$\begin{array}{ll}\\ 41.&\textrm{Penyelesaian pertidaksamaan eksponen}\\ &\left ( \displaystyle \frac{1}{3} \right )^{2x+1}>\sqrt{\displaystyle \frac{27}{3^{x-1}}} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x>\displaystyle \frac{6}{5}\\ \textrm{b}.&x<-\displaystyle \frac{6}{5}\\ \textrm{c}.&x>\displaystyle \frac{5}{6}\\ \color{red}\textrm{d}.&x<-2\\ \textrm{e}.&x<2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}\left ( \displaystyle \frac{1}{3} \right )^{2x+1}&>\sqrt{\displaystyle \frac{27}{3^{x-1}}}\\ 3^{-(2x+1)}&>3^{\frac{1}{2}(3-(x-1))}\\ -(2x+1)&>\displaystyle \frac{1}{2}(3-(x-1))\\ -4x-2&>4-x\\ -4x+x&>4+2\\ -3x&>6\\ 3x&<-6\\ x&\color{red}<-2 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 42.&(\textbf{UMPTN 01})\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &4^{x^{2}-x-2}.2^{x^{2}+3x-10}<\displaystyle \frac{1}{16} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x<-5\: \: \textrm{atau}\: \: x>-2\\ \textrm{b}.&x<-2\: \: \textrm{atau}\: \: x>\displaystyle \frac{5}{3}\\ \textrm{c}.&-2<x<-1\\ \color{red}\textrm{d}.&-2<x<\displaystyle \frac{5}{3}\\ \textrm{e}.&-5<x<2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}4^{x^{2}-x-2}.2^{x^{2}+3x-10}&<\displaystyle \frac{1}{16}\\ 2^{2\left ( x^{2}-x-2 \right )+\left ( x^{2}+3x-10 \right )}&<2^{-4}\\ 2\left ( x^{2}-x-2 \right )+\left ( x^{2}+3x-10 \right )&<-4\\ 3x^{2}-2x+3x-4-10+4&<0\\ 3x^{2}+x-10+&<0\\ (x+2)(3x-5)&<0\\ \therefore \qquad-2<x&<\displaystyle \frac{5}{3} \end{aligned} \end{array}$
$\begin{array}{l}\\ 43.&(\textbf{SPMB 04 Mat IPA})\textrm{Himpunan Penyelesaian}\\ & \textrm{pertidaksamaan eksponen}\\ &2\sqrt{4^{x^{2}-3x+2}}<\sqrt[3]{\left (\displaystyle \frac{1}{2} \right )^{3-6x}} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&\left \{ x|x> 4 \right \}\\ \textrm{b}.&\left \{ x|x> 2 \right \}\\ \textrm{c}.&\left \{ x|x<1 \right \}\\ \color{red}\textrm{d}.&\left \{ x|1<x<4 \right \}\\ \textrm{e}.&\left \{ x|2\leq x\leq 3 \right \} \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}2\sqrt{4^{x^{2}-3x+2}}&<\sqrt[3]{\left (\displaystyle \frac{1}{2} \right )^{3-6x}}\\ 2^{1+\frac{2}{2}\left ( x^{2}-3x+2 \right )}&<2^{- \frac{3-6x}{3}}\\ 1+\left ( x^{2}-3x+2 \right )&<-\displaystyle \frac{3-6x}{3}\\ x^{2}-3x+3&<-1+2x\\ x^{2}-5x+4&<0\\ (x-1)(x-4)&<0\\ 1<x&<4 \end{aligned} \end{array}$
$\begin{array}{l}\\ 44.&(\textbf{SBMPTN 2015 Mat IPA})\\ &\textrm{Nilai}\: \: c\: \: \textrm{yang memenuhi}\\ &(0,12)^{4x^{2}+8x+c}<(0,0144)^{x^{2}+4x+4} \: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&c>0\\ \textrm{b}.&c>2\\ \textrm{c}.&c>4\\ \textrm{d}.&c>6\\ \color{red}\textrm{e}.&c>8 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}(0,12)^{4x^{2}+8x+c}&<(0,0144)^{x^{2}+4x+4}\\ (0,12)^{4x^{2}+8x+c}&<(0,12)^{2\left (x^{2}+4x+4 \right )}\\ 4x^{2}+8x+c&>2\left (x^{2}+4x+4 \right )\\ 2x^{2}+c-8&>0\quad \color{magenta}\textrm{haruslah definit positif}\\ \textrm{Syaratnya}&\begin{cases} a &=2>0 \\ D &=b^{2}-4ac<0\\ \end{cases}\\ \textrm{Maka}\quad D&=b^{2}-4ac<0\\ \textbf{ambil dari}&\: \: \: 2x^{2}-c-8=0\begin{cases} a &=2 \\ b &=0 \\ c &=c-8 \end{cases}\\ &=0^{2}-4.2(c-8)<0\\ -8c&+64<0\\ -8c&<-64\\ 8c&>64\\ c&\color{red}>8 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 45.&\textrm{Nilai}\: \: x\: \: \textrm{yang memenuhi}\\ &9^{2x}-10.9^{x}+9>0, \: \: x\in \mathbb{R}\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \textrm{a}.&x<1\: \: \textrm{atau}\: \: x>0\\ \color{red}\textrm{b}.&x<0\: \: \textrm{atau}\: \: x>1\\ \textrm{c}.&x<-1\: \: \textrm{atau}\: \: x>2\\ \textrm{d}.&x<1\: \: \textrm{atau}\: \: x>2\\ \textrm{e}.&x<-1\: \: \textrm{atau}\: \: x>1 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}9^{2x}-10.9^{x}+9&>0\\ \left (9^{x} \right )^{2}-10.\left ( 9^{x} \right )+9&>0\\ \left ( 9^{x}-1 \right )\left ( 9^{x}-9 \right )&>0\\ 9^{x}<1\: \: \textrm{atau}\: \: 9^{x}&>9\\ 9^{x}<9^{0}\: \: \textrm{atau}\: \: 9^{x}&>9^{1}\\ x<0\: \: \textrm{atau}\: \: x&\color{red}>1 \end{aligned} \end{array}$
