Lanjutan 3 (Fungsi Eksponen Dan logaritma)

\begin{array}{ll}\\ 11.&\textrm{Nyatakanlah dengan pangkat positif kemudian sederhankanlah} \end{array}\\  \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle 3^{-4}&d.&\displaystyle \left ( \frac{c}{ab} \right )^{-1}&g.&\displaystyle \left ( x+y \right )^{-1}&j.&\displaystyle \frac{a^{-1}+b^{-1}}{a+b}\\ &b.&\left ( a^{-3}b \right )^{-5}&e.&\displaystyle \left ( \frac{3a^{-2}b^{-3}}{a^{-5}b} \right )^{-2}&h.&\displaystyle \frac{1+a^{-1}}{1+a}&k.&\displaystyle \left ( \frac{1}{1+\displaystyle \frac{1}{a}} \right ):\left ( \frac{1}{1+a} \right )\\ &c.&\displaystyle \frac{3^{0}a^{-6}b^{-3}}{a^{5}b^{-1}}&f.&a^{-1}+b^{-1}&i.&\left ( a+b \right )^{-1}\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&3^{-4}=\displaystyle \frac{1}{3^{4}}&d&\displaystyle \left ( \frac{c}{ab} \right )^{-1}=\left ( \frac{ab}{c} \right )&g&\left ( x+y \right )^{-1}=\displaystyle \frac{1}{\left ( x+y \right )}&j&\begin{aligned}\displaystyle \frac{a^{-1}+b^{-1}}{a+b}&\\ =\displaystyle \frac{\frac{1}{a}+\frac{1}{b}}{a+b}&\\ =\displaystyle \frac{\left ( \frac{a+b}{ab} \right )}{a+b}&\\ =\displaystyle \frac{1}{ab} \end{aligned}\\\hline f.&\multicolumn{3}{|c|}{\begin{aligned}a^{-1}+b^{-1}&=\displaystyle \frac{1}{a}+\frac{1}{b}\\ &=\displaystyle \frac{a+b}{ab}\\ & \end{aligned}}&k.&\multicolumn{3}{|c|}{\begin{aligned}\left ( \displaystyle \frac{1}{1+\displaystyle \frac{1}{a}} \right ):\left ( \displaystyle \frac{1}{1+a} \right )&=\displaystyle \frac{1+a}{1+\displaystyle \frac{1}{a}}\\ &=\displaystyle \frac{1+a}{\left ( \displaystyle \frac{a+1}{a} \right )}\\ &=a \end{aligned}}\\\hline c&\multicolumn{7}{|c|}{\begin{aligned}\displaystyle \frac{3^{0}a^{-6}b^{-3}}{a^{5}b^{-1}}&=\displaystyle \frac{3^{0}}{a^{5}.a^{6}.b^{-1}.b^{3}}\\ &=\displaystyle \frac{1}{a^{11}b^{2}} \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 12.&\textrm{Nyatakanlah dengan pangkat positif kemudian sederhankanlah} \end{array}\\ \begin{array}{lllllllll}\\ .\quad\quad&a.&\displaystyle \frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}&b.&\displaystyle \frac{x^{-2}y-xy^{-2}}{x^{-1}y-xy^{-1}}&c.&\displaystyle \frac{1-x^{-2}y^{2}}{1-x^{-1}y}\\ &&&&&&&\\ &d.&\displaystyle \frac{3x^{-1}-y^{-2}}{x^{-2}+2y^{-1}}&e.