Contoh Soal 10 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 46.&\textrm{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-2&&&\\ \textrm{b}.&-1\\ \textrm{c}.&0\\ \color{red}\textrm{d}.&1\\ \textrm{e}.&2 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{blue}\begin{aligned}5^{x+1}+5^{2-x}-30&=0\\ \left (5^{x} \right ).5^{1}+\displaystyle \frac{5^{2}}{5^{x}}-30&=0\\ 5\left ( 5^{x} \right )^{2}+25-30\left ( 5^{x} \right )&=0\\ \textrm{Persamaan kuadrat}&\: \textrm{dalam}\: \: 5^{x},\: \textrm{maka}\\ 5(5^{x})^{2}-30(5^{x})+25&=0\begin{cases} a & =5 \\ b & =-30 \\ c & =25 \end{cases}\\ (5^{x_{1}}).\left ( 5^{x_{2}} \right )&=\displaystyle \frac{c}{a}\\ 5^{x_{1}+x_{2}}&=\displaystyle \frac{25}{5}=5\\ 5^{x_{1}+x_{2}}&=5^{1}\\ x_{1}+x_{2}&=1 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 47.&\textrm{Jumlah akar-akar persamaan}\\ &2020^{x^{2}-7x+7}=2021^{x^{2}-7x+7}\: \: \textrm{adalah}\: ....\\\\ &\begin{array}{lllllllll}\\ \textrm{a}.&-7\\ \textrm{b}.&-5\\ \textrm{c}.&-3\\ \textrm{d}.&5\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\color{blue}\begin{aligned}2020^{x^{2}-7x+7}&=2021^{x^{2}-7x+7}\\ \textrm{Karena basis}&\: \textrm{tidak sama},\\ \textrm{maka harusl}&\textrm{ah pangkatnya}=0,\\ x^{2}-7x+7&=0\\ \textrm{dan jumlah}\: &\textrm{akar-akarnya adalah}:\\ x_{1}+x_{2}&=-\displaystyle \frac{b}{a}, \: \: \textrm{dari persamaan}\\ x^{2}-7x+7&=0\begin{cases} a &=1 \\ b &=-7 \\ c &=7 \end{cases}\\ \textrm{maka}\: \: x_{1}+x_{2}&=-\displaystyle \frac{b}{a}=-\frac{-7}{1}=7 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 48.&\textrm{Nilai dari}\: \: \displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}} \: \: \textrm{adalah... .}\\ &\begin{array}{llll}\\ \textrm{a}.&2\\ \color{red}\textrm{b}.&5\\ \textrm{c}.&10\\ \textrm{d}.&20\\ \textrm{e}.&40 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{b}\\ &\color{blue}\begin{aligned}\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}&=\displaystyle \frac{2^{4}.2^{2016}+2^{2}.2^{2016}}{2^{2}.2^{2018}+2^{2016}}\\ &=\displaystyle \frac{2^{2016}\left ( 2^{4}+2^{2} \right )}{2^{2016}\left ( 2^{2}+1 \right )}\\ &=\displaystyle \frac{16+4}{4+1}\\ &=\displaystyle \frac{20}{5}\\ &=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 49.&(\textbf{UM IPB})\textrm{Jika}\: \: ab=a^{b} \: \: \textrm{dan}\: \: \displaystyle \frac{a}{b}=a^{3b}\\ &\textrm{maka nilai}\: \: a\: \: \textrm{adalah}....\\ &\begin{array}{llll}\\ \color{red}\textrm{a}.&0\\ \textrm{b}.&0,5\\ \textrm{c}.&1\\ \textrm{d}.&0,25\\ \textrm{e}.&0,75 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{a}\\ &\color{blue}\begin{aligned}\textrm{Diketahui}&\\ ab&=a^{b}\\ b&=\displaystyle \frac{a^{b}}{a}=a^{b-1}.....\textbf{1}\\ \textrm{maka}&\\ \displaystyle \frac{a}{b}&=a^{3b}...............\textbf{2}\\ \textbf{1}&\: \: ke\: \: \textbf{2}\\ \displaystyle \frac{a}{a^{b-1}}&=a^{3b}\\ a^{2-b}&=a^{3b}\\ 2-b&=3b\\ -4b&=-2\\ b&=\displaystyle \frac{1}{2}................\textbf{3}\\ \textbf{3}&\: \: ke\: \: \textbf{1}\\ a\left ( \displaystyle \frac{1}{2} \right )&=a^{\frac{1}{2}}\\ \displaystyle \frac{1}{4}a^{2}&=a\\ a^{2}-4a&=0\\ a(a-4)&=0\\ a=0\: \: &\textrm{atau}\: \: a=4 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 50.&\textrm{Jika}\: \: \displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}=4\: ,\: \textrm{maka}\: \: x=\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&\displaystyle 1,1\\ \textrm{b}.&\displaystyle 1,2\\ \color{red}\textrm{c}.&1,3\\ \textrm{d}.&\displaystyle 1,4\\ \textrm{e}.&1,5\\\\ &&&(\textbf{SAT Test Math Level 2})\\ \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{c}\\ &\color{blue}\begin{aligned}\displaystyle 3.x^{^{^{^{ \frac{3}{2}}}}}&=4\\ \left (3.x^{^{^{^{ \frac{3}{2}}}}} \right )^{2}&=4^{2}\\ 3^{2}.x^{3}&=4^{2}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\\ x^{3}&=\displaystyle \frac{4^{2}}{3^{2}}\times \frac{3}{3}\\ x^{3}&\leq \displaystyle \frac{4^{2}}{3^{2}}\times \frac{4}{3}\\ x^{3}&\leq \left ( \displaystyle \frac{4^{3}}{3^{3}} \right )\\ x^{3}&\leq \left ( \displaystyle \frac{4}{3} \right )^{3}\\ x&\leq \displaystyle \frac{4}{3}\\ x&\leq 1,\overline{333}\\ x&\approx 1,3 \end{aligned} \end{array}$

