Lanjutan Materi : Persamaan Eksponen

A. Persamaan Eksponen

Berikut bentuk persamaan eksponen yang sering digunakan terangkum dalam tabel berikut beserta cara penyelesaiannya

$\begin{array}{|cll|}\hline \mathbf{\text{No}}& \mathbf{\text{Persamaan Eksponen}}& \mathbf{\text{Penyelesaian}}\\ 1& a^{f(x)}=1,a>0,a\ne 1& f(x)=0\\ 2& a^{f(x)}=a^p,a>0,a\ne 1& f(x)=p\\ 3& a^{f(x)}=a^{g(x)},a>0,a\ne 1& f(x)=g(x)\\ 4& a^{f(x)}=b^{f(x)},a>0,a\ne 1& f(x)=0\\ & \text{dan}b>0,b\ne 1& \\ 5& h(x{)}^{f(x)}=h(x{)}^{g(x)}& \begin{array}{rl}(1)& f(x)=g(x)\\ (2)& h(x)=1\\ (3)& h(x)=0\\ & \text{dengan syarat}\\ & f(x)>0\text{dan}\\ & g(x)>0\\ (4)& h(x)=-1\\ & \text{dengan syarat}\\ & f(x)\text{dan}g(x)\\ & \text{keduanya}\\ & \text{genap atau}\\ & \text{keduanya}\\ & \text{ganjil}\\ & \text{atau}\\ & \text{dapat juga}\\ & \text{ditunjukkan}\\ & (-1{)}^{f(x)}=(-1{)}^{g(x)}\end{array}\\ 6& g(x{)}^{f(x)}=h(x{)}^{f(x)}& \begin{array}{rl}(1)& g(x)=h(x)\\ (2)& f(x)=0\\ & \text{dengan syarat}\\ & g(x)\ne 0\text{dan}\\ & h(x)\ne 0\end{array}\\ 7& f(x{)}^{g(x)}=1& \begin{array}{rl}(1)& f(x)=1\\ (2)& f(x)=-1\\ & \text{dengan syarat}\\ & g(x)\text{genap}\\ (3)& g(x)=0\\ & \text{dengan syarat}\\ & f(x)\ne 0\end{array}\\ 8& A{\left(a^{f(x)}\right)}^2+B\left(a^{f(x)}\right)+C=0& \begin{array}{rl}& \text{ubah}{a}^{f(x)}=y\\ & \text{sehingga}\\ & A{y}^2+By+C=0\\ & \text{selanjutnya}\\ & \text{substitusikan}\\ & \text{nilai}y\text{ke}\\ & \text{persamaan}\\ & a^{f(x)}=y\end{array}\\ \hline\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.2^{2x-2021}=1\\ & \text{b}.{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}=1\\ & \text{c}.{\sqrt{2}}^{2x-2021}=1\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|ccc|}\hline \text{a}& \text{b}& \text{c}\\ \begin{array}{rl}{2}^{2x-2021}& =1\\ 2^{2x-2021}& =2^0\\ 2x-2021& =0\\ 2x& =2021\\ x& ={\displaystyle \frac{2021}{2}}\\ & \end{array}& \begin{array}{rl}{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& =1\\ {\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& ={\left(\frac{1}{2}\right)}^0\\ 2x-2021& =0\\ 2x& =2021\\ x& ={\displaystyle \frac{2021}{2}}\end{array}& \begin{array}{rl}{\sqrt{2}}^{2x-2021}& =1\\ {\sqrt{2}}^{2x-2021}& ={\sqrt{2}}^0\\ 2x-2021& =0\\ 2x& =2021\\ x& ={\displaystyle \frac{2021}{2}}\\ & \end{array}\\ \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2021}{2}}\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2021}{2}}\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2021}{2}}\right\}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.2^{2x-2021}=128\\ & \text{b}.{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}=128\\ & \text{c}.{\sqrt{2}}^{2x-2021}=128\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{|ccc|}\hline \text{a}& \text{b}& \text{c}\\ \begin{array}{rl}{2}^{2x-2021}& =128\\ 2^{2x-2021}& =2^7\\ 2x-2021& =7\\ 2x=7& +2021\\ x={\displaystyle \frac{2028}{2}}& =1014\\ & \end{array}& \begin{array}{rl}{\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& =128\\ {\left({\displaystyle \frac{1}{2}}\right)}^{2x-2021}& ={\left({\displaystyle \frac{1}{2}}\right)}^{-7}\\ 2x-2021& =-7\\ 2x=2021& -7\\ x={\displaystyle \frac{2014}{2}}& =1007\end{array}& \begin{array}{rl}{\sqrt{2}}^{2x-2021}& =128\\ {\sqrt{2}}^{2x-2021}& ={\sqrt{2}}^{256}\\ 2x-2021& =256\\ 2x=2021& +256\\ x& ={\displaystyle \frac{2277}{2}}\\ & \end{array}\\ \mathbf{\text{HP}}=\left\{1014\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle 1007}\right\}& \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2277}{2}}\right\}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& (\mathbf{\text{SPMB 04}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{27}{3^{2x-1}}={81}^{-0,125}\text{adalah... .}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\displaystyle \frac{27}{3^{2x-1}}}& ={81}^{-0,125}\\ 3^{3-(2x-1)}& =3^{4(\frac{1}{8})}\\ 3-2x+1& =-{\displaystyle \frac{1}{2}}\\ -2x+4& =-{\displaystyle \frac{1}{2}}\\ -x+2& =-{\displaystyle \frac{1}{4}}\\ -x& =-2-{\displaystyle \frac{1}{4}}\\ -x& =-2{\displaystyle \frac{1}{4}}\\ x& =2{\displaystyle \frac{1}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& (\mathbf{\text{UMPTN 00}})\\ & \text{Bentuk}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ & \text{Jika}{x}_0\text{memenuhi persamaan, maka nilai}\\ & 1-{\displaystyle \frac{3}{4}{x}_0=....}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}& ={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ 3^{-5x}& =3^{2(1-(x-2))}{.3}^{-\frac{2}{3}}\\ -5x& =2(1-(x-2))+(-{\displaystyle \frac{2}{3}}),\text{dikali}3\\ -15x& =6(3-x)+(-2)\\ -15x& =18-6x-2\\ 6x-15x& =16\\ -9x& =16\\ x& ={\displaystyle \frac{16}{-9}}\\ x_0& =-{\displaystyle \frac{16}{9},\text{selanjutnya}}\\ 1-{\displaystyle \frac{3}{4}{x}_0}& =1-{\displaystyle \frac{3}{4}\times (-\frac{16}{9})}\\ & =1+\frac{4}{3}\\ & =1+1{\displaystyle \frac{1}{3}}\\ & =2{\displaystyle \frac{1}{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{5}^{x+1}+5^{2-x}-30& =0\\ \left(5^x\right){.5}^1+{\displaystyle \frac{5^2}{5^x}-30}& =0\\ 5{\left(5^x\right)}^2+25-30\left(5^x\right)& =0\\ \text{Persamaan kuadrat}& \text{dalam}{5}^x,\text{maka}\\ 5(5^x{)}^2-30(5^x)+25& =0\{\begin{array}{ll}a& =5\\ b& =-30\\ c& =25\end{array}\\ (5^{x_1}).\left(5^{x_2}\right)& ={\displaystyle \frac{c}{a}}\\ 5^{x_1+x_2}& ={\displaystyle \frac{25}{5}=5}\\ 5^{x_1+x_2}& =5^1\\ x_1+x_2& =1\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Jumlah akar-akar persamaan}\\ & {2023}^{x^2-7x+7}={2024}^{x^2-7x+7}\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}{2023}^{x^2-7x+7}& ={2024}^{x^2-7x+7}\\ \text{Karena basis}& \text{tidak sama},\\ \text{maka harusl}& \text{ah pangkatnya}=0,\\ x^2-7x+7& =0\\ \text{dan jumlah}& \text{akar-akarnya adalah}:\\ x_1+x_2& =-{\displaystyle \frac{b}{a},\text{dari persamaan}}\\ x^2-7x+7& =0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =7\end{array}\\ \text{maka}{x}_1+x_2& =-{\displaystyle \frac{b}{a}=-\frac{-7}{1}=7}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tentukan himpunan penyelesaian dari}\\ & (x-2{)}^{x^2-7x+6}=1\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat bentuk}f(x{)}^{g(x)}=1\{\begin{array}{ll}f(x)& =x-2\\ g(x)& =x^2-7x+5\end{array}\\ & \begin{array}{|lll|}\hline f(x)=1& f(x)=-1& g(x)=0\\ & \text{Syarat}g(x)\text{genap}& \text{Syarat}f(x)\ne 0\\ \begin{array}{rl}x& -2=1\\ x& =3\\ & \end{array}& \begin{array}{rl}x& -2=-1\\ x& =2-1=1\\ & \end{array}& \begin{array}{rl}{x}^2& -7x+6=0\\ \iff & (x-1)(x-6)\\ \iff & x=1\text{atau}x=6\end{array}\\ & \begin{array}{rl}\text{S}& \text{yaratnya}x\\ \text{u}& \text{ntuk}x=1\\ g& (1)=1^2-7+6\\ & =0(\text{memenuhi})\end{array}& \begin{array}{rl}f(1)& =1-2=-1\ne 0\\ f(6)& =6-2=4\ne 0\\ & \\ & \end{array}\\ & \mathbf{\text{Catatan}}:0& \\ & \text{paritasnya genap}& \\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{1,3,6\}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Tentukan himpunan penyelesaian dari}\\ & (x^2-9x+19{)}^{2x+3}=(x^2-9x+19{)}^{x-1}\text{adalah}....\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat bentuk}h(x{)}^{f(x)}=h(x{)}^{g(x)}\{\begin{array}{ll}h(x)& =x^2-9x+19\\ f(x)& =2x+3\\ g(x)& =x-1\end{array}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}& \text{Syarat-syaratnya}\\ & \bullet f(x)=g(x)\\ & \iff 2x+3=x-1\\ & \iff x=-4\\ & \bullet h(x)=1\\ & \iff x^2-9x+19=1\\ & \iff x^2-9x+18=0\\ & \iff (x-3)(x-6)=0\\ & \iff x=3\text{atau}x=6\\ & \bullet h(x)=0\\ & \iff x^2-9x+19=0\\ & \iff x_{1,2}={\displaystyle \frac{9\pm \sqrt{5}}{2}}\\ & \text{gunakan rumus ABC}\\ & \text{Setelah diuji keduanya}\\ & \text{positif, maka}\\ & x={\displaystyle \frac{9\pm \sqrt{5}}{2}\text{merupakan}}\\ & \mathbf{\text{penyelesaian}}\end{array}& \begin{array}{rl}& \text{lanjutannya}\\ & \bullet h(x)=-1\\ & \iff x^2-9x+19=-1\\ & \iff x^2-9x+20=0\\ & \iff (x-4)(x-5)=0\\ & \iff x=4\text{atau}x=5\\ & \text{Uji nilanya}\\ & \text{untuk}x=4\\ & ⧫f(4)=2(4)+3\text{ganjil}\\ & ⧫g(4)=4-1\text{ganjil}\\ & \text{karena}f(4),g(4)\text{keduanya }\\ & \text{ganjil, maka}x=4\\ & \text{adalah}\mathbf{\text{penyelesaian}}\\ & \text{untuk}x=5\\ & ⧫f(5)=2(5)+3\text{ganjil}\\ & ⧫g(5)=5-1\text{genapl}\\ & \text{karena}f(4)\ne g(4),\text{maka}x=5\\ & \text{adalah}\mathbf{\text{bukan penyelesaian}}\\ & \end{array}\\ \hline\end{array}\\ & \mathbf{\text{HP}}=\{-4,3,4,6,{\displaystyle \frac{9-\sqrt{5}}{2},\frac{9+\sqrt{5}}{2}}\}\end{array}$.