&\displaystyle \frac{x^{-1}y-xy^{-1}}{x^{-1}-y^{-1}}\end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline 12.a&\begin{aligned}\displaystyle \frac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}&=\frac{\displaystyle \frac{1}{x^{2}}+\frac{1}{y^{2}}}{\displaystyle \frac{1}{x}+\frac{1}{y}}\times \displaystyle \frac{x^{2}y^{2}}{x^{2}y^{2}}\\ &=\frac{\displaystyle y^{2}+x^{2}}{\displaystyle xy^{2}+x^{2}y}\\ &=\displaystyle \frac{\left ( x+y \right )^{2}-2xy}{xy\left ( x+y \right )}\\ &=\displaystyle \frac{x+y}{xy}-\frac{2}{x+y} \end{aligned}&12.c&\begin{aligned}\displaystyle \frac{1-x^{-2}y^{2}}{1-x^{-1}y}&=\frac{1-\displaystyle \frac{y^{2}}{x^{2}}}{1-\displaystyle \frac{y}{x}}\times \displaystyle \frac{x^{2}}{x^{2}}\\ &=\displaystyle \frac{x^{2}-y^{2}}{x^{2}-xy}\\ &=\displaystyle \frac{\left ( x+y \right )\left ( x-y \right )}{x\left ( x-y \right )}\\ &=\displaystyle \frac{x+y}{x} \end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 13.&\textrm{sederhankanlah bilangan di bawah tanda akar berikut ini!}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle \sqrt{20}&b.&\sqrt{48}&c.&\sqrt{96}\\ &&&&&&&\\ d.&\sqrt{120}&e.&\sqrt{200}&f.&\displaystyle \frac{2}{5}\sqrt{500}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|l|l|l|l|}\hline a.&\sqrt{20}=\sqrt{4.5}=\sqrt{4}.\sqrt{5}=2\sqrt{5}&b.&\sqrt{48}=\sqrt{16.3}=4\sqrt{3}&c.&\sqrt{96}=\sqrt{16.6}=4\sqrt{6}\\\hline d.&\sqrt{120}=\sqrt{4.30}=2\sqrt{30}&e.&\sqrt{200}=\sqrt{100.2}=10\sqrt{2}&f.&\begin{aligned}\displaystyle &\frac{2}{5}\sqrt{500}=\displaystyle \frac{2}{5}\sqrt{100.5}\\ &=\displaystyle \frac{2}{5}.10\sqrt{5}=4\sqrt{5} \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 14.&\textrm{sederhankanlah bentuk akar berikut ini jika terdefinisi}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle \sqrt{x^{5}}&b.&\sqrt{9a^{2}b}&c.&\sqrt{18a^{3}}\\ &&&&&&&\\ d.&\sqrt{49a^{4}b}&e.&\sqrt{72c^{6}d^{10}}&f.&\displaystyle \sqrt{28p^{8}q^{6}r^{4}}\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|l|l|l|l|}\hline a.&\begin{aligned}\sqrt{x^{5}}&=\sqrt{x^{4}.x^{1}}=x^{\frac{4}{2}}.x^{\frac{1}{2}}\\ &=x^{2}.x^{\frac{1}{2}}=x\sqrt{x} \end{aligned}&b.&\sqrt{9a^{2}b}=\sqrt{3^{2}.a^{2}.b}=3a\sqrt{b}&c.&\begin{aligned}\sqrt{18a^{3}}&=\sqrt{9.2.a^{2}.a}\\ &=3a\sqrt{2a} \end{aligned}\\\hline d.