Lanjutan Contoh Soal (Bentuk Pangkat, Akar, dan Logaritma (KTSP)) 3

\begin{array}{ll}\\ \fbox{21}.&\textrm{Hitunglah}\:\\\\ &\qquad\qquad \displaystyle \sqrt{\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}}\\ \end{array}\\.

Jawab:

\begin{aligned}&\begin{cases} & \displaystyle \frac{1}{1+\sqrt{2}}\: \: \: \, =\sqrt{2}-1 \\ & \displaystyle \frac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2} \\ & \displaystyle \frac{1}{\sqrt{3}+\sqrt{4}}=\sqrt{4}-\sqrt{3} \\ & \: ..............\: \: \: \: \: \: \: =... \\ & \displaystyle \frac{1}{\sqrt{99}+\sqrt{100}}=\sqrt{100}-\sqrt{99} \end{cases}\\ &\\ &\displaystyle \sqrt{\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}}\\ &=\displaystyle \sqrt{\left ( \sqrt{2}-1 \right )+\left ( \sqrt{3}-\sqrt{2} \right )+\left ( \sqrt{4}-\sqrt{3} \right )+...+\left ( \sqrt{100}-\sqrt{99} \right )}\\ &=\displaystyle \sqrt{\sqrt{100}-1}\\ &=\displaystyle \sqrt{10-1}\\ &=\displaystyle \sqrt{9}\\ &=3 \end{aligned}.

\begin{array}{ll}\\ \fbox{22}&\textrm{Hitunglah} \end{array}\\ \begin{array}{lll}\\ .\quad\quad &a.&\displaystyle \frac{1}{\sqrt[3]{2}+1}\\ &b.&\displaystyle \frac{1}{\sqrt[3]{2}-1} \end{array}.

Jawab:

Sebelumnya

\begin{array}{|lll|}\hline \textbf{Ingat}&&\\ &&\\ &&1.\quad a+ b=\left (\sqrt[3]{a}+\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}-\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &&\\ &&2.\quad a- b=\left (\sqrt[3]{a}-\sqrt[3]{b} \right )\left ( \sqrt[3]{a^{2}}+\sqrt[3]{ab}+\sqrt[3]{b^{2}} \right )\\ &&\\\hline \end{array}.

\begin{aligned}\displaystyle a.\qquad \frac{1}{\sqrt[3]{2}+1}&=\frac{1}{\sqrt[3]{2}+1}\times \frac{\sqrt[3]{4}-\sqrt[3]{2}+1}{\sqrt[3]{4}-\sqrt[3]{2}+1}\\ &=\displaystyle \frac{\sqrt[3]{4}-\sqrt[3]{2}+1}{2-1}\\ &=\sqrt[3]{4}-\sqrt[3]{2}+1 \end{aligned}.

\begin{array}{ll}\\ \fbox{23}&\textrm{Carilah nilai x yang memenuhi}\\ &\\ &\qquad \sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{...}}}}}=\sqrt{4x+\sqrt{4x+\sqrt{4x+\sqrt{4x+...}}}} \end{array}.

Jawab:

Sebelumnya dapat kita lihat di sini bahwa  untuk x > 0,   \sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{...}}}}}=x .

Sehingga

\begin{aligned}\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{...}}}}}&=\sqrt{4x+\sqrt{4x+\sqrt{4x+\sqrt{4x+\sqrt{...}}}}}\\ x&=\sqrt{4x+\sqrt{4x+\sqrt{4x+\sqrt{4x+\sqrt{...}}}}},\: \textrm{(kuadratkan masing-masing ruas)}\\ x^{2}&=4x+\sqrt{4x+\sqrt{4x+\sqrt{4x+\sqrt{4x+\sqrt{...}}}}}\\ x^{2}&=4x+x\\ x^{2}&=5x\\ x^{2}&-5x=0\\ x(x-5)&=0\\ x=&0\quad \textrm{atau}\quad x=5\\ &\\ \therefore &\textrm{harga x yang memenuhi adalah 5}\end{aligned}.

\begin{array}{ll}\\ \fbox{24}.&\textrm{Nyatakanlah dalam bentuk logaritma} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle 3^{3}=27&b.&\displaystyle 3^{-3}=\displaystyle \frac{1}{27}&c.&\displaystyle \sqrt{3}^{3}=3\sqrt{3}&d.&\displaystyle 0,001=10^{-3}\\ &&&&&&&\\ &e.&\displaystyle 2^{x}=y&f.&\displaystyle 10^{2x}=t&g.&\displaystyle \sqrt[3]{4}=\displaystyle 2^{\frac{2}{3}}&h.&\displaystyle \displaystyle 27^{\frac{2}{3}}=9\\ \end{array}.