DAFTAR PUSTAKA

1. Kurnia, N, dkk. 2016. Jelajah Matematika I SMA Kelas X Peminatan MIPA. Jakarta: YUDHISTIRA.

Lanjutan : Fungsi Eksponensial

1. Pengertian Fungsi Eksponen

Sebuah fungsi adalah relasi khusus dengan aturan tertentu. Fungsi adalah sebuah pemetaan yang memetakan setiap anggota domoain dengan tepat satu anggota kodomain. Jika suatu himpunan A sebagai domain yang setiap anggota himpunannya dipetakan ke tepat satu anggota himpunan B sebagai kodomain selanjutnya disebut fungsi dari himpunan A ke B atau  $f:A\to B$.

Perhatikanlah gambar berikut

Pada gambar di atas terlihat jelas bahwa setiap bilangan riil  $x$ dipetakan dengan tepat ke bilangan riil  $y$. Sehingga fungsi $f$  memtakan  $x\in A$  ke  $y$  atau  $f:x\to y$ dan aturan dari fungsi  $f$ ini sendiri ini biasanya sering dituliskan dalam notasi  $y=f(x)$. Selanjutnya untuk ilustrasi fungsi eksponensial adalah sebagai berikut:

Tampak jelas bahwa  $y\in B$  adalah  $y=f(x)=k.a^x$,  dengan  $k$ konstanta,  $x$ sebagai variabel bebas, serta  $a$ adalah bilangan basis atau bilangan pokok, dengan  $a>0$  dan   $a\ne 1$.