&\begin{aligned}\sqrt{49a^{4}b}&=\sqrt{7^{2}.\left ( a^{2} \right )^{2}b}\\ &=7a^{2}\sqrt{b} \end{aligned}&e.&\begin{aligned}\sqrt{72c^{6}d^{10}}&=\sqrt{36.2.\left ( c^{3} \right )^{2}\left ( d^{5} \right )^{2}}\\ &=6c^{3}d^{5}\sqrt{2} \end{aligned}&f.&\begin{aligned}\displaystyle &\sqrt{28p^{8}q^{6}r^{4}}\\ &=\sqrt{4.7\left ( p^{4} \right )^{2}\left ( q^{3} \right )^{2}\left ( r^{2} \right )^{2}}\\ &=2p^{4}q^{3}r^{2}\sqrt{7}\end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 15.&\textrm{sederhankanlah bentuk akar berikut ini}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle 2\sqrt{3}+4\sqrt{3}&b.&8\sqrt{3}+4\sqrt{3}+\sqrt{3}&c.&4\sqrt{5}+3\sqrt{500}\\ &&&&&&&\\ d.&8\sqrt{5}-2\sqrt{5}&e.&6\sqrt{5}-3\sqrt{5}-\sqrt{5}&f.&\displaystyle 3\sqrt{45}-\sqrt{80}\\ &&&&&&&\\ g.&2\sqrt{3}+3\sqrt{4}+4\sqrt{5}&h.&\displaystyle \sqrt{3}-\frac{1}{2}\sqrt{3}-\frac{1}{3}\sqrt{3}&i.&3\sqrt{24}+2\sqrt{72}-\sqrt{108}\\&&&&&&&\\ j.&\sqrt{2}+\sqrt{8}+\sqrt{32}+\sqrt{128}&k.&\sqrt{8}-\sqrt{20}+3\sqrt{45}-\sqrt{72}&l.&\sqrt{52}-2\sqrt{13}+\sqrt{1024}+\sqrt{32}\end{array} \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.&\begin{aligned}2\sqrt{3}+4\sqrt{3}&=\left ( 2+4 \right )\sqrt{3}\\ &=6\sqrt{3} \end{aligned}&b.&\begin{aligned}8\sqrt{3}&+4\sqrt{3}+\sqrt{3}\\ &=\left ( 8+4+1 \right )\sqrt{3}\\ &=13\sqrt{3} \end{aligned}&c.&\begin{aligned}4\sqrt{5}+3\sqrt{500}&=4\sqrt{5}+3\sqrt{100\times 5}\\ &=4\sqrt{5}+3\left ( 10\sqrt{5} \right )\\ &=4\sqrt{5}+30\sqrt{5}\\ &=\left ( 4+30 \right )\sqrt{5}\\ &=34\sqrt{5} \end{aligned}&d.&\begin{aligned}8\sqrt{5}-2\sqrt{5}=6\sqrt{5} \end{aligned}\\\hline k.&\multicolumn{3}{|l|}{\begin{aligned}\sqrt{8}-\sqrt{20}+&3\sqrt{45}-\sqrt{72}\\ &=\sqrt{4.2}-\sqrt{4.5}+3\sqrt{9.5}-\sqrt{36.2}\\ &=2\sqrt{2}-2\sqrt{5}+3.3\sqrt{5}-6\sqrt{2}\\ &=\left (2-6 \right )\sqrt{2}+\left ( 9-2 \right )\sqrt{5}\\ &=-4\sqrt{2}+7\sqrt{5} \end{aligned}}&l.&\multicolumn{3}{|l|}{\begin{aligned}\sqrt{52}-2\sqrt{13}+&\sqrt{1024}+\sqrt{32}\\ &=\sqrt{4.13}-2\sqrt{13}+\sqrt{32\times 32}+\sqrt{16.2}\\ &=2\sqrt{13}-2\sqrt{13}+32-4\sqrt{2}\\ &=0+32-4\sqrt{2}\\ &=32-4\sqrt{2} \end{aligned}}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 16.&\textrm{sederhankanlah bentuk akar berikut ini}\\ &\begin{array}{lllllllll}\\ a.&\displaystyle \sqrt{3}\left ( 7+\sqrt{3} \right )&b.