Tambahan Contoh Soal 5 (Bentuk Pangkat, Akar, dan Logaritma)

Lihat sebelumnya di sini

\begin{array}{ll}\\ \fbox{41}.&\textrm{Nyatakanlah bentuk berikut ke bentuk pangkat}\\ &\begin{array}{llllllll}\\ a.&\sqrt{3}&b.&\sqrt[3]{\displaystyle \frac{1}{8}}&c.&\sqrt{27}&d.&\sqrt{\displaystyle \frac{1}{5}}\\ e.&\sqrt{\displaystyle \frac{1}{\sqrt{3}}}&f.&\sqrt[3]{\sqrt{\displaystyle \frac{1}{8}}}&g.&\sqrt{\sqrt[3]{27}}&h.&\sqrt[5]{\displaystyle \frac{1}{\sqrt{\sqrt[3]{5}}}}\\ i.&\sqrt[3]{x^{2}.y^{3}}&j.&\sqrt{x^{2}+y^{2}}&k.&\sqrt{x\sqrt[3]{x^{2}}}&l.&\sqrt{x\sqrt[3]{x^{2}\sqrt[5]{x^{3}}}} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{llll}\\ a.&\sqrt{3}=3^{\frac{1}{2}}&b.&\sqrt[3]{\displaystyle \frac{1}{8}}=\left ( \displaystyle \frac{1}{8} \right )^{\frac{1}{3}}=\left ( 2^{-3} \right )^{\frac{1}{3}}=2^{-1}\\ c.&\sqrt{27}=\left ( 3^{3} \right )^{\frac{1}{2}}=3^{\frac{3}{2}}&d.&\sqrt{\displaystyle \frac{1}{5}}=\left ( 5^{-1} \right )^\frac{1}{2}=5^{-\frac{1}{2}}\\ e.&\sqrt{\displaystyle \frac{1}{\sqrt{3}}}=\left ( \displaystyle \frac{1}{3^{\frac{1}{2}}} \right )^{\frac{1}{2}}=\left ( 3^{-\frac{1}{2}} \right )^{\frac{1}{2}}=3^{-\frac{1}{4}}&f.&\sqrt[3]{\sqrt{\displaystyle \frac{1}{8}}}=\left ( \left ( 2^{-3} \right )^{\frac{1}{2}} \right )^{\frac{1}{3}}=2^{-\frac{3}{6}}=2^{-\frac{1}{2}}\\ g.&\sqrt{\sqrt[3]{27}}=\left ( \left ( 3^{3} \right )^{\frac{1}{3}} \right )^{\frac{1}{2}}=3^\frac{1}{2}&h.&\sqrt[5]{\displaystyle \frac{1}{\sqrt{\sqrt[3]{5}}}}=\left ( \displaystyle \frac{1}{\left ( \left ( 5 \right )^{\frac{1}{3}} \right )^{\frac{1}{2}}} \right )^{\frac{1}{5}}=5^{-\frac{1}{30}} \end{array}.

Silahkan dilanjutkan yang belum dibahas

\begin{array}{ll}\\ \fbox{42}.&\textrm{Nyatakanlah bentuk berikut ke bentuk pangkat}\\ &\begin{array}{llllllll}\\ a.&\displaystyle \left ( \frac{1}{\sqrt[3]{x}} \right )^{8}&b.&\displaystyle \frac{x^{5}.y^{2}.z^{-1}}{x^{2}.y^{4}.z^{3}}&c.&\sqrt{y.\sqrt[3]{x^{2}.y}}&d.&\sqrt[3]{x^{2}.\sqrt{x.\sqrt[5]{x^{2}}}}\\ e.&\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}&f.&\displaystyle \sqrt{\frac{2}{\sqrt{\frac{2}{\sqrt{\frac{2}{\sqrt{2}}}}}}}&g.&\displaystyle \sqrt[3]{\sqrt[4]{\sqrt[5]{\sqrt[6]{2}}}}&h.&\displaystyle \left ( \sqrt{\frac{\sqrt{\frac{\sqrt{\frac{\sqrt{\frac{1}{x}}}{x}}}{x}}}{x}} \right )^{-\frac{16}{45}}\end{array} \end{array}.