2. Garfik Fungsi Eksponen

a. Grafik fungsi eksponen  $y=f(x)=k.a^x$, dengan  $a>1$  dan   $x\in \mathbb{R}$.

Perhatikan ilustrasi berikut

a. Grafik fungsi eksponen  $y=f(x)=k.a^x$, dengan  $0<a<1$,  $a\in \mathbb{Q}$  dan   $x\in \mathbb{R}$.

Ilustrasi lain dari garfik fungsi eksponen adalah sebagai berikut

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Lengkapilah tabel berikut}\\ & \begin{array}{|ccccccccc|}\hline \text{fungsi}& -3& -2& -1& 0& 1& 2& 3& 4\\ f(x)=2^x& & & & & & & & \\ f(x)=2^{-x}& & & & & & & & \\ f(x)=3^x& & & & & & & & \\ f(x)=3^{-x}& & & & & & & & \\ \hline\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{|ccccccccc|}\hline \text{fungsi}& -3& -2& -1& 0& 1& 2& 3& 4\\ f(x)=2^x& {\displaystyle \frac{1}{8}}& {\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{2}}& 1& 2& 4& 8& 16\\ f(x)=2^{-x}& 8& 4& 2& 1& {\displaystyle \frac{1}{2}}& {\displaystyle \frac{1}{4}}& {\displaystyle \frac{1}{8}}& {\displaystyle \frac{1}{16}}\\ f(x)=3^x& {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& 81\\ f(x)=3^{-x}& 27& 9& 3& 1& {\displaystyle \frac{1}{3}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{81}}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Gambarlah grafik fungsi eksponen berikut}\\ & \text{a}.f(x)=3^{x+1}\\ & \text{b}.f(x)=3^x+1\\ \\ & \mathbf{\text{Jawab}}\\ & \text{a. Untuk fungsi}f(x)=3^{x+1}\text{sebagai berikut}:\\ & \begin{array}{|cccccccccccc|}\hline \text{fungsi/titik}& -\mathrm{\infty }& \cdots & -3& -2& -1& 0& 1& 2& 3& \cdots & \mathrm{\infty }\\ f(x)=3^x& 0& \cdots & {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& \cdots & \text{TD}\\ f(x)=3^{x+1}& 0& \cdots & {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& 81& \cdots & \text{TD}\\ (x,f(x))& (-\mathrm{\infty },0)& \cdots & (-3,{\displaystyle \frac{1}{9}})& (-2,{\displaystyle \frac{1}{3}})& (-1,1)& (0,3)& (1,9)& (2,27)& (3,81)& \cdots & \\ \hline\end{array}\\ & \text{b. Dan untuk fungsi}f(x)=3^x+1\text{sebagai berikut}:\\ & \begin{array}{|cccccccccccc|}\hline \text{fungsi/titik}& -\mathrm{\infty }& \cdots & -3& -2& -1& 0& 1& 2& 3& \cdots & \mathrm{\infty }\\ f(x)=3^x& 0& \cdots & {\displaystyle \frac{1}{27}}& {\displaystyle \frac{1}{9}}& {\displaystyle \frac{1}{3}}& 1& 3& 9& 27& \cdots & \text{TD}\\ f(x)=3^x+1& 1& \cdots & 1{\displaystyle \frac{1}{27}}& 1{\displaystyle \frac{1}{9}}& 1{\displaystyle \frac{1}{3}}& 2& 4& 10& 28& \cdots & \text{TD}\\ (x,f(x))& (-\mathrm{\infty },1)& \cdots & (-3,1{\displaystyle \frac{1}{27}})& (-2,1{\displaystyle \frac{1}{9}})& (-1,1{\displaystyle \frac{1}{3}})& (0,2)& (1,4)& (2,10)& (3,28)& \cdots & \\ \hline\end{array}\end{array}$

.

Lanjutan 2 Fungsi Eksponen

 C. 2. 2  Merasionalkan penyebut

Jika suatu pecahan penyebutnya mengandung bilangan irasional atau bentuk akar, maka penyebut ini dapat dibuat menjadi bilangan rasional. Perhatikanlah langkah berikut

$\begin{array}{rl}1.& {\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{\left(\sqrt{b^2}\right)}=\frac{a}{b}\sqrt{b}}\\ 2.& {\displaystyle \frac{a}{\sqrt[3]{b}}=\frac{a}{\sqrt[3]{b}}\times \frac{\sqrt[3]{b^2}}{\sqrt[3]{b^2}}=\frac{a\sqrt[3]{b^2}}{\left(\sqrt[3]{b^3}\right)}=\frac{a}{b}\sqrt[3]{b^2}}\\ 3.& {\displaystyle \frac{a}{\sqrt[5]{b^3}}={\displaystyle \frac{a}{\sqrt[5]{b^3}}\times \frac{\sqrt[5]{b^2}}{\sqrt[5]{b^2}}=\frac{a\sqrt[5]{b^2}}{\sqrt[5]{b^5}}=\frac{a}{b}\sqrt[5]{b^2}}}\end{array}$

Merasionalkan di atas adalah contoh bebrapa contoh model merasionalkan jika berjenis tunggal tetapi jika nanti jenisnya lebih dari itu, maka perhatikanlah simulasi contoh berikut

$\begin{array}{rl}& \\ 1.& {\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c(a-\sqrt{b})}{a^2-b}}\\ 2.& {\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c(a+\sqrt{b})}{a^2-b}}\\ 3.& {\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c(\sqrt{a}-\sqrt{b})}{a-b}}\end{array}$