Jawab:

\begin{array}{|ll||ll||ll||ll|}\hline a.&^{3}\log 27=3&b.&^{3}\log \displaystyle \frac{1}{27}=-3&c.&^{\sqrt{3}}\log 3\sqrt{3}=3&d.&^{10}\log 0,001=-3\\ &&&&&&&\\\hline &&&&&&&\\ e.&^{2}\log y=x&f.&^{10}\log t=2x&g.&^{2}\log \sqrt[3]{4}=\displaystyle \frac{2}{3}&h.&^{27}\log 9=\displaystyle \frac{2}{3}\\ &&&&&&&\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{25}.&\textrm{Nyatakanlah bentuk logaritma berikut ke bentuk pangkat} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle ^{3}\log 9=2&b.&\displaystyle ^{3}\log 9\sqrt{9}=3&c.&\displaystyle ^{3\sqrt{3}}\log 9=\frac{4}{3}&d.&^{\frac{1}{3}}\displaystyle \log \displaystyle \frac{1}{9}=2\\ &&&&&&&\\ &e.&\displaystyle ^{\frac{1}{9}}\log \displaystyle \frac{1}{3}=\frac{1}{2}&f.&\displaystyle ^{3}\log 1=0&g.&\displaystyle ^{3}\log a=b^{3}&h.&\displaystyle \displaystyle ^{\frac{1}{8}}\log 2=-\frac{1}{3}\\ \end{array}.

Jawab:

\begin{array}{|ll||ll||ll||ll|}\hline a.&3^{2}=9&b.&3^{3}=9\sqrt{9}&c.&\left (3\sqrt{3} \right )^{\frac{4}{3}}=9&d.&\left (\displaystyle \frac{1}{3} \right )^{2}=\displaystyle \frac{1}{9}\\ &&&&&&&\\\hline &&&&&&&\\ e.&\left ( \displaystyle \frac{1}{9} \right )^{\frac{1}{2}}=\displaystyle \frac{1}{3}&f.&3^{0}=1&g.&\displaystyle 3^{\left ( \displaystyle b^{3} \right )}=a&h.&\left ( \displaystyle \frac{1}{8} \right )^{-\frac{1}{3}}=2\\ &&&&&&&\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{26}.&\textrm{Tentukanlah nilai x yang memenuhi} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&^{5}\log x=2&b.&\displaystyle ^{2}\log 32=x&c.&\displaystyle ^{6}\log \, (x+1)=2&d.&^{x}\displaystyle \log 8=\displaystyle \frac{1}{3}\\ &&&&&&&\\ &e.&\displaystyle ^{x}\log \displaystyle \frac{1}{125}=3&f.&\displaystyle ^{\frac{1}{2}}\log x=2&g.&\displaystyle ^{x}\log 11\sqrt{11}=\displaystyle \frac{3}{2}&h.&\displaystyle \displaystyle ^{\frac{1}{8}}\log 2=x\\ \end{array}.

Jawab:

