Induksi Matematika (Kelas XI Matematika Wajib)

$\text{A. Pendahuluan}$

Misalkan kita menjumlahkan 100 bilangan ganjil pertama (anggap saja sebagai penjumlahan suku pertama sampai suku ke seratus) yaitu : 1+3+5+...+199, maka untuk memudahkannya kita dapat menentukan cara menjumlahkan dengan atau menurut pola tertentu sebagaimana ilstrasi berikut ini

$\begin{array}{rl}1& =1^2=S_1\\ 1+3& =2^2=S_2\\ 1+3+5& =3^2=S_3\\ 1+3+5+7& =4^2=S_4\\ 1+3+5+7+9& =5^2=S_5\\ ⋮& \\ 1+3+5+7+9+\cdots +(2n-1)& =n^2=S_n\\ ⋮& \\ 1+3+5+7+9+\cdots +199& ={100}^2=S_{100}\end{array}$

$\text{B. Induksi Matematika}$

Pola bilangan tertentu dalam matematika sebagaimana misal contoh di atas dapat ditarik suatu bentuk umum. Selanjutnya untuk membuktikan bahwa suatu bentuk umum dari sebuah pernyataan berlaku, kita dapat menggunakan Induksi Matematika ini. Tentunya semunya dari pernyataan tersebut harus memenuhi kriteria tertentu. Sehingga Induksi Matematika dapat juga disebutkan sebagai proses pembuktian pernyataan (teorema) dari kejadian-kejadian khusus yang berlaku untuk setiap bilangan asli.

Dalam pembuktian dengan Induksi Matematika, perhatikanlah beberapa langkah-langkah ini

Misalkan $P(n)$ adalah suatu pernyataan yang akan dibuktikan kebenarannya untuk semua bilangan asli $n$, maka

$\begin{array}{rl}& \mathbf{\text{Langkahnya:}}\\ & (1)\text{buktikan}P(1)\text{benar untuk}n=1\\ & \text{Selanjutnya disebut Langkah Basis}\\ & (2)\text{Asumsikan pernyataan berlaku untuk}n=k,\\ & \text{yaitu}P(k)\text{benar, dengan}k\in A,\\ & \text{maka untuk}n=k+1\text{bahwa}P(k+1)\\ & \text{juga benar}\\ & \text{Selanjutnya disebut Langkah Induksi}\\ & (3)\text{Setelah langkah}(1)\text{dan}(2)\text{terpenuhi}\\ & \text{atau benar, maka dapat disimpulkan bahwa}\\ & P(n)\text{benar untuk setiap}n\\ & \text{Selanjutnya disebut Konkulsi}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{rl}(1)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 1+2+3+...+n={\displaystyle \frac{n(n+1)}{2}}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 1+2+3+...+n={\displaystyle \frac{n(n+1)}{2}}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}1={\displaystyle \frac{1(1+1)}{2}}\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv 1+2+3+...+k={\displaystyle \frac{k(k+1)}{2},\text{maka}}\\ & \text{untuk}n=k+1:\\ & 1+2+3+\cdots +(k)+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}\\ & \underset{{\displaystyle \frac{k(k+1)}{2}}}{\underbrace{1+2+3+\cdots +(k)}}+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}\\ & {\displaystyle \frac{k(k+1)}{2}+(k+1)={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle (k+1)(\frac{k}{2}+1)={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle (k+1)\frac{k+2}{2}={\displaystyle \frac{(k+1)(k+2)}{2}}}\\ & {\displaystyle \frac{(k+1)(k+2)}{2}={\displaystyle \frac{(k+1)(k+2)}{2}\equiv P(k+1)}}\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(2)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 1+3+5+7+...+(2n-1)=n^2\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 1+3+5+...+(2n-1)=n^2\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}2.1-1=1^2\iff 1=1\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv 1+3+5+...+(2k-1)=k^2,\text{maka}\\ & \text{untuk}n=k+1:\\ & 1+3+5+\cdots +(2k-1)+(2(k+1)-1)=(k+1{)}^2\\ & \underset{{\displaystyle k^2}}{\underbrace{1+3+5+\cdots +(2k-1)}}+(2(k+1)-1)=(k+1{)}^2\\ & k^2+(2(k+1)-1)=(k+1{)}^2\\ & k^2+(2k+2-1)=(k+1{)}^2\\ & k^2+2k+1=(k+1{)}^2\\ & (k+1{)}^2=P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(3)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & 2n-3<2^{n-2}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv 2n-3<2^{n-2}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}2.1-3<2^{1-2}\\ & \text{demikian pula untuk}n=2\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv (2k-3)<2^{k-2},\text{maka}\\ & \text{untuk}n=k+1:\\ & 2(k+1)-3=2k+2-3=(2k-3)+2<2^{k-2}+2\\ & \text{sehingga}\\ & (2k-3)+2<2^{k-2}+2<2^{k-2}+2^{k-2},\text{untuk}k\ge 3\\ & (2k-3)+2<{2.2}^{k-2}\\ & (2k-3)+2<2^{(k+1)-2}\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}(4)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & (1+h{)}^n\ge 1+nh\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv (1+h{)}^n\ge 1+nh\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena}(1+h{)}^1\ge 1+1h\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv (1+h{)}^k\ge 1+kh,\text{maka}\\ & \text{untuk}n=k+1:\\ & (1+h{)}^{k+1}\ge (1+kh)(1+h)\\ & (1+h{)}^{k+1}\ge (1+(k+1)h+k{h}^2)\\ & (1+h{)}^{k+1}\ge 1+(k+1)h\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\begin{array}{rl}.& \mathbf{\text{Catatan}}\\ & \text{untuk}k=2\\ & (1+h{)}^2=1+2h+h^2\ge 1+2h,\text{maka}\\ & (1+h{)}^n\ge 1+nh\end{array}$