Perhatikanlah simulasi contoh di atas, bentuk $a+\sqrt{b}$ memiliki bentuk sekawan (irasional juga) $a-\sqrt{b}$, demikian juga bentuk $\sqrt{a}+\sqrt{b}$ memiliki sekawan $\sqrt{a}-\sqrt{b}$. Disamping itu ada bentuk khusus yatu bentuk  $\sqrt[3]{a}+\sqrt[3]{b}$ memiliki bentuk sekawan $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Rasionalkanlah penyebut pecahan berikut}\\ & \text{dan serderhankanlah hasilnya}\\ & \text{a}.{\displaystyle \frac{2}{\sqrt{5}}\text{d}.{\displaystyle \frac{\sqrt{2}}{\sqrt{5}}}}\\ & \text{b}.{\displaystyle \frac{2}{5\sqrt{2}}\text{e}.{\displaystyle \frac{p}{\sqrt{q}}}}\\ & \text{c}.{\displaystyle \frac{6}{3\sqrt{5}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{2}{\sqrt{5}}={\displaystyle \frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}={\displaystyle \frac{2\sqrt{5}}{\sqrt{25}}={\displaystyle \frac{2}{5}\sqrt{5}}}}}\\ \text{b}.& {\displaystyle \frac{2}{5\sqrt{2}}={\displaystyle \frac{2}{5\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}={\displaystyle \frac{2\sqrt{2}}{5\sqrt{4}}=\frac{2\sqrt{2}}{5.2}={\displaystyle \frac{1}{5}\sqrt{2}}}}}\\ \text{c}.& {\displaystyle \frac{6}{3\sqrt{5}}={\displaystyle \frac{6}{3\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}={\displaystyle \frac{6\sqrt{5}}{3\sqrt{25}}=\frac{6\sqrt{5}}{3.5}={\displaystyle \frac{2}{5}\sqrt{5}}}}}\\ \text{d}.& {\displaystyle \frac{\sqrt{2}}{\sqrt{5}}={\displaystyle \frac{\sqrt{2}}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{\sqrt{10}}{\sqrt{25}}=\frac{\sqrt{10}}{5}={\displaystyle \frac{1}{5}\sqrt{10}}}}\\ \text{e}.& {\displaystyle \frac{p}{\sqrt{q}}={\displaystyle \frac{p}{\sqrt{q}}\times \frac{\sqrt{q}}{\sqrt{q}}={\displaystyle \frac{p\sqrt{q}}{\sqrt{q^2}}=\frac{p\sqrt{q}}{q}={\displaystyle \frac{p}{q}\sqrt{q}}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Rasionalkanlah penyebut pecahan berikut}\\ & \text{dan serderhankanlah hasilnya}\\ & \text{a}.{\displaystyle \frac{3}{6-\sqrt{5}}\text{f}.{\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}}}\\ & \text{b}.{\displaystyle \frac{3}{6+\sqrt{5}}\text{g}.\frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}}\\ & \text{c}.{\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\text{h}.\frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}}\\ & \text{d}.{\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\text{i}.\frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}}\\ & \text{e}.{\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\text{j}.\frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{3}{6-\sqrt{5}}={\displaystyle \frac{3}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}={\displaystyle \frac{3(6+\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{18+3\sqrt{5}}{36-5}=\frac{18+3\sqrt{5}}{31}={\displaystyle \frac{1}{31}(18+3\sqrt{5})}}\\ \text{b}.& {\displaystyle \frac{3}{6+\sqrt{5}}={\displaystyle \frac{3}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}={\displaystyle \frac{3(6-\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{18-3\sqrt{5}}{36-5}=\frac{18-3\sqrt{5}}{31}={\displaystyle \frac{1}{31}(18-3\sqrt{5})}}\\ \text{c}.& {\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{6-\sqrt{5}}\times \frac{6+\sqrt{5}}{6+\sqrt{5}}={\displaystyle \frac{\sqrt{3}(6+\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{6\sqrt{3}+\sqrt{15}}{36-5}=\frac{6\sqrt{3}+\sqrt{15}}{31}={\displaystyle \frac{1}{31}(6\sqrt{3}+\sqrt{15})}}\\ \text{d}.& {\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{6+\sqrt{5}}\times \frac{6-\sqrt{5}}{6-\sqrt{5}}={\displaystyle \frac{\sqrt{3}(6-\sqrt{5})}{6^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{6\sqrt{3}-\sqrt{15}}{36-5}=\frac{6\sqrt{3}-\sqrt{15}}{31}={\displaystyle \frac{1}{31}(6\sqrt{3}-\sqrt{15})}}\\ \text{e}.& {\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{3}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{3(\sqrt{6}+\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{3(\sqrt{6}+\sqrt{5})}{6-5}=\frac{3(\sqrt{6}+\sqrt{5})}{1}=3(\sqrt{6}+\sqrt{5})}\\ \text{f}.& {\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{3}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{3(\sqrt{6}-\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{3(\sqrt{6}-\sqrt{5})}{6-5}=\frac{3(\sqrt{6}-\sqrt{5})}{1}=3(\sqrt{6}-\sqrt{5})}\\ \text{g}.& {\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{\sqrt{6}-\sqrt{5}}\times \frac{\sqrt{6}+\sqrt{5}}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{\sqrt{3}(\sqrt{6}+\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{\sqrt{18}+\sqrt{15}}{6-5}=\frac{\sqrt{9.2}+\sqrt{15}}{1}=(3\sqrt{2}+\sqrt{15})}\\ \text{h}.& {\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}={\displaystyle \frac{\sqrt{3}}{\sqrt{6}+\sqrt{5}}\times \frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}-\sqrt{5}}={\displaystyle \frac{\sqrt{3}(\sqrt{6}-\sqrt{5})}{{\sqrt{6}}^2-{\sqrt{5}}^2}}}}\\ & ={\displaystyle \frac{\sqrt{18}-\sqrt{15}}{6-5}=\frac{\sqrt{9.2}-\sqrt{15}}{1}=(3\sqrt{2}-\sqrt{15})}\\ \text{i}.& \frac{\sqrt{3}}{\sqrt{6-2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1-2\sqrt{5.1}}}={\displaystyle \frac{\sqrt{3}}{\sqrt{5}-\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}-1}}\\ & =\frac{\sqrt{3}}{\sqrt{5}-1}\times \frac{\sqrt{5}+1}{\sqrt{5}+1}={\displaystyle \frac{\sqrt{3.5}+\sqrt{3.1}}{{\sqrt{5}}^2-1^2}=\frac{\sqrt{15}+\sqrt{3}}{5-1}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{15}+\sqrt{3})}\\ \text{j}.& \frac{\sqrt{3}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{3}}{\sqrt{5+1+2\sqrt{5.1}}}={\displaystyle \frac{\sqrt{3}}{\sqrt{5}+\sqrt{1}}=\frac{\sqrt{3}}{\sqrt{5}+1}}\\ & =\frac{\sqrt{3}}{\sqrt{5}+1}\times \frac{\sqrt{5}-1}{\sqrt{5}-1}={\displaystyle \frac{\sqrt{3.5}-\sqrt{3.1}}{{\sqrt{5}}^2-1^2}=\frac{\sqrt{15}-\sqrt{3}}{5-1}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{15}-\sqrt{3})}\end{array}\end{array}$.