Jawab:

\begin{array}{|l|l|}\hline \begin{aligned}a.\quad \displaystyle \left ( \frac{1}{\sqrt[3]{x}} \right )^{8}&=\displaystyle \left ( \frac{1}{x^{\frac{1}{3}}} \right )^{8}\\ &=\left ( x^{-\frac{1}{3}} \right )^{8}\\ &=x^{-\frac{8}{3}} \end{aligned}&\begin{aligned}b.\quad \displaystyle \frac{x^{5}.y^{2}.z^{-1}}{x^{2}.y^{4}.z^{3}}&=x^{5-2}.y^{2-4}.z^{-1-3}\\ &=x^{3}.y^{-2}.z^{-4}\\ &=\displaystyle \frac{x^{3}}{y^{2}.z^{4}} \end{aligned}\\\hline \begin{aligned}e.\quad \sqrt{\sqrt{\sqrt{\sqrt{\sqrt{2}}}}}&=\displaystyle \sqrt[2.2.2.2.2]{2}\\ &=\sqrt[32]{2}\\ &=2^{\frac{1}{32}} \end{aligned}&\begin{aligned}g.\quad \sqrt[3]{\sqrt[4]{\sqrt[5]{\sqrt[6]{2}}}}&=\displaystyle \sqrt[3.4.5.6]{2}\\ &=\sqrt[360]{2}\\ &=2^{\frac{1}{360}} \end{aligned}\\\hline \begin{aligned}f.\quad \sqrt{\frac{2}{\sqrt{\frac{2}{\sqrt{\frac{2}{\sqrt{2}}}}}}}&=\sqrt{2:\sqrt{2:\sqrt{2:\sqrt{2}}}}\\ &=\sqrt{2:\left ( 2^{\frac{1}{2}}:\left ( 2^{\frac{1}{4}}:2^{\frac{1}{8}} \right ) \right )}\\ &=\sqrt{2^{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}}}\\ &=\sqrt{2^{\frac{8-4+2-1}{8}}}\\ &=\sqrt{2^{\frac{5}{8}}}\\ &=2^{\frac{5}{16}}\\ &\\ &\\ & \end{aligned}&\begin{aligned}h.\quad &\left (\sqrt{\frac{\sqrt{\frac{\sqrt{\frac{\sqrt{\frac{1}{x}}}{x}}}{x}}}{x}} \right )^{-\frac{16}{45}}\\ &=\left (\sqrt{x^{-1}.\sqrt{x^{-1}.\sqrt{x^{-1}.\sqrt{x^{-1}}}}} \right )^{-\frac{16}{45}}\\ &=\left ( \sqrt{x^{-1}.x^{-\frac{1}{2}}.x^{-\frac{1}{4}}.x^{-\frac{1}{8}}} \right )^{-\frac{16}{45}}\\ &=\left ( \sqrt{x^{-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}}} \right )^{-\frac{16}{45}}=\left ( \sqrt{x^{-\frac{15}{8}}} \right )^{-\frac{16}{45}}\\ &=x^{-\frac{15}{8}.\frac{1}{2}.-\frac{16}{45}}\\ &=x^{-\frac{1}{3}} \end{aligned}\\\hline \end{array}.

Bagi yang ingin tahu pembahasan No. 42. c  dan 42. d silahkan klik di sini

\begin{array}{ll}\\ \fbox{43}.&\textrm{Bentuk sederhana dari}\\ &abc.\displaystyle \sqrt[3]{\frac{ab}{c^{5}}}.\sqrt[3]{\frac{ac}{b^{5}}}.\sqrt[3]{\frac{bc}{a^{5}}} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}abc.\displaystyle \sqrt[3]{\frac{ab}{c^{5}}}.\sqrt[3]{\frac{ac}{b^{5}}}.\sqrt[3]{\frac{bc}{a^{5}}} &=\displaystyle abc\sqrt[3]{\frac{ab}{c^{5}}.\frac{ac}{b^{5}}.\frac{bc}{a^{5}}}\\ &=\displaystyle abc\sqrt[3]{\frac{(abc)^{2}}{(abc)^{5}}}\\ &=abc\sqrt[3]{(abc)^{2-5}}\\ &=abc\sqrt[3]{(abc)^{-3}}\\ &=abc.(abc)^{-1}\\ &=(abc)^{0}\\ &=1 \end{aligned}.