$\begin{array}{rl}(5)& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ & \sum _{h=1}^n{3}^h={\displaystyle \frac{3}{2}(3^n-1)}\\ \\ & \mathbf{\text{Bukti}}\end{array}$

$\begin{array}{rl}.\text{Diket}& \text{ahui}P(n)\equiv \sum _{h=1}^n{3}^h={\displaystyle \frac{3}{2}(3^n-1)}\\ (\text{a})& \text{Langkah basis}\\ & P(1)\text{benar, karena untuk}n=1\\ & 3^1={\displaystyle \frac{3}{2}(3^1-1)}\\ (\text{b})& \text{Langkah induksi}\\ & \text{Misalkan rumus berlaku untuk}n=k,\text{yaitu}\\ & P(k)\equiv \sum _{h=1}^k{3}^h={\displaystyle \frac{3}{2}(3^k-1),\text{maka}}\\ & \text{untuk}n=k+1:\\ & \sum _{h=1}^{k+1}{3}^h=\sum _{h=1}^k{3}^h+\sum _{h=k+1}^{k+1}{3}^h\\ & ={\displaystyle \frac{3}{2}(3^k-1)+3^{k+1}}\\ & ={\displaystyle \frac{1}{2}(3^{k+1}-3+{2.3}^{k+1})}\\ & ={\displaystyle \frac{1}{2}({3.3}^{k+1}-3)}\\ & ={\displaystyle \frac{3}{2}({.3}^{k+1}-1)}\\ & \text{maka rumus berlaku untuk}P(k+1)\\ & \text{Jadi},P(n)\text{benar untuk}n=k+1\\ (\text{c})& \text{Kesimpulan/Konklusi}\\ & \text{Jadi},P(n)\text{berlaku untuk}\mathrm{\forall }n\in \mathbb{N}.\end{array}$

$\boxed{\text{LATIHAN SOAL}}$

$\begin{array}{rl}.& \text{Buktikanlah dengan induksi matematika}\\ & \text{untuk}n\text{bilangan asli berlaku}\\ \end{array}$