$\begin{array}{l}\\ 3& \text{Rasionalkan penyebut dan sederhanakanlah}\\ & \text{a}.{\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}}\\ & \text{b}.{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}}\\ & ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{5}+\sqrt{7}}\times {\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{\sqrt{2}+\sqrt{5}-\sqrt{7}}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{{(\sqrt{2}+\sqrt{5})}^2-{\left(\sqrt{7}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{(2+2\sqrt{10}+5)-7}={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{5}-\sqrt{7}}{2\sqrt{10}}\times {\displaystyle \frac{\sqrt{10}}{\sqrt{10}}={\displaystyle \frac{\sqrt{20}+\sqrt{50}-\sqrt{70}}{2\times 10}}}}\\ & ={\displaystyle \frac{2\sqrt{5}+5\sqrt{2}+\sqrt{70}}{20}}\end{array}\\ & \begin{array}{rl}\text{b}.& {\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}}\\ & ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}\times \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{{(\sqrt{2}+\sqrt{3})}^2-{\left(\sqrt{5}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{(2+2\sqrt{6}+3)-5}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}}\\ & =\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{2\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}\\ & ={\displaystyle \frac{\sqrt{12}+\sqrt{18}+\sqrt{30}}{2\times 6}}\\ & ={\displaystyle \frac{2\sqrt{3}+3\sqrt{2}+\sqrt{30}}{12}}\end{array}\end{array}$

Lanjutan Fungsi Eksponen

$\text{C.2 Operasi Bilangan Bentuk Akar}$.

C. 2. 1  Sifat-sifat yang berlaku pada operasi bilangan bentuk akar

$\begin{array}{rl}& \\ 1.& a\sqrt[n]{c}+b\sqrt[n]{c}=(a+b)\sqrt[n]{c}\\ 2.& a\sqrt[n]{c}-b\sqrt[n]{c}=(a-b)\sqrt[n]{c}\\ 3.& \sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.& \sqrt[n]{a^n}=a\\ 5.& a\sqrt[n]{c}xb\sqrt[n]{d}=ab\sqrt[n]{cd}\\ 6.& \frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.& \sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.& \sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}\end{array}$.