mungkin pembaca dapat membandingkan dengan bentuk berikut ini

\begin{array}{ll}\\ \fbox{44}.&\textrm{Bentuk sederhana dari}\\ &\displaystyle \frac{\left ( 1+ \left ( \displaystyle \frac{x}{y} \right )^{2}\right )^{-\frac{1}{2}}.\left ( 1-\left ( \displaystyle \frac{y}{x} \right )^{2} \right )^{-\frac{1}{2}}}{\left ( \left ( \displaystyle \frac{x}{y} \right )^{2}-1 \right )^{-\frac{1}{2}}.\left ( \left ( \displaystyle \frac{y}{x} \right )^{2}+1 \right )^{-\frac{1}{2}}} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\displaystyle \frac{\left ( 1+ \left ( \displaystyle \frac{x}{y} \right )^{2}\right )^{-\frac{1}{2}}.\left ( 1-\left ( \displaystyle \frac{y}{x} \right )^{2} \right )^{-\frac{1}{2}}}{\left ( \left ( \displaystyle \frac{x}{y} \right )^{2}-1 \right )^{-\frac{1}{2}}.\left ( \left ( \displaystyle \frac{y}{x} \right )^{2}+1 \right )^{-\frac{1}{2}}}&=\displaystyle \frac{\left ( 1+ \left ( \displaystyle \frac{x}{y} \right )^{2}\right )^{-\frac{1}{2}}.\left ( 1-\left ( \displaystyle \frac{x}{y} \right )^{-2} \right )^{-\frac{1}{2}}}{\left ( \left ( \displaystyle \frac{x}{y} \right )^{2}-1 \right )^{-\frac{1}{2}}.\left ( \left ( \displaystyle \frac{x}{y} \right )^{-2}+1 \right )^{-\frac{1}{2}}}\: \: ,&\textnormal{misalkan}\quad \frac{x}{y}=z,\: \: \textrm{maka}\\ &=\displaystyle \left ( \frac{\left ( 1+z^{2} \right ).\left ( 1-z^{-2} \right )}{\left ( z^{2}-1 \right ).\left ( z^{-2}+1 \right )} \right )^{-\frac{1}{2}}\\ &=\displaystyle \left ( \frac{1-z^{-2}+z^{2}-1}{1+z^{2}-z^{-2}-1} \right )^{-\frac{1}{2}}\\ &=\displaystyle \left ( \frac{z^{2}-z^{-2}}{z^{2}-z^{-2}} \right )^{-\frac{1}{2}}=1^{-\frac{1}{2}}\\ &=1 \end{aligned}.

\begin{array}{ll}\\ \fbox{45}.&\textrm{Hitunglah nilai dari}\\ &\begin{array}{ll}\\ \textrm{a}.&\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\\ \textrm{b}.&\sqrt{3+\sqrt{8}}+\sqrt{3-\sqrt{8}}\\ \textrm{c}.&\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\\ \textrm{d}.&\left ( 2\sqrt[3]{5}-\sqrt[3]{10} \right )\times \left ( 4\sqrt[3]{25}+2\sqrt[3]{50}+\sqrt[3]{100} \right )\\ \textrm{e}.&\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\\ \textrm{f}.&\sqrt{2-\sqrt{3}}\times \sqrt{2-\sqrt{2-\sqrt{3}}}\times \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}}\times \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{3}}}} \end{array} \end{array}.

Jawab:

\begin{array}{l}\\ \begin{aligned}\textrm{a}.\quad \sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}&=\sqrt{2+2.\frac{1}{2}\sqrt{3}}+\sqrt{2-2.\frac{1}{2}.\sqrt{3}}=\sqrt{2+2\sqrt{\frac{1}{4}.3}}+\sqrt{2-2\sqrt{\frac{1}{4}.3}}\\ &=\sqrt{2+2\sqrt{\frac{3}{2}.\frac{1}{2}}}+\sqrt{2-2\sqrt{\frac{3}{2}.\frac{1}{2}}}\\ &=\sqrt{\left ( \sqrt{\frac{3}{2}} \right )^{2}+\left ( \sqrt{\frac{1}{2}} \right )^{2}+3\sqrt{\frac{3}{2}.\frac{1}{2}}}+\sqrt{\left ( \sqrt{\frac{3}{2}} \right )^{2}+\left ( \sqrt{\frac{1}{2}} \right )^{2}-3\sqrt{\frac{3}{2}.\frac{1}{2}}}\\ &=\sqrt{\left ( \sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}} \right )^{2}}+\sqrt{\left ( \sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}} \right )^{2}}=\left ( \sqrt{\frac{3}{2}}+\sqrt{\frac{1}{2}} \right )+\left ( \sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}} \right )\\ &=2\sqrt{\frac{3}{2}}=\sqrt{4.\frac{3}{2}}=\sqrt{6} \end{aligned} \end{array}.

