle 6^{1}\times 6^{\frac{1}{2}}\\ &=\displaystyle 6^{1+\frac{1}{2}}\\ &=\displaystyle 6^{1\frac{1}{2}}=6^{\frac{3}{2}} \end{aligned}&c.&\begin{aligned}\displaystyle \sqrt[2\times 2]{6\times 6^{\frac{1}{3}}}&=\displaystyle \sqrt[4]{6^{1\frac{1}{3}}}\\ &=\sqrt[4]{6^{\frac{4}{3}}}\\ &=\displaystyle 6^{\frac{4}{3}\times \frac{1}{4}}\\ &=\displaystyle 6^{\frac{1}{3}} \end{aligned}&d.&\begin{aligned}\displaystyle \sqrt{6\sqrt{6\sqrt{6\sqrt{6}}}}&=\displaystyle \left ( 6\left ( 6\left ( 6\left ( 6 \right )^{\frac{1}{2}} \right )^{\frac{1}{2}} \right )^{\frac{1}{2}} \right )^{\frac{1}{2}}\\ &=\displaystyle 6^{\frac{1}{2}}\times 6^{\frac{1}{4}}\times 6^{\frac{1}{8}}\times 6^{\frac{1}{16}}\\ &=\displaystyle 6^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}\\ &=\displaystyle 6^{\frac{8+4+2+1}{16}}\\ &=\displaystyle 6^{\frac{15}{16}}\end{aligned} \\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 10.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\: \textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}: \end{aligned}.

.\qquad\: \, \textbf{Misalkan}\: \:x=\: \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}} \\\\\\ \qquad\qquad \begin{array}{l|l}\\ \begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=47 \end{aligned}&\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\quad \textrm{persamaan kuadrat dalam x, gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=2 \end{cases} \end{aligned} \end{array}.

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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