$\Large\textrm{A. Bilangan Pangkat Positif}$
Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $m$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat dedefinisikan dengan:
$\color{blue}\LARGE a^{m}=\underset{m}{\underbrace{a\times a\times \times a\times ...\times a}}$
$\color{purple}\begin{aligned}\textrm{Bilangan}&:\\ \color{blue}a&\: \: \textrm{disebut basis atau bilangan pokok}\\ \color{blue}n&\: \: \textrm{disebut sebagai bilangan pangkat/eksponen} \end{aligned}$.
$\LARGE\colorbox{yellow}{ CONTOH SOAL}$.
$\color{purple}(1).\quad 3^{4}=3\times 3\times 3\times 3=81$
$\color{purple}(2).\quad 5^{4}=5\times 5\times 5\times 5=625$
$\color{purple}(3).\quad 2^{6}=2\times 2\times 2\times 2\times 2\times 2=64$
$\color{purple}(4).\quad 6^{7}=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$
$\color{purple}(5).\quad (-3)^{3}=(-3)\times (-3)\times (-3)=-27$
$\color{purple}(6).\quad (-2)^{4}=(-2)\times (-2)\times (-2)\times (-2)=16$
$\color{purple}(7).\quad \left ( \displaystyle \frac{1}{5} \right )^{3}=\left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )\times \left ( \displaystyle \frac{1}{5} \right )= \displaystyle \frac{1}{125}$
$\color{purple}(8).\quad \left ( -\displaystyle \frac{1}{2} \right )^{3}=\left ( -\displaystyle \frac{1}{2} \right )\times \left (- \displaystyle \frac{1}{2} \right )\times \left ( -\displaystyle \frac{1}{2} \right )=- \displaystyle \frac{1}{8}$
$\Large\textrm{B. Sifat-Sifat Bilangan Pangkat Positif}$
$\begin{aligned}\\ 1.\quad&a^{m}.a^{n}=a^{m+n}\\ 2.\quad&a^{m}:a^{n}=a^{m-n}\\ 3.\quad&\left ( a^{m} \right )^{n}=a^{m.n},\: \: \textrm{syarat}\: \: a\neq 0\\ 4.\quad&\left ( ab \right )^{n}=a^{n}.b^{n}\\ 5.\quad&\left ( \frac{a}{b} \right )^{n}=\frac{a^{n}}{b^{n}},\: \: \textrm{syarat}\: \: b\neq 0 \end{aligned}$
Beberpa hal yang perlu diketahui juga, yaitu
$\color{blue}\begin{aligned}\\ 1.\quad&(a+b)^{2}=a^{2}+2ab+b^2\\ 2.\quad&(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}\\ 3.\quad&\left ( a+\frac{1}{a} \right )^{2}=a^{2}+2+\displaystyle \frac{1}{a^{2}},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$
$\LARGE\colorbox{yellow}{ CONTOH SOAL}$
$\color{blue}\begin{aligned}\\ (1).\quad&2^{6} \times 2^{4} \times 2^{7} = 2^{6+4+7}=2^{17}\\ (2).\quad&2^{5} \times 3^{5} \times 7^{5} = \left ( 2 . 3 . 7 \right )^{5}=\left ( 42 \right )^{5}\\ (3).\quad&\displaystyle \frac{a^{3}.a^{7}.a^{6}}{a^{9}}=\displaystyle \frac{a^{3+7+6}}{a^{9}}=\frac{a^{16}}{a^{9}}=a^{16-9}=a^{7},\: \: \textrm{syarat}\: \: a\neq 0\\ \end{aligned}$
$\begin{aligned}(4).\quad\displaystyle \frac{3^{7}.7^{3}.2}{\left ( 42 \right )^{3}}&=\frac{2^{1}.3^{7}.7^{3}}{\left ( 2.3.7 \right )^{3}}=\frac{2^{1}.3^{7}.7^{3}}{2^{3}.3^{3}.7^{3}}\\ &=2^{1-3}.3^{7-3}.7^{3-3}=2^{-2}.3^{4}.7^{0}\\ &=\frac{1}{2^{2}}.3^{4}.1=\frac{3^{4}}{2^{2}} \end{aligned}$
$\color{blue}\begin{aligned}(5).\quad\displaystyle \frac{2^{2020}+2^{2021}+2^{2022}}{7}&=\displaystyle \frac{1.2^{2020}+2^{1}.2^{2020}+2^{2}.2^{2020}}{7}\\ &=\frac{\left ( 1+2+4 \right ).2^{2020}}{7}\\ &=\frac{7.2^{2020}}{7}\\ &=2^{2020} \end{aligned}$
$\color{blue}\begin{aligned}(6)\quad \displaystyle \frac{\left ( 2^{n+2} \right )^{2}-2^{2}.2^{2n}}{2^{n}.2^{n+2}}&=\displaystyle \frac{2^{2(n+2)}-2^{2}.2^{2n}}{2^{n}.2^{n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}.2^{2.2}-2^{2}.2^{2n}}{2^{n+n}.2^{2}}\\ &=\displaystyle \frac{2^{2n}(2^{4}-2^{2})}{2^{2n}.2^{2}}\\ &=\displaystyle \frac{(2^{4}-2^{2})}{2^{2}}=\frac{16-4}{4}\\ &=\displaystyle \frac{12}{4}=3 \end{aligned}$
$\LARGE\textrm{C. Bentuk Akar}$
Bilangan bentuk akar di sini adalah kebalikan dari bilangan bentuk pangkat. Bilangan bentuk akar selanjutnya disebut bilangan irasional. Sebagai contoh $\sqrt{2}$, $\sqrt{3}$, $\sqrt{8}$, $\sqrt[3]{3}$, $\sqrt[3]{4}$, $\sqrt[3]{7}$ dan tapi ingat $\sqrt{4}$ dan $\sqrt[3]{8}$ serta $\sqrt[3]{27}$ adalah bukan bentuk akar, karena nantinya akan menghasilkan masing-masing 2 dan 3 serta 3.
