Fungsi Eksponen

  $\text{A. Bilangan Pangkat Positif}$

Misalkan diketahui bahwa $a$ adalah suatu bilangan tidak nol dan $m$ adalah bilangan asli, maka bilangan ekponen atau bilangan berpangkat dedefinisikan dengan:

$a^m=\underset{m}{\underbrace{a\times a\times \times a\times ...\times a}}$

$\begin{array}{rl}\text{Bilangan}& :\\ a& \text{disebut basis atau bilangan pokok}\\ n& \text{disebut sebagai bilangan pangkat/eksponen}\end{array}$.

$\text{ CONTOH SOAL}$.

$(1).3^4=3\times 3\times 3\times 3=81$

$(2).5^4=5\times 5\times 5\times 5=625$

$(3).2^6=2\times 2\times 2\times 2\times 2\times 2=64$

$(4).6^7=6\times 6\times 6\times 6\times 6\times 6\times 6=279936$

$(5).(-3{)}^3=(-3)\times (-3)\times (-3)=-27$

$(6).(-2{)}^4=(-2)\times (-2)\times (-2)\times (-2)=16$

$(7).{\left({\displaystyle \frac{1}{5}}\right)}^3=\left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)\times \left({\displaystyle \frac{1}{5}}\right)={\displaystyle \frac{1}{125}}$

$(8).{(-{\displaystyle \frac{1}{2}})}^3=(-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{1}{8}}$

$\text{B. Sifat-Sifat Bilangan Pangkat Positif}$

$\begin{array}{r}\\ 1.& a^m.a^n=a^{m+n}\\ 2.& a^m:a^n=a^{m-n}\\ 3.& {\left(a^m\right)}^n=a^{m.n},\text{syarat}a\ne 0\\ 4.& {\left(ab\right)}^n=a^n.b^n\\ 5.& {\left(\frac{a}{b}\right)}^n=\frac{a^n}{b^n},\text{syarat}b\ne 0\end{array}$

Beberpa hal yang perlu diketahui juga, yaitu

$\begin{array}{r}\\ 1.& (a+b{)}^2=a^2+2ab+b^2\\ 2.& (a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3\\ 3.& {(a+\frac{1}{a})}^2=a^2+2+{\displaystyle \frac{1}{a^2},\text{syarat}a\ne 0}\end{array}$

$\text{ CONTOH SOAL}$

$\begin{array}{r}\\ (1).& 2^6\times 2^4\times 2^7=2^{6+4+7}=2^{17}\\ (2).& 2^5\times 3^5\times 7^5={(2.3.7)}^5={\left(42\right)}^5\\ (3).& {\displaystyle \frac{a^3.a^7.a^6}{a^9}={\displaystyle \frac{a^{3+7+6}}{a^9}=\frac{a^{16}}{a^9}=a^{16-9}=a^7,\text{syarat}a\ne 0}}\end{array}$

$\begin{array}{rl}(4).{\displaystyle \frac{3^7{.7}^3.2}{{\left(42\right)}^3}}& =\frac{2^1{.3}^7{.7}^3}{{\left(2.3.7\right)}^3}=\frac{2^1{.3}^7{.7}^3}{2^3{.3}^3{.7}^3}\\ & =2^{1-3}{.3}^{7-3}{.7}^{3-3}=2^{-2}{.3}^4{.7}^0\\ & =\frac{1}{2^2}{.3}^4.1=\frac{3^4}{2^2}\end{array}$

$\begin{array}{rl}(5).{\displaystyle \frac{2^{2020}+2^{2021}+2^{2022}}{7}}& ={\displaystyle \frac{{1.2}^{2020}+2^1{.2}^{2020}+2^2{.2}^{2020}}{7}}\\ & =\frac{(1+2+4){.2}^{2020}}{7}\\ & =\frac{{7.2}^{2020}}{7}\\ & =2^{2020}\end{array}$

$\begin{array}{rl}(6){\displaystyle \frac{{\left(2^{n+2}\right)}^2-2^2{.2}^{2n}}{2^n{.2}^{n+2}}}& ={\displaystyle \frac{2^{2(n+2)}-2^2{.2}^{2n}}{2^n{.2}^n{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}{.2}^{2.2}-2^2{.2}^{2n}}{2^{n+n}{.2}^2}}\\ & ={\displaystyle \frac{2^{2n}(2^4-2^2)}{2^{2n}{.2}^2}}\\ & ={\displaystyle \frac{(2^4-2^2)}{2^2}=\frac{16-4}{4}}\\ & ={\displaystyle \frac{12}{4}=3}\end{array}$

$\text{C. Bentuk Akar}$

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.