&4\sqrt{2}\left ( 3\sqrt{14}-\sqrt{18} \right )&c.&\left ( 5+2\sqrt{6} \right )\left ( 5+2\sqrt{6} \right )\\ &&&&&&&\\ d.&\left ( 7-\sqrt{11} \right )\left ( 7-\sqrt{11} \right )&e.&\left ( 5\sqrt{3}+2\sqrt{7} \right )^{2}&f.&\left ( 5\sqrt{2}-6 \right )^{2} \end{array}\\\\ &\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.&\begin{aligned}&\sqrt{3}\left ( \sqrt{7}+\sqrt{3} \right )\\ &=\sqrt{3}\sqrt{7}+\sqrt{3}.\sqrt{3}\\ &=\sqrt{3.7}+\sqrt{3^{2}}\\ &=\sqrt{21}+3 \end{aligned}&b.&\begin{aligned}&4\sqrt{2}\left ( 3\sqrt{14}-\sqrt{18} \right )\\ &=4.3.\sqrt{2.14}-4\sqrt{2.18}\\ &=12\sqrt{2.2.7}-4\sqrt{2.2.9}\\ &=12.2\sqrt{7}-4.2.3\\ &=24\sqrt{7}-24\end{aligned}&c.&\begin{aligned}&\left ( 5+2\sqrt{6} \right )\left ( 5+2\sqrt{6} \right )\\ &=\left ( 5+2\sqrt{6} \right )^{2}\\ &\textrm{ingat bentuk}\\ &\left ( a+b \right )^{2}=a^{2}+2ab+b^{2}\\ &=5^{2}+2\left ( 5.2\sqrt{6} \right )+\left ( 2\sqrt{6} \right )^{2}\\ &=25+20\sqrt{6}+4.6\\ &=25+24+20\sqrt{6}\\ &=49+20\sqrt{6} \end{aligned}&d.&\begin{aligned}&\left ( 7-\sqrt{11} \right )\left ( 7-\sqrt{11} \right )\\ &=.............\\ &=.............\\ &=.............\\ &=.............\\ &=.............\\ &=....... \end{aligned}\\\hline \end{array} \end{array}.

\begin{array}{ll}\\ 17.&\textrm{sederhankanlah bentuk akar berikut ini}\\ &\begin{array}{lllllllll}\\ a.&\sqrt{8+2\sqrt{15}}&b.&\sqrt{3-2\sqrt{2}}&c.&\sqrt{9+6\sqrt{2}}\\ &&&&&&&\\ d.&\sqrt{9-4\sqrt{5}}&e.&\sqrt{7+\sqrt{40}}&f.&\sqrt{\displaystyle \frac{5}{2}+\sqrt{6}}\\ \end{array}\\ \end{array}.

.\quad\: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a.&\begin{aligned}\sqrt{8+2\sqrt{15}}&=\sqrt{5+3+2\sqrt{5.3}}\\ &=\sqrt{\left ( 5 \right )^{2}+\left ( 3 \right )^{2}+2\sqrt{5.3}}\\ &=\sqrt{\left ( \sqrt{5}+\sqrt{3} \right )^{2}}\\ &=\sqrt{5}+\sqrt{3} \end{aligned}&b.&\begin{aligned}\sqrt{3-2\sqrt{2}}&=\sqrt{2+1-2\sqrt{2.1}}\\ &=\sqrt{\left ( 2 \right )^{2}+1^{2}-2\sqrt{2.1}}\\ &=\sqrt{\left ( \sqrt{2}-1 \right )^{2}}\\ &=\sqrt{2}-1 \end{aligned}&c.&\begin{aligned}\sqrt{9+6\sqrt{2}}&=\sqrt{9+2.3\sqrt{2}}\\ &=\sqrt{9+2\sqrt{9}.\sqrt{2}}\\ &=\sqrt{9+2\sqrt{18}}\\ &=\sqrt{6+3+2\sqrt{6.3}}\\ &=\sqrt{\left ( \sqrt{6}+\sqrt{3} \right )^{2}}\\ &=\sqrt{6}+\sqrt{3} \end{aligned}\\\hline d.