$\begin{array}{ll}\\ 51.&\textrm{Hitunglah}\:\\\\ &\quad\quad\qquad \displaystyle \sqrt[8]{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{2207-\displaystyle \frac{1}{...}}}}}\\\\ &\textrm{nyatakan jawabannya dalam bentuk }\: \displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textrm{dengan a, b, c, dan d bilangan-bilangan bulat}\\ \end{array}$

Pembahasan:

$\color{blue}\begin{aligned}x^{8}&=2207-\displaystyle \underset{x^{8}}{\underbrace{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}\\ x^{8}&=2207-\displaystyle \frac{1}{x^{8}}\\ x^{8}+\displaystyle \frac{1}{x^{8}}&=2207\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )^{2}&=2207+2\\ \left ( x^{4}+\displaystyle \frac{1}{x^{4}} \right )&=\sqrt{2209}=47 \end{aligned}$

$\color{blue}\begin{aligned}x^{4}+\displaystyle \frac{1}{x^{4}}&=47\\ \left ( x^{2}+\displaystyle \frac{1}{x^{2}} \right )^{2}&=47+2\\ x^{2}+\displaystyle \frac{1}{x^{2}}&=\sqrt{49}=7\\ \left ( x+\displaystyle \frac{1}{x} \right )^{2}&=7+2\\ x+\displaystyle \frac{1}{x}&=\sqrt{9}=3\\ x^{2}-3x+1&=0,\\ &\textrm{persamaan kuadrat dalam x,}\\ & \textbf{gunakan rumus abc}\\ x_{1,2}=&\displaystyle \frac{3\pm \sqrt{5}}{2}=\displaystyle \frac{3\pm 1\sqrt{5}}{2}=\displaystyle \frac{a\pm b\sqrt{c}}{d}\\ &\textbf{Sehingga},\quad \begin{cases} & a=3 \\ & b=1 \\ & c=5 \\ & d=2 \end{cases} \end{aligned}$

DAFTAR PUSTAKA

1. Enung, S., Untung, W. 2009. Mandiri Matematika SMA Jilid I untuk Kelas X. Jakarta: ERLANGGA.
2. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA
3. Kanginan, M., Terzalgi, Y. 2013. Matematika untuk SMA-MA/SMK Kelas X Wajib. Bandung: SEWU.
4. Susianto, B. 2011. Olimpiade Matematika dengan Proses Berpikir Aljabar dan Bilangan. Jakarta: GRASINDO.