$\text{ CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Sederhanakanlah bentuk akar berikut}\\ & \begin{array}{llllll}\text{a}.& \sqrt{8}& \text{f}.& \sqrt[3]{16}& \text{k}.& \sqrt{8x^5},x\ge 0\\ \text{b}.& \sqrt{12}& \text{g}.& \sqrt[3]{32}& \text{l}.& \sqrt{48x^6{y}^{11}},y\ge 0\\ \text{c}.& \sqrt{27}& \text{h}.& \sqrt[3]{54}& \text{m}.& 2\sqrt{8}\times \sqrt{3}\\ \text{d}.& \sqrt{28}& \text{i}.& \sqrt[3]{81}& \text{n}.& 3\sqrt{6}\times 2\sqrt{2}\\ \text{e}.& \sqrt{32}& \text{j}.& \sqrt[3]{625}& \text{o}.& 2\sqrt[3]{6}\times 6\sqrt[3]{9}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& \sqrt{8}=\sqrt{4\times 2}=\sqrt{2^2}\times \sqrt{2}=2\sqrt{2}\\ \text{b}.& \sqrt{12}=\sqrt{4\times 3}=\sqrt{2^2}\times \sqrt{3}=2\sqrt{3}\\ \text{c}.& \sqrt{27}=\sqrt{9\times 3}=\sqrt{3^2}\times \sqrt{3}=3\sqrt{3}\\ \text{d}.& \sqrt{28}=\sqrt{4\times 7}=\sqrt{2^2}\times \sqrt{7}=2\sqrt{7}\\ \text{e}.& \sqrt{32}=\sqrt{16\times 2}=\sqrt{4^2}\times \sqrt{2}=4\sqrt{2}\\ \text{f}.& \sqrt[3]{16}=\sqrt[3]{8\times 2}=\sqrt[3]{2^3}\times \sqrt[3]{2}=2\sqrt[3]{2}\\ \text{g}.& \sqrt[3]{32}=\sqrt[3]{8\times 4}=\sqrt[3]{2^3}\times \sqrt[3]{4}=2\sqrt[3]{4}\\ \text{h}.& \sqrt[3]{54}=\sqrt[3]{27\times 2}=\sqrt[3]{3^3}\times \sqrt[3]{2}=3\sqrt[3]{2}\\ \text{i}.& \sqrt[3]{81}=\sqrt[3]{27\times 3}=\sqrt[3]{3^3}\times \sqrt[3]{3}=3\sqrt[3]{3}\\ \text{j}.& \sqrt[3]{625}=\sqrt[3]{125\times 5}=\sqrt[3]{5^3}\times \sqrt[3]{5}=5\sqrt[3]{5}\\ \text{k}.& \sqrt{8x^5}=\sqrt{4.2.x^4.x^1}=\sqrt{2^2}\times \sqrt{2}\times \sqrt{x^4}\times \sqrt{x}\\ & =2.\sqrt{2}.x^2.\sqrt{x}=2x^2\sqrt{2x},x\ge 0\\ \text{l}.& \sqrt{48x^6{y}^{11}}=\sqrt{16.3.x^6.y^{10}.y^1}\\ & =\sqrt{4^2}\times \sqrt{3}\times \sqrt{x^6}\times \sqrt{y^{10}}\times \sqrt{y}\\ & =4\sqrt{3}.x^3.y^5.\sqrt{y}=4x^3{y}^5\sqrt{3y},\ge 0\\ \text{m}.& 2\sqrt{8}\times \sqrt{3}=2\sqrt{4\times 2}\times \sqrt{3}\\ & =2\sqrt{2^2}\times \sqrt{2}\times \sqrt{3}=2.2.\sqrt{2.3}\\ & =4\sqrt{6}\\ \text{n}.& 3\sqrt{6}\times 2\sqrt{2}=3\sqrt{2\times 3}\times 2\sqrt{2}\\ & =3\times 2\times \sqrt{2^2\times 3}=6\times \sqrt{2^2}\times \sqrt{3}\\ & =6\times 2\times \sqrt{3}=12\sqrt{3}\sqrt{6}\\ \text{o}.& 2\sqrt[3]{6}\times 6\sqrt[3]{9}=2.6.\sqrt[3]{6\times 9}=12\times \sqrt[3]{2.3.3.3}\\ & =12\times \sqrt[3]{{2.3}^3}=12\times \sqrt[3]{2}\times \sqrt[3]{3^3}\\ & =12\times \sqrt[3]{2}\times 3\\ & =36\sqrt[3]{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah pangkat rasional dari}\\ & \text{a}.\sqrt{y\sqrt[3]{x^{2}y}}\\ & \text{b}.\sqrt[3]{x^3\sqrt[5]{x^3\sqrt{x^3}}}\\ & \text{c}.\sqrt[3]{x^2\sqrt{x\sqrt[5]{x^2}}}\\ & \text{d}.xyz\sqrt[3]{{\displaystyle \frac{xy}{z^5}}}\sqrt[3]{{\displaystyle \frac{xz}{y^5}}}\sqrt[3]{{\displaystyle \frac{yz}{x^5}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& \sqrt{y\sqrt[3]{x^{2}y}}=\sqrt{y{\left(x^{2}y\right)}^{\frac{1}{3}}}={\left(y{\left(x^{2}y\right)}^{\frac{1}{3}}\right)}^{\frac{1}{2}}\\ & =y^{{.}^{\frac{1}{2}}}.x^{{.}^{\frac{2}{3}.\frac{1}{2}}}.y^{{.}^{\frac{1}{3}.\frac{1}{2}}}=y^{{.}^{\frac{1}{2}+\frac{1}{6}}}{x}^{{.}^{\frac{1}{3}}}=x^{{.}^{\frac{1}{3}}}.y^{{.}^{\frac{4}{6}}}\\ & =x^{{.}^{\frac{1}{3}}}.y^{{.}^{\frac{2}{3}}}\\ \text{b}.& \sqrt[3]{x^3\sqrt[5]{x^3\sqrt{x^3}}}=\sqrt[3]{x^3\sqrt[5]{x^3.x^{{.}^{\frac{3}{2}}}}}=\sqrt[3]{x^3.x^{{.}^{\frac{3}{5}}}{x}^{{.}^{\frac{3}{2.5}}}}\\ & =x^{{.}^{\frac{3}{3}}}.x^{{.}^{\frac{3}{5.3}}}.x^{{.}^{\frac{3}{2.5.3}}}=x^1+x^{{.}^{\frac{1}{5}}}.x^{{.}^{\frac{1}{10}}}\\ & =x^{{.}^{1+\frac{1}{5}+\frac{1}{10}}}=x^{{.}^{\frac{10+2+1}{10}}}=x^{{.}^{\frac{13}{10}}}\\ \text{c}.& \sqrt[3]{x^2\sqrt{x\sqrt[5]{x^2}}}=\sqrt[3]{x^2\sqrt{x.x^{{.}^{\frac{2}{5}}}}}=\sqrt[3]{x^2.x^{{.}^{\frac{1}{2}}}.x^{{.}^{\frac{2}{5.2}}}}\\ & =x^{{.}^{\frac{2}{3}}}.x^{{.}^{\frac{1}{2.3}}}.x^{{.}^{\frac{2}{5.2.3}}}=x^{{.}^{\frac{2}{3}}}.x^{{.}^{\frac{1}{6}}}.x^{{.}^{\frac{1}{15}}}\\ & =x^{{.}^{\frac{20+5+2}{30}}}=x^{{.}^{\frac{27}{30}}}=x^{{.}^{\frac{9}{10}}}\\ \text{d}.& xyz\sqrt[3]{{\displaystyle \frac{xy}{z^5}}}\sqrt[3]{{\displaystyle \frac{xz}{y^5}}}\sqrt[3]{{\displaystyle \frac{yz}{x^5}}}=xyz\sqrt[3]{{\displaystyle \frac{x^2{y}^2{z}^2}{x^5{y}^5{z}^5}}}\\ & =xyz\sqrt[3]{{\displaystyle \frac{1}{x^{(5-2)}{y}^{(5-2)}{z}^{(5-2)}}}}\\ & =xyz\sqrt[3]{{\displaystyle \frac{1}{x^3{y}^3{z}^3}}}=xyz\sqrt[3]{{\displaystyle \frac{1}{(xyz{)}^3}}}\\ & =xyz.{\displaystyle \frac{1}{xyz}={\displaystyle \frac{xyz}{xyz}=1}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Jika}a,b\text{bilangan positif dan}\\ & \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=a^x.b^y,\text{tentukan nilai}x-y\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=a^x.b^y\\ & \text{perhatikan cara menguraikannya}\\ & \sqrt{a^{2}b\sqrt[3]{a{b}^2\sqrt{ab}}}=\sqrt{a^{2}b\sqrt[3]{a{b}^2.a^{{.}^{\frac{1}{2}}}{b}^{{.}^{\frac{1}{2}}}}}\\ & =\sqrt{a^{2}b\sqrt[3]{a^{{.}^{1+\frac{1}{2}}}{b}^{{.}^{2+\frac{1}{2}}}}}=\sqrt{a^{2}b\sqrt[3]{a^{{.}^{\frac{3}{2}}}{b}^{{.}^{\frac{5}{2}}}}}\\ & =\sqrt{a^{2}b.a^{{.}^{\frac{3}{2.3}}}{b}^{{.}^{\frac{5}{2.3}}}}=\sqrt{a^{2}b.a^{{.}^{\frac{1}{2}}}{b}^{{.}^{\frac{5}{6}}}}\\ & =\sqrt{a^{{.}^{2+\frac{1}{2}}}.b^{{.}^{1+\frac{5}{6}}}}=\sqrt{a^{{.}^{\frac{5}{2}}}{b}^{{.}^{\frac{11}{6}}}}=a^{{.}^{\frac{5}{2.2}}}{b}^{{.}^{\frac{11}{6.2}}}\\ & =a^{{.}^{\frac{5}{4}}}{b}^{{.}^{\frac{11}{12}}}\\ & =a^x{b}^y\\ & \text{maka}x={\displaystyle \frac{5}{4},\text{dan}y=\frac{11}{12}}\\ & x-y={\displaystyle \frac{5}{4}-\frac{11}{12}={\displaystyle \frac{15-11}{12}=\frac{4}{12}={\displaystyle \frac{1}{3}}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& (\mathbf{\text{Matematika Dasar UM UGM 2008}})\\ & \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\sqrt[6]{x^2}\sqrt[3]{x^2\sqrt{x+1}}}{x\sqrt[6]{x+1}}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt[6]{x^2}\sqrt[3]{x^2\sqrt{x+1}}}{x\sqrt[6]{x+1}}}\\ & ={\displaystyle \frac{\sqrt[6]{x^2}.\sqrt[3.2]{{(x^2.\sqrt{x+1})}^2}}{\sqrt[6]{x^6}.\sqrt[6]{x+1}}={\displaystyle \frac{\sqrt[6]{x^2}.\sqrt[6]{x^4.(x+1)}}{\sqrt[6]{x^6(x+1)}}}}\\ & ={\displaystyle \frac{\sqrt[6]{x^{(2+4)}(x+1)}}{\sqrt[6]{x^6(x+1)}}={\displaystyle \frac{\sqrt[6]{x^6(x+1)}}{\sqrt[6]{x^6(x+1)}}}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Nilai dari}\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}=....