$\begin{array}{rl}& \\ 1.& a^{\frac{1}{n}}=\sqrt[n]{a}\\ 2.& a^{\frac{m}{n}}=\sqrt[n]{a^m}\\ 3.& a^{\frac{1}{2}}=\sqrt[2]{a^1}=\sqrt{a}\end{array}$.

$\text{Cara membaca}$.

$\begin{array}{rl}1.& \sqrt[n]{p}\mathbf{\text{dibaca}}\text{akar pangkat n dari p}\\ 2.& \sqrt[n]{p^2}\mathbf{\text{dibaca}}\text{akar pangkat n dari p kuadrat}\\ 3.& \sqrt[n]{p^3}\mathbf{\text{dibaca}}\text{akar pangkat n dari p pangkat tiga}\\ 4.& \sqrt{p}\mathbf{\text{dibaca}}\text{akar dari p}\text{atau}\\ & \text{akar kuadrat dari p}\\ & \text{ingat bahwa}:\sqrt{p}=\sqrt[2]{p}\end{array}$.

$\begin{array}{rl}\text{Defini}& \text{si}\\ \text{Jika}& a\text{dan}b\text{bilangan real dan}\\ & n\text{bilangan bulat positif, maka}:\\ & a^n=b\iff \sqrt[n]{b}=a\\ \text{keter}& \text{angan}:\\ \sqrt[n]{b}& \text{disebut}\mathbf{\text{akar (radikal)}}\\ b& \text{disebut}\mathbf{\text{radikan}}\\ & \text{(bilangan pokok yang ditarik akarnya)}\\ n& \text{disebut}\mathbf{\text{indeks}}\\ & (\text{pangkat akar})\end{array}$.

$\text{C.1 Bilangan Pangkat Pecahan}$.