\begin{array}{|l|l||l|l||l|l||l|l|}\hline a.&x=5^{2}=25&b.&\begin{aligned}2^{x}&=32\\ 2^{x}&=2^{5}\\ x&=5\\ & \end{aligned}&c.&\begin{aligned}(x+1)&=6^{2}\\ (x+1)&=36\\ (x+1)&=35+1\\ x&=35 \end{aligned}&d.&\begin{aligned}x^{\frac{1}{3}}&=8\\ x&=8^{\frac{3}{1}}\\ x&=8^{3}\\ x&=512 \end{aligned}\\ &&&&&&&\\\hline &&&&&&&\\ e.&\begin{aligned}x^{3}&=\displaystyle \frac{1}{125}\\ x^{3}&=\displaystyle \left ( \frac{1}{5} \right )^{3}\\ x&=\displaystyle \frac{1}{5}\\ &\\ & \end{aligned}&f.&x=\left ( \displaystyle \frac{1}{2} \right )^{2}=\displaystyle \frac{1}{4}&g.&\begin{aligned}x^{\frac{3}{2}}&=11\sqrt{11}\\ x^{\frac{3}{2}}&=11^{\frac{3}{2}}\\ x&=11\\ &\\ &\\ & \end{aligned}&h.&\begin{aligned}\left ( \frac{1}{8} \right )^{x}&=2\\ \left ( 2^{-3} \right )^{x}&=2^{1}\\ 2^{-3x}&=2^{1}\\ -3x&=1\\ x&=-\displaystyle \frac{1}{3} \end{aligned}\\ &&&&&&&\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{27}.&\textrm{Tentukanlah nilai logaritma berikut} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&^{5}\log 25\sqrt{5}&b.&\displaystyle ^{2}\log \displaystyle \frac{1}{32}&c.&\displaystyle ^{\sqrt{3}}\log 81&d.&^{\frac{1}{3}}\displaystyle \log \displaystyle \frac{1}{243}\\ &&&&&&&\\ &e.&\displaystyle ^{\sqrt{2}}\log 16&f.&\displaystyle ^{\sqrt{5}}\log \sqrt{125}&g.&\displaystyle ^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}&h.&\displaystyle \displaystyle ^{0,333...}\log \left (0,111... \right )\\ \end{array}.

Jawab:

\begin{array}{|l|l||l|l||l|l||l|l|}\hline a.&\begin{aligned}^{5}\log 25\sqrt{5}&=\: ^{5^{1}}\log 5^{2}.5^{\frac{1}{2}}\\ &=\: ^{5^{1}}\log 5^{\frac{5}{2}}\\ &=\displaystyle \frac{\frac{5}{2}}{1}\times \: ^{5}\log 5\\ &=\displaystyle \frac{5}{2} \end{aligned}&b.&\begin{aligned}^{2}\log \displaystyle \frac{1}{32}&=\: ^{2^{1}}\log 2^{-5}\\ &=\displaystyle \frac{-5}{1}\times \: ^{2}\log 2\\ &=-5 \end{aligned}&c.&\begin{aligned}^{\sqrt{3}}\log 81&=\: ^{\displaystyle 3^{\frac{1}{2}}}\log 3^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{3}\log 3\\ &=8 \end{aligned}&d.&\begin{aligned}^{\frac{1}{3}}\log \displaystyle \frac{1}{243}&=\: ^{\left (\frac{1}{3} \right )^{1}}\log \displaystyle \left (\frac{1}{3} \right )^{5}\\ &=\displaystyle \frac{5}{1}\times \: ^{\frac{1}{3}}\log \frac{1}{3}\\ &=5 \end{aligned}\\\hline e.&\begin{aligned}^{\sqrt{2}}\log 16&=\: ^{\displaystyle 2^{\frac{1}{2}}}\log 2^{4}\\ &=\displaystyle \frac{4}{\frac{1}{2}}\times \: ^{2}\log 2\\ &=8\\ &\\ &\\ & \end{aligned} &f.&\begin{aligned}^{\sqrt{5}}\log \sqrt{125}&=\: ^{\sqrt{5}^{1}}\log \left ( \sqrt{5} \right )^{3}\\ &=\displaystyle \frac{3}{1}\times \: ^{\sqrt{5}}\log \sqrt{5}\\ &=3\\ &\\ &\\ & \end{aligned} &g.&\begin{aligned}&^{\sqrt{\sqrt{2}}}\log \sqrt{8\sqrt{8}}\\ &=\: ^{\sqrt[4]{2}}\log \left ( 8\left ( 8 \right )^{\frac{1}{2}} \right )^{\frac{1}{2}}\\ &=\: ^{2^{\frac{1}{4}}}\log 8^{\left (\frac{1}{2}+\frac{1}{4} \right )}\\ &=\: ^{2^{\frac{1}{4}}}\log 2^{3\left ( \frac{3}{4} \right )}\\ &=\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}\times \: ^{2}\log 2\\ &=9 \end{aligned}&h.&\begin{aligned}&^{0,333...}\log 0,111...\\ &=\: ^{\frac{1}{3}}\log \frac{1}{9}\\ &=\: ^{\frac{1}{3}}\log \left (\frac{1}{3} \right )^{2}\\ &=2\\ &\\ &\\ & \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ \fbox{28}.&\textrm{Hitunglah} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle ^{6}\log 9+ ^{6}\log 8-^{6}\log 2&f.&\displaystyle \frac{\log ^{2}5-\log ^{2}2}{\log \sqrt[3]{2,5}}\\ &&&&&&&\\ &b.&\displaystyle \log 2+\log 4+\log 125&g.&\displaystyle \frac{\left ( ^{3}\log 36 \right )^{2}-\left ( ^{3}\log 4 \right )^{2}}{^{3}\log \sqrt{12}}\\ &&&&&&&\\ &c.&\displaystyle \frac{\log \sqrt{2}+\log 4\sqrt{32}+\log 4}{\log 8}&h.&\displaystyle \frac{\left ( ^{3}\log 8+^{5}\log 8 \right )\times ^{15}\log 8}{^{3}\log 8\times ^{5}\log 8}\\ &&&&&&&\\ &d.&\displaystyle \frac{\log x\sqrt{x}+\log \sqrt{y}+\log xy^{2}}{\log xy}&i.&\displaystyle 4^{{^{2}\log 6}}+\displaystyle 3^{{^{9}\log 5}}+\displaystyle 5^{{^{0,2}\log 2}}\\ &&&&&&&\\ &e.&\sqrt{\displaystyle \frac{8\log 512}{3\log 2+7\log 4-5\log 8}}&j.&^{5}\log \sqrt{27}\times ^{9}\log 125+^{16}\log 32 \end{array}.