&\begin{aligned}\sqrt{9-4\sqrt{5}}&=\sqrt{9-2.2\sqrt{5}}\\ &=\sqrt{9-2\sqrt{4}.\sqrt{5}}\\ &=\sqrt{5+4-2\sqrt{5.4}}\\ &=\sqrt{\left ( \sqrt{5}+\sqrt{4} \right )^{2}}\\ &=\sqrt{5}+\sqrt{4}\\ &=\sqrt{5}+2 \end{aligned}&e.&\begin{aligned}\sqrt{7+\sqrt{40}}&=\sqrt{7+\sqrt{4.10}}\\ &=\sqrt{7+2\sqrt{10}}\\ &=\sqrt{5+2+2\sqrt{5.2}}\\ &=\sqrt{\left ( \sqrt{5}+\sqrt{2} \right )^{2}}\\ &=\sqrt{5}+\sqrt{2}\end{aligned}&f.&\begin{aligned}\sqrt{\displaystyle \frac{5}{2}+\sqrt{6}}&=\sqrt{\displaystyle \frac{(3+2)}{2}+2.\frac{1}{2}\sqrt{3.2}}\\ &=\sqrt{\displaystyle \frac{(3+2)}{2}+2\sqrt{\displaystyle \frac{3.2}{4}}}\\ &=\sqrt{\displaystyle \frac{3}{2}+\frac{2}{2}+2\sqrt{\displaystyle \frac{3}{2}.\frac{2}{2}}}\\ &=\sqrt{\left ( \displaystyle \sqrt{\frac{3}{2}}+1 \right )^{2}}\\ &=\sqrt{\displaystyle \frac{3}{2}}+1\\ &=\sqrt{\displaystyle \frac{2.3}{2.2}}+1\\ &=\displaystyle \frac{1}{2}\sqrt{6}+1 \end{aligned} \\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 18.&\textrm{Tentukanlah nilai x yang memenuhi persamaan berikut} \end{array}\\ \begin{array}{lllllll}\\ .\quad\quad&a.&\displaystyle 2^{x}=8&j.&\displaystyle 5^{x^{2}-8x+12}=1&s.&\displaystyle \sqrt{8^{2x-1}}-4\sqrt[3]{2^{1-x}}=0\\ &b.&\displaystyle 2^{x+1}=32&k.&\displaystyle 3^{x^{2}+3x}=3^{x+8}&t.&\displaystyle 3^{5-x}=\frac{1}{27}\sqrt[3]{3^{x}}=0\\ &c.&\displaystyle 2^{3x-4}=32&l.&\displaystyle 25^{x^{2}-5x+7}=125^{x-2}&u.&\displaystyle \sqrt[3]{8^{x+2}}=\left ( \frac{1}{32} \right )^{2-x}\\ &d.&\displaystyle 3^{2x-1}=\frac{1}{27}&m.&\displaystyle 2^{x^{2}+x}=4^{x+1}&v.&\displaystyle 4^{1-2x}=\frac{16^{2-x}}{32^{1-x}}\\ &e.&\displaystyle 3^{5x+2}=9^{x+4}&n.&\displaystyle 4^{x+3}=\sqrt[3]{8^{x+5}}&w.&\displaystyle \left ( \frac{1}{2} \right )^{2x+1}=\sqrt{\frac{2^{4x-1}}{128}}\\ &f.&\displaystyle 5^{x-9}=25^{3-x}&o.&\displaystyle 9^{3x+2}=\frac{1}{81^{2x-5}}&x.&\displaystyle \left ( \frac{2}{3} \right )^{2x-3}=\left ( \frac{27}{8} \right )^{3-2x}\\ &g.&\displaystyle 4^{2x-1}=1&p.&\displaystyle \sqrt{2^{x-5}}=2\sqrt{2}&y.&\displaystyle x^{x^{x^{x^{...}}}}=2\\ &h.&\displaystyle 5^{x-1}=\sqrt{5}&q.&\displaystyle \left ( \frac{1}{4} \right )^{x-1}=\sqrt[3]{2^{3x+1}}&z.&\displaystyle 2^{x+5}+2^{5-x}=64\\ &i.&\displaystyle 7^{4x}=\frac{7}{\sqrt[4]{7}}&r.&\displaystyle \frac{2^{x}}{8^{x+2}}=64.4^{x}& \end{array}.