Operasi Bilangan pangkat pecahan sama dengan operasi pangkat bilangan bulat.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& a^{{.}^{\frac{1}{2}}}\times a^{{.}^{\frac{1}{3}}}=a^{{.}^{\frac{1}{2}+\frac{1}{3}}}=a^{{.}^{\frac{5}{6}}}\\ 2.& a^{{.}^{\frac{1}{5}}}:a^{{.}^{\frac{1}{3}}}=a^{{.}^{\frac{1}{5}-\frac{1}{3}}}=a^{{.}^{-\frac{2}{15}}}\\ 3.& {\left(a^{{.}^{\frac{2}{5}}}\right)}^{\frac{4}{7}}=a^{{.}^{\frac{8}{35}}}\\ 4.& {81}^{{.}^{\frac{1}{2}}}={\left(9^2\right)}^{{.}^{\frac{1}{2}}}=9^1=9\\ 5.& {27}^{{.}^{-\frac{2}{3}}}={\left(3^3\right)}^{{.}^{-\frac{2}{3}}}={\left(3\right)}^{-2}={\displaystyle \frac{1}{3^2}=\frac{1}{9}}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakanlah bentuk berikut dan}\\ & \text{nyatakan hasilnya dalam pangkat positif}\\ \\ & \text{a}.\left(3p^{{.}^{\frac{5}{3}}}{q}^{{.}^{-\frac{3}{4}}}\right)\left(2p^{{.}^{-\frac{2}{3}}}{q}^{{.}^{\frac{5}{4}}}\right)\\ \\ & \text{b}.{\displaystyle \frac{\left(8p^{{.}^{\frac{2}{3}}}{q}^0{r}^{{.}^{-\frac{1}{2}}}\right)}{\left(4p^{{.}^{-\frac{1}{2}}}{q}^{{.}^{-\frac{1}{3}}}r\right)}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& \left(3p^{{.}^{\frac{5}{3}}}{q}^{{.}^{-\frac{3}{4}}}\right)\left(2p^{{.}^{-\frac{2}{3}}}{q}^{{.}^{\frac{5}{4}}}\right)\\ & =3.2.p^{{.}^{\frac{5}{3}+(-\frac{2}{3})}}.q^{{.}^{-\frac{3}{4}+\frac{5}{4}}}\\ & =6.p^{{.}^{\frac{3}{3}}}{q}^{{.}^{\frac{2}{4}}}\\ & =6p{q}^{{.}^{\frac{1}{2}}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\displaystyle \frac{\left(8p^{{.}^{\frac{2}{3}}}{q}^0{r}^{{.}^{-\frac{1}{2}}}\right)}{\left(4p^{{.}^{-\frac{1}{2}}}{q}^{{.}^{-\frac{1}{3}}}r\right)}}\\ & =2.p^{{.}^{\frac{2}{3}-(-\frac{1}{2})}}{q}^{{.}^0-(-\frac{1}{3})}{r}^{{.}^{-\frac{1}{2}-1}}\\ & =2p^{{.}^{\frac{2}{3}+\frac{1}{2}}}{q}^{{.}^{\frac{1}{3}}}{r}^{{.}^{-\frac{3}{2}}}\\ & =2p^{{.}^{\frac{4+3}{6}}}{q}^{{.}^{\frac{1}{3}}}.r^{{.}^{-\frac{3}{2}}}\\ & =2p^{{.}^{\frac{7}{6}}}{q}^{{.}^{\frac{1}{3}}}.r^{{.}^{-\frac{3}{2}}}\\ & ={\displaystyle \frac{2p^{{.}^{\frac{7}{6}}}{q}^{{.}^{\frac{1}{3}}}}{r^{{.}^{\frac{3}{2}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Sederhanakanlah bentuk berikut dan}\\ & \text{nyatakan hasilnya dalam pangkat positif}\\ & \text{a}.{\left({\displaystyle \frac{p^{3n+1}{q}^n}{p^{3n+4}{q}^{4n}}}\right)}^{\frac{1}{3}}\\ & \text{b}.{\left({\displaystyle \frac{p^{-2}{q}^3}{p^4{q}^{-3}}}\right)}^{-\frac{1}{2}}{\left({\displaystyle \frac{p^4{q}^{-5}}{pq}}\right)}^{-\frac{1}{3}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\left({\displaystyle \frac{p^{3n+1}{q}^n}{p^{3n+4}{q}^{4n}}}\right)}^{\frac{1}{3}}\\ & ={\left(p^{(3n+1)-(3n+4)}{q}^{n-4n}\right)}^{\frac{1}{3}}\\ & ={\left(p^{-3}{q}^{-3n}\right)}^{\frac{1}{3}}\\ & =p^{-3.\frac{1}{3}}{q}^{-3n.\frac{1}{3}}\\ & =p^{-1}{q}^{-n}\\ & ={\displaystyle \frac{1}{p{q}^n}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\left({\displaystyle \frac{p^{-2}{q}^3}{p^4{q}^{-3}}}\right)}^{-\frac{1}{2}}{\left({\displaystyle \frac{p^4{q}^{-5}}{pq}}\right)}^{-\frac{1}{3}}\\ & =\left({\displaystyle \frac{p^{-2.(-\frac{1}{2})}{q}^{3.(-\frac{1}{2})}}{p^{4.(-\frac{1}{2})}{q}^{-3.(-\frac{1}{2})}}}\right)\left({\displaystyle \frac{p^{4.(-\frac{1}{3})}{q}^{-5.(-\frac{1}{3})}}{p^{{.}^{-\frac{1}{3}}}{q}^{{.}^{-\frac{1}{3}}}}}\right)\\ & ={\displaystyle \frac{p^1{q}^{{.}^{-\frac{3}{2}}}}{p^{-2}{q}^{{.