Jawab:

\begin{array}{|l|l|l|l|}\hline a.&\begin{aligned}\displaystyle ^{6}\log 9+ ^{6}\log 8-^{6}\log 2&=\: ^{6}\log \displaystyle \frac{9\times 8}{2}\\ &=\: ^{6}\log 36\\ &=\: ^{6}\log 6^{2}\\ &=2\\ & \end{aligned}&b.&\begin{aligned}\displaystyle \log 2+\log 4+\log 125&=\displaystyle ^{10}\log 2+^{10}\log 4+^{10}\log 125\\ &=\: ^{10}\log 2\times 4\times 125\\ &=\: ^{10}\log 1000\\ &=\: ^{10}\log 10^{3}\\ &=\: 3 \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|l|l|}\hline c.&\begin{aligned}\displaystyle \frac{\log \sqrt{2}+\log 4\sqrt{32}+\log 4}{\log 8}&=\: ^{8}\log \left (\sqrt{2}\times 4\sqrt{32}\times 4 \right )\\ &=\: ^{2^{3}}\log 16\sqrt{64}\\ &=\: ^{2^{3}}\log 2^{4}\times 2^{3}\\ &=\: ^{2^{3}}\log 2^{7}\\ &=\displaystyle \frac{7}{3} \end{aligned}&d.&\begin{aligned}\displaystyle \frac{\log x\sqrt{x}+\log \sqrt{y}+\log xy^{2}}{\log xy}&=\: ^{xy}\log \left ( x\sqrt{x}.\sqrt{y}.xy^{2} \right )\\ &=\: ^{xy}\log x^{2}y^{2}\sqrt{xy}\\ &=\: ^{xy}\log \left ( xy \right )^{\frac{5}{2}}\\ &=\displaystyle \frac{5}{2} \end{aligned} \\\hline \end{array}.