.\quad \: \, \begin{aligned}&\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|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}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline a&\begin{aligned}\displaystyle 2^{x}&=8\\ \displaystyle 2^{x}&=2^{3}\\ x&=3 \end{aligned}&b&\begin{aligned}\displaystyle 2^{x+1}&=32\\ \displaystyle 2^{x+1}&=2^{5}\\ x+1&=5\\ x&=4 \end{aligned}&c&\begin{aligned}\displaystyle 2^{3x-4}&=32\\ \displaystyle 2^{3x-4}&=2^{5}\\ 3x-4&=5\\ 3x&=9\\ x&=3 \end{aligned}&d&\begin{aligned}3^{2x-1}&=\displaystyle \frac{1}{27}\\ \displaystyle 3^{2x-1}&=\displaystyle \frac{1}{3^{3}}\\ \displaystyle 3^{2x-1}&=3^{-3}\\ 2x-1&=-3\\ 2x&=-2\\ x&=-1 \end{aligned}&e&\begin{aligned}\displaystyle 3^{5x+2}&=9^{x+4}\\ \displaystyle 3^{5x+2}&=3^{2\left ( x+4 \right )}\\ 5x+2&=2x+8\\ 5x-2x&=8-2\\ 3x&=6\\ x&=2 \end{aligned}&f&\begin{aligned}\displaystyle 5^{x-9}&=25^{3-x}\\ \displaystyle 5^{x-9}&=5^{2\left ( 3-x \right )}\\ x-9&=6-2x\\ x+2x&=6+9\\ 3x&=15\\ x&=5 \end{aligned}\\\hline \end{array} \end{aligned}.

.\quad \: \, \begin{aligned} &\begin{array}{|c|l|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}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline g&\begin{aligned}\displaystyle 4^{2x-1}&=1\\ \displaystyle 4^{2x-1}&=4^{0}\\ 2x-1&=0\\ 2x&=1\\ x&=\displaystyle \frac{1}{2} \end{aligned}&h&\begin{aligned}\displaystyle 5^{x-1}&=\sqrt{5}\\ \displaystyle 5^{x-1}&=5^{\frac{1}{2}}\\ x-1&=\displaystyle \frac{1}{2}\\ x&=\displaystyle 1\frac{1}{2}\\ x&=\displaystyle \frac{3}{2} \end{aligned}&i&\begin{aligned}\displaystyle 7^{4x}&=\displaystyle \frac{7}{\sqrt[4]{7}}\\ \displaystyle 7^{4x}&=\displaystyle \frac{7^{1}}{7^{\frac{1}{4}}}\\ \displaystyle 7^{4x}&=7^{\left (1-\frac{1}{4} \right )}\\ 4x&=\displaystyle \frac{3}{4}\\ x&=\displaystyle \frac{3}{16}\end{aligned}&j&\begin{aligned}\displaystyle 5^{x^{2}-8x+12}&=1\\ \displaystyle 5^{x^{2}-8x+12}&=5^{0}\\ \displaystyle x^{2}-8x+12&=0\\ \left ( x-2 \right )\left ( x-6 \right )&=0\\ x=2\quad \textrm{atau}\quad x=6& \end{aligned}&k&\begin{aligned}\displaystyle 3^{x^{2}+3x}&=3^{x+8}\\ x^{2}+3x&=x+8\\ x^{2}+3x-x-8&=0\\ x^{2}+2x-8&=0\\ \left ( x+4 \right )\left ( x-2 \right )&=0\\ x=-4\quad \textrm{atau}\quad x=2& \end{aligned}\\\hline \end{array} \end{aligned}.

