Tambahan Contoh Soal 6 (Bentuk Pangkat, Akar, dan Logaritma)

\begin{array}{ll}\\ \fbox{46}&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}\\ &\textrm{b}.\quad ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}\\ &\textrm{c}.\quad \displaystyle \frac{1}{6}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &\textrm{d}.\quad ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )\\ &\textrm{e}.\quad^{3}\log 54+\: ^{3}\log 6-2.\, ^{3}\log 2\\ &\textrm{f}.\quad 2.\, ^{5}\log 15+\: ^{5}\log 4-2.\, ^{5}\log 6\end{array}\\\\\\ \textrm{Jawab}:\\\\.

\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad ^{6}\log 8 +\: ^{6}\log \displaystyle \frac{9}{2}&=\: ^{6}\log 8\times \frac{9}{2}\\ &=\: ^{6}\log 36\\ &=\: ^{6}\log 6^{2}\\ &=2\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad ^{2}\log \displaystyle \frac{4}{3}+\: 2.\, ^{2}\log \sqrt{12}&=\: ^{2}\log \frac{4}{3}+\: ^{2}\log \left ( \sqrt{12} \right )^{2}\\ &=\: ^{2}\log \frac{4}{3}+\: ^{2}\log 12\\ &=\: ^{2}\log \frac{4}{3}\times 12\\ &=\: ^{2}\log 16\\ &=\: ^{2}\log 2^{4}\\ &=4 \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad \displaystyle \frac{1}{6}.\, ^{2}\log &25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\displaystyle \frac{1}{3}.\frac{1}{2}.\, ^{2}\log 25-\: \frac{1}{3}.\, ^{2}\log 10\\ &=\frac{1}{3}\left ( ^{2}\log 25^{\frac{1}{2}} -\: ^{2}\log 10\right )\\ &=\frac{1}{3}\left ( ^{2}\log 5-\: ^{2}\log 10 \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{5}{10} \right )\\ &=\frac{1}{3}\left ( ^{2}\log \frac{1}{2} \right )\\ &=\frac{1}{3}\left ( ^{2}\log 2^{-1} \right )\\ &=-\frac{1}{3} \end{aligned}&\begin{aligned}\textrm{d}.\quad ^{2}\log \left ( ^{2}\log \sqrt{\sqrt{2}} \right )&=\: ^{2}\log \left ( ^{2}\log 2^{\frac{1}{4}} \right )\\ &=\: ^{2}\log \frac{1}{4}\\ &=\: ^{2}\log \left (2 \right )^{-2}\\ &=-2\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.

\begin{array}{ll}\\ \fbox{47}&\textrm{Tentukanlah nilai dari}\\ &\textrm{a}.\quad \left ( 2^{\: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &\textrm{b}.\quad \left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &\textrm{c}.\quad \left ( ^{625}\log 19 \right )\left ( ^{7}\log \displaystyle \frac{1}{25} \right )\left ( ^{19}\log 7 \right )\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad &\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: ^{^{^{9}}}\log 5} \right )\left ( 5^{\: ^{^{^{\frac{1}{5}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3^{2}}}}\log 5} \right )\left ( 5^{\: \: ^{^{^{5^{-1}}}}\log 2} \right )\\ &=\left ( 2^{\: \: ^{^{^{2}}}\log 6} \right )\left ( 3^{\: \: ^{^{^{3}}}\log 5^{^{\frac{1}{2}}}} \right )\left ( 5^{\: \: ^{^{^{5}}}\log 2^{-1}} \right )\\ &=\left ( 2^{\: ^{^{2}}\log 6} \right )\left ( 3^{\: ^{^{3}}\log \sqrt{5}} \right )\left ( 5^{\: ^{^{5}}\log \frac{1}{2}} \right )\\ &=6\times \sqrt{5}\times \frac{1}{2}\\ &=3\sqrt{5} \end{aligned}&\begin{aligned}\textrm{b}.\quad &\left ( ^{27}\log 125 \right )\left ( ^{25}\log \displaystyle \frac{1}{64} \right )\left ( ^{64}\log \frac{1}{9} \right )\\ &=\left ( ^{^{3^{3}}}\log 5^{3} \right )\left ( ^{^{5^{2}}}\log 4^{-3} \right )\left ( ^{^{4^{3}}}\log 3^{-2} \right )\\ &=\displaystyle \frac{3}{3}.\left ( -\frac{3}{2} \right ).\left ( -\frac{2}{3} \right ).\: ^{^{3}}\log 5.\: ^{^{5}}\log 4.\: ^{^{4}}\log 3\\ &=1\\ &\\ &\\ &\\ &\end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ \fbox{48}.&\textrm{Diketahui bahwa}\: \: ^{^{3}}\log 7=m,\: \textrm{maka nilai untuk}\\ &\textrm{a}.\quad ^{^{27}}\log \sqrt[3]{\displaystyle \frac{1}{49}}\\ &\textrm{b}.\quad \displaystyle \frac{^{^{5}}\log \sqrt{27}}{^{^{5}}\log 7\sqrt{7}}\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{l|l}\\ \begin{aligned}\textrm{a}.\quad ^{^{27}}\log \sqrt[3]{\displaystyle \frac{1}{49}}&=\: ^{^{3^{3}}}\log 7^{^{-\frac{2}{3}}}\\ &=\displaystyle \frac{\left ( -\frac{2}{9} \right )}{3}.\: ^{^{3}}\log 7\\ &=-\frac{2}{9}m\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad \displaystyle \frac{^{^{5}}\log \sqrt{27}}{^{^{5}}\log 7\sqrt{7}}&=\: ^{^{7\sqrt{7}}}\log \sqrt{27}\\ &=\: ^{^{7^{\frac{3}{2}}}}\log 3^{^{\frac{3}{2}}}\\ &=\displaystyle \frac{\frac{3}{2}}{\frac{3}{2}}\: ^{7}\log 3\\ &=\displaystyle \frac{1}{^{^{3}}\log 7}\\ &=\displaystyle \frac{1}{m} \end{aligned} \end{array}.