\begin{array}{|l|l|l|l|}\hline e.&\begin{aligned}\displaystyle \sqrt{\frac{8\log 512}{3\log 2+7\log 4-5\log 8}}&=\sqrt{\frac{\log 512^{8}}{\log 2^{3}+\log 4^{7}-\log 8^{5}}}\\ &=\displaystyle \sqrt{\frac{\log \left (2^{9} \right )^{8}}{\log \displaystyle \frac{2^{3}\times \left ( 2^{2} \right )^{7}}{\left ( 2^{3} \right )^{5}}}}\\ &=\displaystyle \sqrt{\frac{\log 2^{72}}{\log 2^{3+14-35}}}\\ &=\displaystyle \sqrt{\frac{72\log 2}{2\log 2}}\\ &=6 \end{aligned}&f.&\begin{aligned}\displaystyle \displaystyle \frac{\log ^{2}5-\log ^{2}2}{\log \sqrt[3]{2,5}}&=\displaystyle \frac{\left ( \log 5+\log 2 \right )\left ( \log 5-\log 2 \right )}{\displaystyle \log \left ( \frac{5}{2} \right )^{\frac{1}{3}}}\\ &=\displaystyle \frac{1.\left ( \log \left ( \frac{5}{2} \right ) \right )}{\frac{1}{3}\log \left ( \frac{5}{2} \right )}\\ &=3\\ &\\ &\\ &\\ &\\ \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|l|l|}\hline i.&\begin{aligned}\displaystyle \displaystyle 4^{{^{2}\log 6}}+\displaystyle 3^{{^{9}\log 5}}+\displaystyle 5^{{^{0,2}\log 2}}&=\displaystyle 2^{2\: {^{2}\log 6}}+\displaystyle 3\: ^{{^{3^{2}}\log 5}}+\displaystyle 5^{{^{\frac{1}{5}}\log 2}}\\ &=\displaystyle 2\: ^{{^{2}\log 6^{2}}}+\displaystyle 3\: ^{{^{3}\log 5^{\frac{1}{2}}}}+\displaystyle 5\: ^{{^{5}\log 2^{-1}}}\\ &=6^{2}+5^{\frac{1}{2}}+2^{-1}\\ &=36+\sqrt{5}+\frac{1}{2}\\ &=36\frac{1}{2}+\sqrt{5} \end{aligned}&j.&\begin{aligned}\displaystyle ^{5}\log \sqrt{27}\times ^{9}\log 125+^{16}\log 32&=\: ^{5}\log 3^{\frac{3}{2}}\times ^{3^{2}}\log 5^{3}+^{2^{4}}\log 2^{5}\\ &=\displaystyle \frac{3}{2}.\frac{3}{2}.^{5}\log 3.^{3}\log 5+\displaystyle \frac{5}{4}.^{2}\log 2\\ &=\displaystyle \frac{9}{4}+\frac{5}{4}\\ &=\displaystyle \frac{14}{4}\\ &=\displaystyle \frac{7}{2} \end{aligned}\\\hline \end{array}.

\fbox{29}.\quad \textrm{jika}\: f(x)=\displaystyle \frac{^{8}\log x}{1-2\times ^{8}\log x}\: ,\: \textrm{tentukanlah}\: f(x)=f\left ( \displaystyle \frac{8}{x} \right ).

Jawab:

\begin{aligned}f(x)&=f\left ( \frac{8}{x} \right )\\ \displaystyle \frac{^{8}\log x}{1-2\times ^{8}\log x}&=\displaystyle \frac{^{8}\log \left ( \frac{8}{x} \right )}{1-2\times ^{8}\log \left ( \frac{8}{x} \right )}\\ ^{8}\log x\left ( 1-2\times ^{8}\log \left (\frac{8}{x} \right ) \right )&=\: ^{8}\log \left ( \frac{8}{x} \right )\left ( 1-2\times ^{8}\log x \right )\\ ^{8}\log x-2\times ^{8}\log x\left ( ^{8}\log 8-^{8}\log x \right )&=\: \left ( ^{8}\log 8-^{8}\log x \right )\left ( 1-2\times ^{8}\log x \right )\\ ^{8}\log x-2\times ^{8}\log x+2\times \left ( ^{8}\log x \right )^{2}&=1-2\times ^{8}\log x-^{8}\log x+2\times \left ( ^{8}\log x \right )^{2}\\ 2\times ^{8}\log x&=1\\ x^{2}&=8^{1}\\ x&=\sqrt{8}\\ x&=2\sqrt{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{30}.&\textbf{(UM IKIP PGRI 2010)}\textrm{Jika diketahui}\\ &\\ &\quad \displaystyle ^{2}\log \,^{2}\log x=\: ^{2}\log \left ( 6-\: ^{2}\log x \right )+1 \\ &\\ &\textrm{maka nilai x adalah ....} \end{array}.

Jawab:

\begin{aligned}\displaystyle ^{2}\log \, ^{2}\log x&=\: ^{2}\log \left ( 6-\: ^{2}\log x \right )+1 \\ \displaystyle ^{2}\log \, ^{2}\log x&=\: ^{2}\log \left ( 6-\: ^{2}\log x \right )+\, ^{2}\log 2\\ \displaystyle ^{2}\log \, ^{2}\log x&=\: ^{2}\log 2\times \left ( 6-\: ^{2}\log x \right )\\ ^{2}\log x&= 12-2.^{2}\log x \\ 2.^{2}\log x&=12\\ ^{2}\log x&=4\\ x&=2^{4}\\ x&=16\\ &\\ \textrm{Jadi},\: \textrm{nilai}\,\: & \textrm{x = 16} \end{aligned}.

 

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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