}^{\frac{3}{2}}}}\times \frac{p^{{.}^{-\frac{4}{3}}}{q}^{{.}^{\frac{5}{3}}}}{p^{{.}^{-\frac{1}{3}}}{q}^{{.}^{-\frac{1}{3}}}}}\\ & =p^{1-(-2)+(-\frac{4}{3})-(-\frac{1}{3})}{q}^{-\frac{3}{2}-\frac{3}{2}+\frac{5}{3}-(-\frac{1}{3})}\\ & =p^{3-\frac{3}{3}}{q}^{-\frac{6}{2}+\frac{6}{3}}\\ & =p^{3-1}{q}^{-3+2}\\ & =p^2{q}^{-1}\\ & ={\displaystyle \frac{p^2}{q}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Jabarkanlah bentuk}\\ & {(2m^{{.}^{\frac{3}{2}}}+n^{{.}^{\frac{3}{4}}})}^2\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {(2m^{{.}^{\frac{3}{2}}}+n^{{.}^{\frac{3}{4}}})}^2\\ & ={\left(2m^{{.}^{\frac{3}{2}}}\right)}^2+2\left(2m^{{.}^{\frac{3}{2}}}\right)\left(n^{{.}^{\frac{3}{4}}}\right)+{\left(n^{{.}^{\frac{3}{4}}}\right)}^2\\ & \text{INGAT}:{(A+B)}^2=A^2+2AB+B^2\\ & =2^2{m}^{{.}^{\frac{3.2}{2}}}+2.2.m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{4}}}+n^{{.}^{\frac{3.2}{4}}}\\ & =4m^3+4m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{4}}}+n^{{.}^{\frac{3}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jabarkanlah bentuk}\\ & {(2m^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{3}{4}}})}^3\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {(2m^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{3}{4}}})}^3\\ & ={\left(2m^{{.}^{\frac{3}{2}}}\right)}^3-3{\left(2m^{{.}^{\frac{3}{2}}}\right)}^2\left(n^{{.}^{\frac{3}{4}}}\right)+3\left(2m^{{.}^{\frac{3}{2}}}\right){\left(n^{{.}^{\frac{3}{4}}}\right)}^2-{\left(n^{{.}^{\frac{3}{4}}}\right)}^3\\ & \text{INGAT}:{(A-B)}^3=A^3-3A^{2}B+3A{B}^2-B^3\\ & =2^3{m}^{{.}^{\frac{3.3}{2}}}-{3.2}^2.m^{{.}^{\frac{3.2}{2}}}{n}^{{.}^{\frac{3}{4}}}+3.2.m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3.2}{4}}}-n^{{.}^{\frac{3.3}{4}}}\\ & =8m^{{.}^{\frac{9}{2}}}-12m^3{n}^{{.}^{\frac{3}{4}}}+6m^{{.}^{\frac{3}{2}}}{n}^{{.}^{\frac{3}{2}}}-n^{{.}^{\frac{9}{2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jabarkanlah bentuk berikut}\\ \\ & \text{a}.(2p^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}})(p^{{.}^{\frac{1}{2}}}+4q^{{.}^{\frac{1}{2}}})\\ \\ & \text{b}.(p^{{.}^{\frac{1}{3}}}-q^{{.}^{\frac{1}{3}}})(p^{{.}^{\frac{2}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{1}{3}}}+q^{{.}^{\frac{2}{3}}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& (2p^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}})(p^{{.}^{\frac{1}{2}}}+4q^{{.}^{\frac{1}{2}}})\\ & =2{\left(p^{{.}^{\frac{1}{2}}}\right)}^2+2.4.p^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-3q^{{.}^{\frac{1}{2}}}.p^{{.}^{\frac{1}{2}}}-3.4.{\left(q^{{.}^{\frac{1}{2}}}\right)}^2\\ & =2p^1+8q^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-3p^{{.}^{\frac{1}{2}}}{q}^{{.}^{\frac{1}{2}}}-12.q^1\\ & =2p+5(pq{)}^{{.}^{\frac{1}{2}}}-12q\end{array}\\ & \begin{array}{rl}\text{b}.& (p^{{.}^{\frac{1}{3}}}-q^{{.}^{\frac{1}{3}}})(p^{{.}^{\frac{2}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{1}{3}}}+q^{{.}^{\frac{2}{3}}})\\ & =p^{{.}^{\frac{1+2}{3}}}+{\left(p^{{.}^{\frac{1}{3}}}\right)}^2{q}^{{.}^{\frac{1}{3}}}+p^{{.}^{\frac{1}{3}}}{q}^{{.}^{\frac{2}{3}}}-p^{{.}^{\frac{2}{3}}}{q}^{{.}^{\frac{1}{3}}}-p^{{.}^{\frac{1}{3}}}{\left(q^{{.}^{\frac{1}{3}}}\right)}^2-q^{{.}^{\frac{1+2}{3}}}\\ & =p^1+0+0-q^1\\ & =p-q\end{array}\end{array}$

DAFTAR PUSTAKA

1. 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.