.\quad \: \, \begin{aligned} &\begin{array}{|c|l|c|l|c|l|}\hline \multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}&\multicolumn{2}{|c|}{\textrm{Pembahasan}}\\\hline l&\begin{aligned}\displaystyle 25^{x^{2}-5x+7}&=125^{x-2}\\ \displaystyle \left ( 5^{2} \right )^{x^{2}-5x+7}&=\left ( 5^{3} \right )^{x-2}\\ 2x^{2}-10x+14&=3x-6\\ 2x^{2}-10x-3x+14+6&=0\\ 2x^{2}-13x+20&=0\\ \left ( 2x-5 \right )\left ( x-4 \right )&=0\\ x=\displaystyle \frac{5}{2}\quad \textrm{atau}\quad x=4&\end{aligned}&m&\begin{aligned}\displaystyle 2^{x^{2}+x}&=4^{x+1}\\ 2^{x^{2}+x}&=\left ( 2^{2} \right )^{x+1}\\ x^{2}+x&=2x+2\\ x^{2}+x-2x-2&=0\\ x^{2}-x-2&=0\\ \left ( x-2 \right )\left ( x+1 \right )&=0\\ x=2\quad \textrm{atau}\quad x=-1&\end{aligned}&n&\begin{aligned}\displaystyle 4^{x+3}&=\sqrt[3]{8^{x+5}}\\ \displaystyle \left ( 2^{2} \right )^{x+3}&=\left ( 2^{3} \right )^{\frac{x+5}{3}}\\ 2x+6&=x+5\\ x&=-1\end{aligned}\\\hline \end{array} \end{aligned}.

.\quad \: \, \begin{aligned} &\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 w&\begin{aligned}\displaystyle \left ( \frac{1}{2} \right )^{2x+1}&=\sqrt{\frac{2^{4x-1}}{128}}\\ \left ( 2^{-1} \right )^{2x+1}&=\left ( 2^{4x-1-7} \right )^{\frac{1}{2}}\\ -2x-1&=2x-4\\ -4x&=-3\\ x&=\displaystyle \frac{3}{4}\end{aligned}&x&\begin{aligned}\displaystyle \left ( \frac{2}{3} \right )^{2x-3}&=\left ( \frac{27}{8} \right )^{3-2x}\\ \displaystyle \left ( \frac{2}{3} \right )^{2x-3}&=\left ( \left ( \frac{2}{3} \right )^{-3} \right )^{3-2x}\\ 2x-3&=-9+6x\\ -4x&=-6\\ x&=\displaystyle \frac{3}{2} \end{aligned}&y&\begin{aligned}\displaystyle x^{x^{x^{x^{x^{...}}}}} &=2\\ \displaystyle x^{\underset{2}{\underbrace{x^{x^{x^{...}}}}}}&=2\\ \displaystyle x^{2}&=2\\ x&=\sqrt{2}\end{aligned}&z&\begin{aligned}\displaystyle 2^{x+5}+2^{5-x}&=64\\ 2^{x}.2^{5}+2^{5}.2^{-x}&=64\\ 32.2^{x}+\displaystyle \frac{32}{2^{x}}&=64\\ 2^{x}+\displaystyle \frac{1}{2^{x}}&=2\\ \left ( 2^{x} \right )^{2}+1&=2\left ( 2^{x} \right )\\ \left ( 2^{x} \right )^{2}-2\left ( 2^{x} \right )+1&=0\\ \left ( 2^{x}-1 \right )\left ( 2^{x}-1 \right )&=0\\ \left ( 2^{x}-1 \right )^{2}&=0\\ 2^{x}-1&=0\\ 2^{x}&=1\\ 2^{x}&=2^{0}\\ x&=0\end{aligned}\\\hline \end{array} \end{aligned}.

\begin{array}{ll}\\ 19&\textrm{Carilah nilai x dan y yang memenuhi persamaan berikut}\\ &a.\quad \begin{cases} 3^{x-2y} & =\displaystyle \frac{1}{81} \\ 2^{x-y} & = 16 \end{cases}\\ &b.\quad \begin{cases} 5^{x-2y+1} & =25^{x-2y} \\ 4^{x-y+2} & =32^{x-2y+1} \end{cases}\\ &c.\quad \begin{cases} 4^{x-2y+1} & =8^{2x-y} \\ 3^{x+y+1} &= 9^{2x-y-4} \end{cases} \end{array}.