$\color{purple}\begin{aligned}&\\ 1.\quad&a^{ \frac{1}{n}}=\sqrt[n]{a}\\ 2.\quad&a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\\ 3.\quad&a^{\frac{1}{2}}=\sqrt[2]{a^{1}}=\sqrt{a} \end{aligned}$.
$\textrm{Cara membaca}$.
$\begin{aligned}1.\quad&\sqrt[n]{p}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar pangkat n dari p}\\ 2.\quad&\sqrt[n]{p^{2}}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar pangkat n dari p kuadrat}\\ 3.\quad&\sqrt[n]{p^{3}}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar pangkat n dari p pangkat tiga}\\ 4.\quad&\sqrt{p}\: \: \: \textbf{dibaca}\: \: \: \color{red}\textrm{akar dari p}\: \: \: \color{black}\textrm{atau}\\ &\qquad\qquad\qquad \color{red}\textrm{akar kuadrat dari p}\\ &\qquad\qquad\qquad \textrm{ingat bahwa}:\: \: \sqrt{p}=\sqrt[2]{p} \end{aligned}$.
$\begin{aligned}\color{blue}\textrm{Defini}&\color{blue}\textrm{si}\\ \textrm{Jika}\: &\: a\: \: \textrm{dan}\: \: b\: \: \textrm{bilangan real dan}\\ &n\: \: \textrm{bilangan bulat positif, maka}:\\ &a^{n}=b\Leftrightarrow \sqrt[n]{b}=a\\ \textrm{keter}&\textrm{angan}:\\ \sqrt[n]{b}&\quad \textrm{disebut}\: \: \textbf{akar (radikal)}\\ b&\quad \textrm{disebut}\: \: \textbf{radikan}\\ &\quad \textrm{(bilangan pokok yang ditarik akarnya)}\\ n&\quad \textrm{disebut}\: \: \textbf{indeks}\\ &\quad (\textrm{pangkat akar}) \end{aligned}$.
$\Large\textrm{C.1 Bilangan Pangkat Pecahan}$.
Operasi Bilangan pangkat pecahan sama dengan operasi pangkat bilangan bulat.
$\LARGE\colorbox{yellow}{ CONTOH SOAL}$.
$\begin{array}{ll}\\ 1.&a^{.^{\frac{1}{2}}}\times a^{.^{\frac{1}{3}}}=a^{.^{\frac{1}{2}+\frac{1}{3}}}=a^{.^{\frac{5}{6}}}\\ 2.&a^{.^{\frac{1}{5}}}: a^{.^{\frac{1}{3}}}=a^{.^{\frac{1}{5}-\frac{1}{3}}}=a^{.^{-\frac{2}{15}}}\\ 3.&\left (a^{.^{\frac{2}{5}}} \right )^{\frac{4}{7}}=a^{.^{\frac{8}{35}}}\\ 4.&81^{.^{\frac{1}{2}}}=\left ( 9^{2} \right )^{.^{\frac{1}{2}}}=9^{1}=9\\ 5.&27^{.^{-\frac{2}{3}}}=\left ( 3^{3} \right )^{.^{-\frac{2}{3}}}=\left (3 \right )^{-2}=\displaystyle \frac{1}{3^{2}}=\frac{1}{9} \end{array}$.