$.\begin{array}{l}\\ 1.& 2+4+6+8+...+2n=n^2+n\\ 2.& 1^2+2^2+3^2+...+n^2={\displaystyle \frac{1}{6}n(n+1)(2n+1)}\\ 3.& 1^3+2^3+3^3+...+n^3={\displaystyle \frac{1}{4}{n}^2(n+1{)}^2}\\ 4.& 1.2+2.3+3.4+...+n(n+1)={\displaystyle \frac{1}{3}n(n+1)(n+2)}\\ 5.& {\displaystyle \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n(n+1)}={\displaystyle \frac{n}{n+1}}}\\ 6.& {\displaystyle \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{(2n-1)(2n+1)}={\displaystyle \frac{n}{2n+1}}}\\ 7.& {\displaystyle \frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{n(n+1)(n+2)}={\displaystyle \frac{n(n+3)}{4(n+1)(n+2)}}}\\ 8.& 1+{\displaystyle \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}<2\sqrt{n+1}-2}\\ 9.& n^3-n\text{habis dibagi oleh}3\\ 10.& n^5-n\text{habis dibagi oleh}5\\ 11.& 5^n+{6.7}^n+1\text{habis dibagi oleh}4\\ 12.& 5^{2n}-1\text{habis dibagi oleh}3\\ 13.& 3^n-1\ge 2^n\\ 14.& 2n+7<(n+3{)}^2\\ 15.& 2+4+6+8+...+2n\le 2^n\\ 16.& {(3+\sqrt{5})}^n+{(3-\sqrt{5})}^n\text{habis dibagi oleh}{2}^n\end{array}$

DAFTAR PUSTAKA

1. Budhi, W.S., 2018. Bupena Mathematika SMA/MA Kelas XI Kelompok Wajib. Jakarta: Erlangga.
2. Kemendikbud. 2017. Matematika untuk SMA/MA/SMK Kelas XI Edisi Revisi. Jakarta: Kementerian Pendidikan Nasional.
3. Tampomas, H. 1999. SeribuPena Matematika Jilid 3 untuk SMU Kelas 3. Jakarta: Erlangga
4. Tim ITB. 2007. Program Pembinaan Kompetensi Siswa Bidang Matematika Tahap 1. Bandung: LPPM ITB

Persamaan Trigonometri (Matematika Peminatan Kelas XI)

$\text{A. Persamaan Trigonometri}$

Ada minimal 3 yang utama untuk persmaan trigonometri sederhana, yaitu:

$\begin{array}{rl}1.\sin x& =\sin \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k{.360}^{\circ }\\ & =({180}^{\circ }-\alpha )+k{.360}^{\circ }\end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$.\text{atau}$

$\begin{array}{rl}.\sin x& =\sin \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k.2\pi \\ & =(\pi -\alpha )+k.2\pi \end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$\begin{array}{rl}2.\cos x& =\cos \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k{.360}^{\circ }\\ & =-\alpha +k{.360}^{\circ }\end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$.\text{atau}$

$\begin{array}{rl}.\cos x& =\cos \alpha \\ x& =\{\begin{array}{ll}& =\alpha +k.2\pi \\ & =-\alpha +k.2\pi \end{array}\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$\begin{array}{rl}3.\tan x& =\tan \alpha \\ x& =\alpha +k{.180}^{\circ }\\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$.\text{atau}$

$\begin{array}{rl}.\tan x& =\tan \alpha \\ x& =\alpha +k.\pi \\ \text{deng}& \text{an}\\ k& \in \mathbb{Z}\end{array}$

$\boxed{\text{CONTOH SOAL}}$

$\begin{array}{l}\\ 1.& \text{Tentukanlah himpunan penyelesaian dari }\\ & \text{persamaan-persamaan trigonometri berikut}\\ & \text{ini untuk}{0}^{\circ }\le x\le {360}^{\circ }\\ & \begin{array}{l}\\ \text{a}.& \sin x={\displaystyle \frac{1}{2}}& \text{f}.& \tan x=-{\displaystyle \frac{1}{3}\sqrt{3}}& \text{k}.& \sin 2x={\displaystyle \frac{1}{2}}\\ \text{b}.& \cos x={\displaystyle \frac{1}{2}\sqrt{3}}& \text{g}& 2\cos x=-\sqrt{3}& \text{l}.& \cos 2x=-{\displaystyle \frac{1}{2}\sqrt{3}}\\ \text{c}.& \tan x=\sqrt{3}& \text{h}& 3\tan x=\sqrt{3}& \text{m}.& \tan 2x=\sqrt{3}\\ \text{d}.& \sin x=-1& \text{i}.& \sin x=\sin {46}^{\circ }& \text{n}.& \sin (2x-{30}^{\circ })=\sin {45}^{\circ }\\ \text{e}.& \cos x=-{\displaystyle \frac{1}{2}\sqrt{2}}& \text{j}.& \cos x=\cos {93}^{\circ }& \text{o}.& \sin (2x+{60}^{\circ })=\sin {90}^{\circ }\end{array}\end{array}$