\begin{array}{ll}\\ \fbox{49}.&\textrm{Diketahui}\: \: ^{^{2}}\log 3=p \: \: \textrm{dan}\: \: ^{^{3}}\log 11=q,\: \: \textrm{maka}\: \: ^{^{44}}\log 66=....\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}^{^{44}}\log 66&=\displaystyle \frac{^{^{^{...}}}\log 66}{^{^{^{...}}}\log 44}\\ &=\displaystyle \frac{^{^{^{...}}}\log \left (2\times 3\times 11 \right )}{^{^{^{...}}}\log \left (2^{2}\times 11 \right )}&\textnormal{titik-titik dipilih dengan memperhatikan}\: \: numerus\\ &=\displaystyle \frac{^{^{^{3}}}\log 2+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{^{^{^{3}}}\log 2^{2}+\: ^{^{^{3}}}\log 11}&\textnormal{dipilih}\: \: numerus\: \: 11\: \: \textrm{yang memiliki bilangan basis 3}\\ &=\displaystyle \frac{\frac{1}{^{^{^{2}}}\log 3}+\: ^{^{^{3}}}\log 3+\:^{^{^{3}}}\log 11}{\frac{2}{^{^{^{3}}}\log 2^{2}}+\: ^{^{^{3}}}\log 11}=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\\ &=\displaystyle \frac{\frac{1}{p}+1+q}{\frac{2}{p}+q}\times \displaystyle \frac{p}{p}\\ &=\frac{1+p+pq}{2+pq} \end{aligned}.

\begin{array}{ll}\\ \fbox{50}.&\textrm{(AIME 1984)Diketahui bahwa}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{bilangan real yang memenuhi}\\\\ &\left\{\begin{matrix} ^{^{8}}\log x+\: ^{^{4}}\log y^{2}=5\\ \\ ^{^{8}}\log y+\: ^{^{4}}\log x^{2}=7 \end{matrix}\right.\\\\ &\textrm{Tentukanlah nilai dari}\: \: xy \end{array}\\\\\\ \textrm{Jawab}:\\\\.

\begin{aligned}&^{^{^{8}}}\log x+\: ^{^{4}}\log y^{2}=5&\Rightarrow \: ^{^{^{2^{3}}}}\log x+\: ^{^{^{2^{2}}}}\log y^{2}=5....(1)\\ &^{^{^{8}}}\log y+\: ^{^{4}}\log x^{2}=7&\Rightarrow \: ^{^{^{2^{3}}}}\log y+\: ^{^{^{2^{2}}}}\log x^{2}=7....(2)\\ &&-----------\: \: \\ &\textrm{selanjutnya},&\\ &\frac{1}{3}\:. ^{^{^{2}}}\log x+\: ^{^{^{2}}}\log y=5\: \: |\times \frac{1}{3}|&\Rightarrow \frac{1}{9}\: .^{^{^{2}}}\log x+\: \frac{1}{3}.\: ^{^{^{2}}}\log y=\frac{5}{3}\\ &^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\: \: |\times 1|&==\Rightarrow ^{^{^{2}}}\log x+\:\frac{1}{3}\: . ^{^{^{2}}}\log y=7\\ &&-------------\: \:\ominus \\ &&-\frac{8}{9}.\: ^{^{^{2}}}\log x \: \: \: \: \: \: \: \: =\frac{5}{3}-7=-\frac{16}{3} \end{aligned}.

\begin{aligned}^{^{^{2}}}\log x&=\left ( -\frac{16}{3} \right )\left ( -\frac{9}{8} \right )\\ ^{^{^{2}}}\log x&=6\\ x&=2^{6}=64........(3)\\ \textrm{Sehingga}&,\: \textrm{dengan menggunakan persamaan 1 kita akan mendapatkan}\\ \frac{1}{3}.\: ^{^{^{2}}}\log x&+\: ^{^{^{2}}}\log y=5\: \: \Rightarrow \: \: \frac{1}{3}.\: ^{^{^{2}}}\log 2^{6}+\: ^{^{^{2}}}\log y=5\\ \frac{1}{3}.6&+\: ^{^{^{2}}}\log y=5\\ &\: \: \: \: \: \: \: ^{^{^{2}}}\log y=5-2=3\\ &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y=2^{3}=8\\ \textrm{Jadi}, \: \: \textrm{nilai dari}&\: \: xy=64\times 8=512\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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