$\begin{array}{l}\\ 46.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \color{red}\textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=\color{red}1 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 47.&\textrm{Jumlah akar-akar persamaan}\\ &2020^{x^{2}-7x+7}=2021^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-7\\ \textrm{b}.&-5\\ \textrm{c}.&-3\\ \textrm{d}.&5\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}2020^{x^{2}-7x+7}&=2021^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=\color{red}7 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 48.&\textrm{Nilai dari}\: \: \displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \color{red}\textrm{b}.&5\\ \textrm{c}.&10\\ \textrm{d}.&20\\ \textrm{e}.&40 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\begin{aligned}\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}&=\displaystyle \frac{2^{4}.2^{2016}+2^{2}.2^{2016}}{2^{2}.2^{2018}+2^{2016}}\\ &=\displaystyle \frac{2^{2016}\left ( 2^{4}+2^{2} \right )}{2^{2016}\left ( 2^{2}+1 \right )}\\ &=\displaystyle \frac{16+4}{4+1}\\ &=\displaystyle \frac{20}{5}\\ &=\color{red}4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 49.&(\textbf{UM IPB})\textrm{Jika}\: \: ab=a^{b} \: \: \textrm{dan}\: \: \displaystyle \frac{a}{b}=a^{3b}\\ &\textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&0,5\\ \textrm{c}.&1\\ \textrm{d}.&0,25\\ \textrm{e}.&0,75 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}\textrm{Diketahui}&\\ ab&=a^{b}\\ b&=\displaystyle \frac{a^{b}}{a}=a^{b-1}.....\textbf{1}\\ \textrm{maka}&\\ \displaystyle \frac{a}{b}&=a^{3b}...............\textbf{2}\\ \textbf{1}&\: \: ke\: \: \textbf{2}\\ \displaystyle \frac{a}{a^{b-1}}&=a^{3b}\\ a^{2-b}&=a^{3b}\\ 2-b&=3b\\ -4b&=-2\\ b&=\displaystyle \frac{1}{2}................\textbf{3}\\ \textbf{3}&\: \: ke\: \: \textbf{1}\\ a\left ( \displaystyle \frac{1}{2} \right )&=a^{\frac{1}{2}}\\ \displaystyle \frac{1}{4}a^{2}&=a\\ a^{2}-4a&=0\\ a(a-4)&=0\\ a=0\: \: &\textrm{atau}\: \: \color{red}a=4 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 50.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 1,1\\ \textrm{b}.&\displaystyle 1,2\\ \color{red}\textrm{c}.&1,3\\ \textrm{d}.&\displaystyle 1,4\\ \textrm{e}.&1,5\\\\ &&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\color{red}\approx 1,3 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 51.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}$
Pembahasan:
$\begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=\color{red}47 \end{aligned}$
$\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\\ &\textrm{persamaan kuadrat dalam x,}\\ & \textbf{gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=\color{red}2 \end{cases} \end{aligned}$
DAFTAR PUSTAKA
- Enung, S., Untung, W. 2009. Mandiri Matematika SMA Jilid I untuk Kelas X. Jakarta: ERLANGGA.
- Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
- Kanginan, M., Terzalgi, Y. 2013. Matematika untuk SMA-MA/SMK Kelas X Wajib. Bandung: SEWU.
- Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO.