.\quad \: \, \begin{aligned}&\textrm{Jawab}:\\ &\textrm{a. Diketahui bahwa}\\ & \quad \begin{cases} 3^{x-2y} \\ & =\displaystyle \frac{1}{81} \\ 2^{x-y} & = 16 \end{cases}\\ &\begin{array}{l|l|l}\\ \begin{aligned}(\ast )\qquad 3^{x-2y}&=\displaystyle \frac{1}{81}\\ 3^{x-2y}&=3^{-4}\\ x-2y&=-4\\ (\ast \ast )\qquad 2^{x-y}&=16\\ 2^{x-y}&=2^{4}\\ x-y&=4 \end{aligned}&\qquad \begin{matrix} x-2y=-4\: \: ..............(\ast )\\ x-y=4\: \: .................(\ast \ast ) \end{matrix}&\textrm{dengan eliminasi kita akan mendapatkan}\begin{cases} & x=12 \\ & y=8 \end{cases}\\ && \end{array} \end{aligned}.

\begin{array}{lll}\\ 20.&\textrm{Sederhanakanlah}\\ &&\\ &a.\quad \displaystyle \frac{3^{m+n}\times 3^{n+1}\times 3^{n-m}}{3^{2n+1}\times 3^{n+1}}&c.\quad \displaystyle \frac{\left ( 2^{m+2} \right )^{2}-2^{2}.2^{2m}}{2^{m}.2^{m+2}}\\ &&\\ &b.\quad \displaystyle \frac{3^{n+1}-3^{n}}{3^{n}+3^{n-1}}\\\\ &\textrm{Jawab}:\\\\ &\begin{aligned}a.\quad \displaystyle \frac{3^{m+n}\times 3^{n+1}\times 3^{n-m}}{3^{2n+1}\times 3^{n+1}}&=\displaystyle \frac{3^{m}.3^{n}\times 3^{n}.3\times 3^{n}.3^{-m}}{3^{2n}.3\times 3^{n}.3}\\ &=\displaystyle \frac{3^{m+n+n+1+n-m}}{3^{2n+1+n+1}}\\ &=\displaystyle \frac{3^{3n+1}}{3^{3n+2}}=\frac{3^{3n}.3}{3^{3n}.3^{2}}\\ &=\displaystyle \frac{3}{9}\\ &=\displaystyle \frac{1}{3}\end{aligned} \\ &\begin{aligned}b.\quad \displaystyle \frac{3^{n+1}-3^{n}}{3^{n}+3^{n-1}}&=\displaystyle \frac{3^{n}.3-3^{n}}{3^{n}+3^{n}.3^{-1}}\\ &=\displaystyle \frac{3^{n}\left ( 3-1 \right )}{3^{n}\left ( 1-3^{-1} \right )}=\displaystyle \frac{3^{n}\left ( 3-1 \right )}{3^{n}\left ( 1-\frac{1}{3} \right )}\\ &=\displaystyle \frac{2}{\frac{2}{3}}\\ &=3\end{aligned} \end{array}.

Sumber Referensi

  1. Kurnianingsih, Sri, Kuntarti & Sulistiyono. 2007. Matematika SMA dan MA Untuk Kelas X Semester 1. Jakarta: ESIS.
  2. Noormandiri, Endar Sucipto. 2000. Buku Pelajaran Matematika SMU untuk Jilid 1 Kelas 1 Catur Wulan 1, 2, dan 3. Jakarta: Erlangga.
  3. Sunardi, Slamet Waluyo, Sutrisno & Subagya. 2004. Matematika 1A untuk SMA Kelas 1. Jakarta: Bumi Aksara.

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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