Alternatif lain:

\textrm{Misalkan}:\\\\ x=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}},\: \: \textrm{maka}\\ \begin{aligned}x^{2}&=\left ( \sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\: \right )^{2}\\ &=\left ( \sqrt{2+\sqrt{3}}\: \right )^{2}+\left ( \sqrt{2-\sqrt{3}}\: \right )^{2}+2\left ( \sqrt{2+\sqrt{3}}\: \right ).\left ( \sqrt{2-\sqrt{3}}\: \right )\\ &=2+\sqrt{3}+2-\sqrt{3}+2\sqrt{2^{2}-3}\\ x^{2}&=4+2\\ x&=\sqrt{6} \end{aligned}.

\begin{array}{l}\\ \begin{aligned}\textrm{b}.\quad \sqrt{3+\sqrt{8}}+\sqrt{3-\sqrt{8}}&=\sqrt{3+\sqrt{4.2.1}}+\sqrt{3-\sqrt{4.2.1}}=\sqrt{3+\sqrt{4}.\sqrt{2.1}}+\sqrt{3-\sqrt{4}.\sqrt{2.1}}\\ &=\sqrt{3+2.\sqrt{2}.\sqrt{1}}+\sqrt{3-2.\sqrt{2}.\sqrt{1}}\\ &=\sqrt{\left ( \sqrt{2} \right )^{2}+1+2.\sqrt{2.1}}+\sqrt{\left ( \sqrt{2} \right )^{2}-1+2.\sqrt{2.1}}\\ &=\sqrt{\left ( \sqrt{2}+1 \right )^{2}}+\sqrt{\left ( \sqrt{2}-1 \right )^{2}}\\ &=\left ( \sqrt{2}+1 \right )+\left ( \sqrt{2}-1 \right )\\ &=2\sqrt{2} \end{aligned} \end{array}.

\begin{array}{l}\\ \begin{aligned}\textrm{e}.\quad \sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}&=\sqrt[3]{\left ( 2+\sqrt{2} \right )^{3}}+\sqrt[3]{\left ( 2-\sqrt{2} \right )^{3}}&\textnormal{,cara coba-coba}\\ &=\left ( 2+\sqrt{2} \right )+\left ( 2-\sqrt{2} \right )\\ &=4 \end{aligned} \end{array}.

\begin{array}{l}\\ \begin{aligned}\textrm{f}.\quad \sqrt{2-\sqrt{3}}&\times \sqrt{2-\sqrt{2-\sqrt{3}}}\times \sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}}\times \sqrt{2+\sqrt{2-\sqrt{2-\sqrt{2-\sqrt{3}}}}}\\ &=\sqrt{2-\sqrt{3}}\times \sqrt{2-\sqrt{2-\sqrt{3}}}\times \sqrt{2^{2}-\left (\sqrt{2-\sqrt{2-\sqrt{3}}} \right )^{2}}\\ &=\sqrt{2-\sqrt{3}}\times \sqrt{2-\sqrt{2-\sqrt{3}}}\times \sqrt{4-\left ( 2-\sqrt{2-\sqrt{3}} \right )}\\ &=\sqrt{2-\sqrt{3}}\times \sqrt{2-\sqrt{2-\sqrt{3}}}\times \sqrt{2+\sqrt{2-\sqrt{3}}}\\ &=\sqrt{2-\sqrt{3}}\times \sqrt{2^{2}-\left ( \sqrt{2-\sqrt{3}} \right )^{2}}\\ &=\sqrt{2-\sqrt{3}}\times \sqrt{4-\left ( 2-\sqrt{3} \right )}\\ &=\sqrt{2-\sqrt{3}}\times \sqrt{2+\sqrt{3}}\\ &=\sqrt{2^{2}-\left ( \sqrt{3} \right )^{2}}\\ &=\sqrt{4-3}=\sqrt{1}\\ &=1 \end{aligned} \end{array}.

Sumber Referensi

  1. Suparmin, Sukino, Theresia Suli Intan, Yohanes Eric Santiago. 2016. Pena Emas Olimpiade Sains Nasional Matematika untuk SMP. Bandung: YRAMA WIDYA.
  2. Tung, Khoe Yao. 2012. Pintar Matematika SMA Kelas X Untuk Olimpiade dan Pengayaan Pelajaran. Yogyakarta: ANDI.
  3. _________, Algebra: Teoria con 8000 Problemas Propuestos y Resueltos.

Tentang ahmadthohir1089

Nama saya Ahmad Thohir, asli orang Purwodadi 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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