\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ & \text{berikut uraiannya}\\ & \text{Misalkan}\\ & x^2=x^2,\text{maka}{x}^2=1+(x^2-1)\\ & x^2=1+(x-1)(x+1)\\ & x^2=1+(x-1)\sqrt{(x+1{)}^2}\\ & x^2=1+(x-1)\sqrt{1+((x+1{)}^2-1)}\\ & x^2=1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ & x^2=1+(x-1)\sqrt{1+x(x+2)}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{(x+2{)}^2}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+((x+2{)}^2-1)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{(x+3{)}^2}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+((x+3{)}^2-1)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+3-1)(x+3+1)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)(x+4)}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{(x+4{)}^2}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+((x+4{)}^2-1)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+4-1)(x+4+1)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)(x+5)}}}}\\ & x^2=1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\sqrt{...}}}}}\\ & x=\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+(x+3)\cdots }}}}}\\ & \text{maka}\\ & \cdots =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}}\\ & \text{Jelas tampak bahwa nilai}x\text{yang memenuhi adalah}\\ & x=3\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakanlah bentuk}\\ & \text{a}.\sqrt{3+2\sqrt{2}}\text{d}.\sqrt{21-4\sqrt{5}}\\ & \text{b}.\sqrt{6-\sqrt{32}}\text{e}.\sqrt{6-2\sqrt{8}}\\ & \text{c}.\sqrt{7+4\sqrt{3}}\text{f}.\sqrt{5+\sqrt{24}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{t bahwa}:\sqrt{a+b\pm 2\sqrt{ab}}=\sqrt{a}\pm \sqrt{b},a\ge b\\ \text{a}.& \sqrt{3+2\sqrt{2}}=\sqrt{2+1+2\sqrt{2.1}}=\sqrt{2}+1\\ \text{b}.& \sqrt{6-\sqrt{32}}=\sqrt{6-\sqrt{4.4.2}}=\sqrt{4+2-2\sqrt{4.2}}\\ & =\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \text{c}.& \sqrt{7+4\sqrt{3}}=\sqrt{4+3+2.2\sqrt{3}}=\sqrt{4+3+2\sqrt{4.3}}\\ & =\sqrt{4}+\sqrt{3}=2+\sqrt{3}\\ \text{d}.& \sqrt{21-4\sqrt{5}}=\sqrt{20+1-2.2\sqrt{5}}=\sqrt{20+1-2\sqrt{4.5}}\\ & =\sqrt{20+1-2\sqrt{20.1}}=\sqrt{20}-1=\sqrt{4.5}-1=2\sqrt{5}-1\\ \text{e}.& \sqrt{6-2\sqrt{8}}=\sqrt{4+2-2\sqrt{4.2}}=\sqrt{4}-\sqrt{2}=2-\sqrt{2}\\ \text{f}.& \sqrt{5+\sqrt{24}}=\sqrt{3+2+\sqrt{4.3.2}}=\sqrt{3+2+2\sqrt{3.2}}\\ & =\sqrt{3}+\sqrt{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Sederhanakan bentuk berikut}\\ & \text{a}.\sqrt{0,3+\sqrt{0,08}}\\ & \text{b}.\sqrt{94+2\sqrt{2013}}\\ & \text{c}.\sqrt{17+4\sqrt{15}}=a\sqrt{3}+b\sqrt{5},\text{tentukan}b-a\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{t bahwa}:\sqrt{p+q\pm 2\sqrt{pq}}=\sqrt{p}\pm \sqrt{q},p\ge q\\ \text{a}.& \sqrt{0,3+\sqrt{0,08}}=\sqrt{0,3+\sqrt{4.(0,02)}}=\sqrt{0,3+2\sqrt{0,02}}\\ & =\sqrt{0,2+0,1+2\sqrt{(0,2).(0,1)}}=\sqrt{0,2}+\sqrt{0,1}\\ \text{b}.& \sqrt{94+2\sqrt{2013}}=\sqrt{61+33+2\sqrt{61.33}}=\sqrt{61}+\sqrt{33}\\ \text{c}.& \sqrt{17+4\sqrt{15}}=\sqrt{17+2.2\sqrt{15}}=\sqrt{17+2\sqrt{4.15}}\\ & =\sqrt{17+2\sqrt{60}}=\sqrt{12+5+2\sqrt{12.5}}=\sqrt{12}+\sqrt{5}\\ & =\sqrt{4.3}+\sqrt{1.5}=2\sqrt{3}+1\sqrt{5}=a\sqrt{3}+b\sqrt{5}\{\begin{array}{ll}a& =2\\ b& =1\end{array}\\ & \text{maka}b-a=1-2=-1\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Bentuk paling sederhana dari}\\ & \sqrt[4]{49-20\sqrt{6}}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sqrt[4]{49-20\sqrt{6}}\\ & =\sqrt{\sqrt{49-2.10\sqrt{6}}}=\sqrt{\sqrt{49-2\sqrt{100.6}}}\\ & =\sqrt{\sqrt{49-2\sqrt{600}}}=\sqrt{\sqrt{25+24-2\sqrt{25.24}}}\\ & =\sqrt{\sqrt{25}-\sqrt{24}}=\sqrt{5-\sqrt{24}}=\sqrt{5-\sqrt{4.6}}\\ & =\sqrt{5-2\sqrt{6}}=\sqrt{3+2-2\sqrt{3.2}}=\sqrt{3}-\sqrt{2}\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Bentuk paling sederhana dari}\\ & {\left(\sqrt{52+6\sqrt{43}}\right)}^3-{\left(\sqrt{52-6\sqrt{43}}\right)}^3\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Inga}& \text{tlah bentuk}:(A-B{)}^3=A^3-B^3-3AB(A-B)\\ & \iff A^3-B^3=(A-B{)}^3+3AB(A-B)\\ \bullet & \sqrt{52+6\sqrt{43}}=\sqrt{52+2.3\sqrt{43}}=\sqrt{52+2\sqrt{43.9}}\\ & =\sqrt{43+9+2\sqrt{43.9}}=\sqrt{43}+\sqrt{9}=\sqrt{43}+3\\ \bullet & \sqrt{52-6\sqrt{43}}=\sqrt{52-2.3\sqrt{43}}=\sqrt{52-2\sqrt{43.9}}\\ & =\sqrt{43+9-2\sqrt{43.9}}=\sqrt{43}-\sqrt{9}=\sqrt{43}-3\\ & \text{misalkan}\{\begin{array}{ll}A& =\sqrt{43}+3\\ B& =\sqrt{43}-3\end{array}\\ & A^3-B^3=(A-B{)}^3+3AB(A-B)\\ & ={(\sqrt{43}+3-(\sqrt{43}-3))}^3+3(\sqrt{43}+3)(\sqrt{43}-3)(\sqrt{43}+3-(\sqrt{43}-3))\\ & ={\left(6\right)}^3+3({\sqrt{43}}^2-3^2)\left(6\right)\\ & =216+18(43-9)\\ & =216+18.34\\ & =216+612\\ & =828\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Sederhanakanlah bentuk akar berikut}\\ & \begin{array}{ll}\text{a}.& 3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ \text{b}.& 5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ \text{c}.& 3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ \text{d}.& (2+\sqrt{2})(4-\sqrt{2})\\ \text{e}.& (3\sqrt{2}+\sqrt{3})(\sqrt{2}-2\sqrt{3})\\ \text{f}.& (4\sqrt{3}-3\sqrt{5})(2\sqrt{3}+\sqrt{5})\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{l}\\ \text{a}.& 3\sqrt{2}+5\sqrt{8}-\sqrt{32}\\ & =3\sqrt{2}+5\sqrt{4.2}-\sqrt{16.2}\\ & =3\sqrt{2}+5.2\sqrt{2}-4\sqrt{2}\\ & =(3+10-4)\sqrt{2}=9\sqrt{2}\\ \text{b}.& 5\sqrt{3}+5\sqrt{27}-2\sqrt{75}\\ & =5\sqrt{3}+5\sqrt{9.3}-2\sqrt{25.3}\\ & =5\sqrt{3}+5.3\sqrt{3}-2.5\sqrt{3}\\ & =(5+15-10)\sqrt{3}=10\sqrt{3}\\ \text{c}.& 3\sqrt{50}-4\sqrt{32}-\sqrt{2}\\ & =3\sqrt{25.2}-4\sqrt{16.2}-\sqrt{1.2}\\ & =3.5\sqrt{2}-4.4\sqrt{2}-1\sqrt{2}\\ & =(15-16-1)\sqrt{2}=-2\sqrt{2}\\ \text{d}.& (2+\sqrt{2})(4-\sqrt{2})\\ & =2.4-2.\sqrt{2}+4.\sqrt{2}-\sqrt{2.2}\\ & =8+(4-2)\sqrt{2}-2\\ & =6+2\sqrt{2}\\ \text{e}.& (3\sqrt{2}+\sqrt{3})(\sqrt{2}-2\sqrt{3})\\ & =3\sqrt{2.2}-3.2.\sqrt{2.3}+\sqrt{3.2}-2\sqrt{3.3}\\ & =3.2-6\sqrt{6}+1\sqrt{6}-3.2\\ & =6-6+(1-6)\sqrt{6}=-5\sqrt{6}\\ \text{f}.& (4\sqrt{3}-3\sqrt{5})(2\sqrt{3}+\sqrt{5})\\ & =4.2.\sqrt{3.3}+4\sqrt{3.5}-3.2.\sqrt{5.3}-3\sqrt{5.5}\\ & =8.3+4\sqrt{15}-6\sqrt{15}-3.5\\ & =24-15+(4-6)\sqrt{15}=9-2\sqrt{15}\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M., Nurdiansyah, H., Akhmad, G. 2016. Matematika untuk Siswa SMA/MA Kelas X Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Bandung: YRAMA WIDYA.
2. 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.