$\begin{array}{ll}\\ 6.&\textrm{Sederhanakanlah bentuk berikut dan}\\ &\textrm{nyatakan hasilnya dalam pangkat positif}\\\\ &\textrm{a}.\quad \left ( 3p^{.^{\frac{5}{3}}}q^{.^{-\frac{3}{4}}} \right )\left ( 2p^{.^{-\frac{2}{3}}}q^{.^{\frac{5}{4}}} \right )\\\\ &\textrm{b}.\quad \displaystyle \frac{\left ( 8p^{.^{\frac{2}{3}}}q^{0}r^{.^{-\frac{1}{2}}} \right )}{\left ( 4p^{.^{-\frac{1}{2}}}q^{.^{-\frac{1}{3}}}r \right )}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( 3p^{.^{\frac{5}{3}}}q^{.^{-\frac{3}{4}}} \right )\left ( 2p^{.^{-\frac{2}{3}}}q^{.^{\frac{5}{4}}} \right )\\ &=3.2.p^{.^{\frac{5}{3}+\left ( -\frac{2}{3} \right )}}.q^{.^{-\frac{3}{4}+\frac{5}{4}}}\\ &=6.p^{.^{\frac{3}{3}}}q^{.^{\frac{2}{4}}}\\ &=6pq^{.^{\frac{1}{2}}} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\displaystyle \frac{\left ( 8p^{.^{\frac{2}{3}}}q^{0}r^{.^{-\frac{1}{2}}} \right )}{\left ( 4p^{.^{-\frac{1}{2}}}q^{.^{-\frac{1}{3}}}r \right )}\\ &=2.p^{.^{\frac{2}{3}-\left ( -\frac{1}{2} \right )}}q^{.^{0}-\left ( -\frac{1}{3} \right )}r^{.^{-\frac{1}{2}-1}}\\ &=2p^{.^{\frac{2}{3}+\frac{1}{2}}}q^{.^{\frac{1}{3}}}r^{.^{-\frac{3}{2}}}\\ &=2p^{.^{\frac{4+3}{6}}}q^{.^{\frac{1}{3}}}.r^{.^{-\frac{3}{2}}}\\ &=2p^{.^{\frac{7}{6}}}q^{.^{\frac{1}{3}}}.r^{.^{-\frac{3}{2}}}\\ &=\displaystyle \frac{2p^{.^{\frac{7}{6}}}q^{.^{\frac{1}{3}}}}{r^{.^{\frac{3}{2}}}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 7.&\textrm{Sederhanakanlah bentuk berikut dan}\\ &\textrm{nyatakan hasilnya dalam pangkat positif}\\ &\textrm{a}.\quad \left ( \displaystyle \frac{p^{3n+1}q^{n}}{p^{3n+4}q^{4n}} \right )^{\frac{1}{3}}\\ &\textrm{b}.\quad \left ( \displaystyle \frac{p^{-2}q^{3}}{p^{4}q^{-3}} \right )^{-\frac{1}{2}}\left ( \displaystyle \frac{p^{4}q^{-5}}{pq} \right )^{-\frac{1}{3}}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( \displaystyle \frac{p^{3n+1}q^{n}}{p^{3n+4}q^{4n}} \right )^{\frac{1}{3}}\\ &=\left ( p^{(3n+1)-(3n+4)}q^{n-4n} \right )^{\frac{1}{3}}\\ &=\left ( p^{-3}q^{-3n} \right )^{\frac{1}{3}}\\ &=p^{-3.\frac{1}{3}}q^{-3n.\frac{1}{3}}\\ &=p^{-1}q^{-n}\\ &=\displaystyle \frac{1}{pq^{n}} \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left ( \displaystyle \frac{p^{-2}q^{3}}{p^{4}q^{-3}} \right )^{-\frac{1}{2}}\left ( \displaystyle \frac{p^{4}q^{-5}}{pq} \right )^{-\frac{1}{3}}\\ &=\left ( \displaystyle \frac{p^{-2.(-\frac{1}{2})}q^{3.(-\frac{1}{2})}}{p^{4.(-\frac{1}{2})}q^{-3.(-\frac{1}{2})}} \right )\left ( \displaystyle \frac{p^{4.(-\frac{1}{3})}q^{-5.(-\frac{1}{3})}}{p^{.^{-\frac{1}{3}}}q^{.^{-\frac{1}{3}}}} \right )\\ &=\displaystyle \frac{p^{1}q^{.^{-\frac{3}{2}}}}{p^{-2}q^{.^{\frac{3}{2}}}}\times \frac{p^{.^{-\frac{4}{3}}}q^{.^{\frac{5}{3}}}}{p^{.^{-\frac{1}{3}}}q^{.^{-\frac{1}{3}}}}\\ &=p^{1-(-2)+(-\frac{4}{3})-(-\frac{1}{3})}q^{-\frac{3}{2}-\frac{3}{2}+\frac{5}{3}-(-\frac{1}{3})}\\ &=p^{3-\frac{3}{3}}q^{-\frac{6}{2}+\frac{6}{3}}\\ &=p^{3-1}q^{-3+2}\\ &=p^{2}q^{-1}\\ &=\displaystyle \frac{p^{2}}{q} \end{aligned} \end{array}$