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.

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:

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

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

$\text{ Contoh Soal}$

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

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

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

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

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

$(6).(-2{)}^6=(-2)\times (-2)\times (-2)\times (-2)\times (-2)\times (-2)=64$

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

$(8).{(-{\displaystyle \frac{1}{2}})}^4=(-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})\times (-{\displaystyle \frac{1}{2}})=-{\displaystyle \frac{1}{32}}$

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

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

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.

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

Sifat-sifat yang berlaku pada operasi bilangan bentuk akar

$\begin{array}{rl}& \\ 1.& a\sqrt[n]{c}+b\sqrt[n]{c}=(a+b)\sqrt[n]{c}\\ 2.& a\sqrt[n]{c}-b\sqrt[n]{c}=(a-b)\sqrt[n]{c}\\ 3.& \sqrt[n]{a}.\sqrt[n]{b}=\sqrt[n]{ab}\\ 4.& \sqrt[n]{a^n}=a\\ 5.& a\sqrt[n]{c}xb\sqrt[n]{d}=ab\sqrt[n]{cd}\\ 6.& \frac{a\sqrt[n]{c}}{b\sqrt[n]{d}}=\frac{a}{b}.\sqrt[n]{\frac{c}{d}}\\ 7.& \sqrt{(a+b)+2\sqrt{ab}}=\sqrt{a}+\sqrt{b}\\ 8.& \sqrt{(a+b)-2\sqrt{ab}}=\sqrt{a}-\sqrt{b}\end{array}$

Merasionalkan penyebut

Jika suatu pecahan penyebutnya mengandung bilangan irasional atau bentuk akar, maka penyebut ini dapat dibuat menjadi bilangan rasional. Perhatikanlah langkah berikut

$\begin{array}{rl}1.& {\displaystyle \frac{a}{\sqrt{b}}=\frac{a}{\sqrt{b}}\times \frac{\sqrt{b}}{\sqrt{b}}=\frac{a\sqrt{b}}{\left(\sqrt{b^2}\right)}=\frac{a}{b}\sqrt{b}}\\ 2.& {\displaystyle \frac{a}{\sqrt[3]{b}}=\frac{a}{\sqrt[3]{b}}\times \frac{\sqrt[3]{b^2}}{\sqrt[3]{b^2}}=\frac{a\sqrt[3]{b^2}}{\left(\sqrt[3]{b^3}\right)}=\frac{a}{b}\sqrt[3]{b^2}}\\ 3.& {\displaystyle \frac{a}{\sqrt[5]{b^3}}={\displaystyle \frac{a}{\sqrt[5]{b^3}}\times \frac{\sqrt[5]{b^2}}{\sqrt[5]{b^2}}=\frac{a\sqrt[5]{b^2}}{\sqrt[5]{b^5}}=\frac{a}{b}\sqrt[5]{b^2}}}\end{array}$

Merasionalkan di atas adalah contoh bebrapa contoh model merasionalkan jika berjenis tunggal tetapi jika nanti jenisnya lebih dari itu, maka perhatikanlah simulasi contoh berikut

$\begin{array}{rl}& \\ 1.& {\displaystyle \frac{c}{a+\sqrt{b}}=\frac{c}{a+\sqrt{b}}.\frac{a-\sqrt{b}}{a-\sqrt{b}}=\frac{c(a-\sqrt{b})}{a^2-b}}\\ 2.& {\displaystyle \frac{c}{a-\sqrt{b}}=\frac{c}{a-\sqrt{b}}.\frac{a+\sqrt{b}}{a+\sqrt{b}}=\frac{c(a+\sqrt{b})}{a^2-b}}\\ 3.& {\displaystyle \frac{c}{\sqrt{a}+\sqrt{b}}=\frac{c}{\sqrt{a}+\sqrt{b}}.\frac{\sqrt{a}-\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\frac{c(\sqrt{a}-\sqrt{b})}{a-b}}\end{array}$

Perhatikanlah simulasi contoh di atas, bentuk $a+\sqrt{b}$ memiliki bentuk sekawan (irasional juga) $a-\sqrt{b}$, demikian juga bentuk $\sqrt{a}+\sqrt{b}$ memiliki sekawan $\sqrt{a}-\sqrt{b}$. Disamping itu ada bentuk khusus yatu bentuk  $\sqrt[3]{a}+\sqrt[3]{b}$ memiliki bentuk sekawan $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$.