$\begin{array}{ll}\\ 8.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}+n^{.^{\frac{3}{4}}} \right )^{2}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}+2\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )+\left ( n^{.^{\frac{3}{4}}} \right )^{2}\\ &\color{red}\textrm{INGAT}\: :\: \: \color{black}\left ( A+B \right )^{2}=A^{2}+2AB+B^{2}\\ &=2^{2}m^{.^{\frac{3.2}{2}}}+2.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3.2}{4}}}\\ &=4m^{3}+4m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{4}}}+n^{.^{\frac{3}{2}}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 9.&\textrm{Jabarkanlah bentuk}\\ &\qquad\qquad\quad \left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\left ( 2m^{.^{\frac{3}{2}}}-n^{.^{\frac{3}{4}}} \right )^{3}\\ &=\left ( 2m^{.^{\frac{3}{2}}} \right )^{3}-3\left ( 2m^{.^{\frac{3}{2}}} \right )^{2}\left ( n^{.^{\frac{3}{4}}} \right )+3\left ( 2m^{.^{\frac{3}{2}}} \right )\left ( n^{.^{\frac{3}{4}}} \right )^{2}-\left ( n^{.^{\frac{3}{4}}} \right )^{3}\\ &\color{red}\textrm{INGAT}\: :\: \: \color{black}\left ( A-B \right )^{3}=A^{3}-3A^{2}B+3AB^{2}-B^{3}\\ &=2^{3}m^{.^{\frac{3.3}{2}}}-3.2^{2}.m^{.^{\frac{3.2}{2}}}n^{.^{\frac{3}{4}}}+3.2.m^{.^{\frac{3}{2}}}n^{.^{\frac{3.2}{4}}}-n^{.^{\frac{3.3}{4}}}\\ &=8m^{.^{\frac{9}{2}}}-12m^{3}n^{.^{\frac{3}{4}}}+6m^{.^{\frac{3}{2}}}n^{.^{\frac{3}{2}}}-n^{.^{\frac{9}{2}}} \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 10.&\textrm{Jabarkanlah bentuk berikut}\\\\ &\textrm{a}.\quad \left ( 2p^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}} \right )\left ( p^{.^{\frac{1}{2}}}+4q^{.^{\frac{1}{2}}} \right )\\\\ &\textrm{b}.\quad \left ( p^{.^{\frac{1}{3}}}-q^{.^{\frac{1}{3}}} \right )\left ( p^{.^{\frac{2}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{1}{3}}}+q^{.^{\frac{2}{3}}} \right )\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{a}.\quad&\left ( 2p^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}} \right )\left ( p^{.^{\frac{1}{2}}}+4q^{.^{\frac{1}{2}}} \right )\\ &=2\left (p^{.^{\frac{1}{2}}} \right )^{2}+2.4.p^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-3q^{.^{\frac{1}{2}}}.p^{.^{\frac{1}{2}}}-3.4.\left (q^{.^{\frac{1}{2}}} \right )^{2}\\ &=2p^{1}+8q^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-3p^{.^{\frac{1}{2}}}q^{.^{\frac{1}{2}}}-12.q^{1}\\ &=2p+5(pq)^{.^{\frac{1}{2}}}-12q \end{aligned}\\ &\begin{aligned}\textrm{b}.\quad&\left ( p^{.^{\frac{1}{3}}}-q^{.^{\frac{1}{3}}} \right )\left ( p^{.^{\frac{2}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{1}{3}}}+q^{.^{\frac{2}{3}}} \right )\\ &=p^{.^{\frac{1+2}{3}}}+\left (p^{.^{\frac{1}{3}}} \right )^{2}q^{.^{\frac{1}{3}}}+p^{.^{\frac{1}{3}}}q^{.^{\frac{2}{3}}}-p^{.^{\frac{2}{3}}}q^{.^{\frac{1}{3}}}-p^{.^{\frac{1}{3}}}\left (q^{.^{\frac{1}{3}}} \right )^{2}-q^{.^{\frac{1+2}{3}}}\\ &=p^{1}+0+0-q^{1}\\ &=p-q \end{aligned} \end{array}$
DAFTAR PUSTAKA
- Sembiring, S., Zulkifli, M., Marsito, Rusdi, I. 2016. Matematika untuk Siswa SMA/MA Kelas X kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: SRIKANDI EMPAT WIDYA UTAMA.
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