$.\text{Jawab}:$

$\begin{array}{rl}.\text{a}.\sin x& ={\displaystyle \frac{1}{2}}\\ \sin x& =\sin {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{30}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ {150}^{\circ }(\text{memenuhi})& \end{array}\\ k=1& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{{30}^{\circ },{150}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{b}.\cos x& ={\displaystyle \frac{1}{2}\sqrt{3}}\\ \cos x& =\cos {30}^{\circ }\\ x& =\{\begin{array}{ll}{30}^{\circ }& +k{.360}^{\circ }\\ -{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}:\\ x& =\{\begin{array}{ll}{30}^{\circ }(\text{memenuhi})& \\ -{30}^{\circ }(\text{tidak memenuhi})& \end{array}\\ k=1& \\ x& =\{\begin{array}{ll}{30}^{\circ }+{360}^{\circ }={390}^{\circ }(\text{tidak memenuhi})& \\ -{30}^{\circ }+{360}^{\circ }={330}^{\circ }(\text{memenuhi})& \end{array}\\ \text{HP}& =\{{30}^{\circ },{330}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{c}.\tan x& =\sqrt{3}\\ \tan x& =\tan {60}^{\circ }\\ x& ={60}^{\circ }+k{.180}^{\circ }\\ k=0& \text{diperoleh}:\\ x& ={60}^{\circ }\text{memenuhi}\\ k=1& \\ x& ={60}^{\circ }+{180}^{\circ }={240}^{\circ }\text{memenuhi}\\ k=2& \\ x& ={60}^{\circ }+{360}^{\circ }={420}^{\circ }\text{tidak memenuhi}\\ \text{HP}& =\{{60}^{\circ },{240}^{\circ }\}\end{array}$

$\begin{array}{rl}.\text{d}.\sin x& =-1\\ \sin x& =\sin {270}^{\circ }\\ x& =\{\begin{array}{ll}{270}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{270}^{\circ })& +k{.360}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}{270}^{\circ }& \text{memenuhi}\\ -{90}^{\circ }& \text{tidak memenuhi}\end{array}\\ k=1& \text{tidak memenuhi semuanya}\\ \text{HP}& =\left\{{270}^{\circ }\right\}\end{array}$

$\begin{array}{rl}.\text{n}.\sin (2x-{30}^{\circ })& =\sin {45}^{\circ }\\ (2x-{30}^{\circ })& =\{\begin{array}{ll}{45}^{\circ }& +k{.360}^{\circ }\\ ({180}^{\circ }-{45}^{\circ })& +k{.360}^{\circ }\end{array}\\ 2x& =\{\begin{array}{ll}{45}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\\ {135}^{\circ }+{30}^{\circ }& +k{.360}^{\circ }\end{array}\\ x& =\{\begin{array}{ll}37,5^{\circ }& +k{.180}^{\circ }\\ 82,5^{\circ }& +k{.180}^{\circ }\end{array}\\ k=0& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }& \\ 82,5^{\circ }& \end{array}\\ k=1& \text{diperoleh}\\ x& =\{\begin{array}{ll}37,5^{\circ }+{180}^{\circ }& =217,5^{\circ }\\ 82,5^{\circ }+{180}^{\circ }& =262,5^{\circ }\end{array}\\ k=2& \text{tidak ada yang memenuhi}\\ \text{HP}& =\{37,5^{\circ },82,5^{\circ },217,5^{\circ },262,5^{\circ }\}\end{array}$

Contoh Soal 10 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 46.& \text{Jumlah akar-akar persamaan}\\ & 5^{x+1}+5^{2-x}-30=0\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -2& & & \\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& 2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{5}^{x+1}+5^{2-x}-30& =0\\ \left(5^x\right){.5}^1+{\displaystyle \frac{5^2}{5^x}-30}& =0\\ 5{\left(5^x\right)}^2+25-30\left(5^x\right)& =0\\ \text{Persamaan kuadrat}& \text{dalam}{5}^x,\text{maka}\\ 5(5^x{)}^2-30(5^x)+25& =0\{\begin{array}{ll}a& =5\\ b& =-30\\ c& =25\end{array}\\ (5^{x_1}).\left(5^{x_2}\right)& ={\displaystyle \frac{c}{a}}\\ 5^{x_1+x_2}& ={\displaystyle \frac{25}{5}=5}\\ 5^{x_1+x_2}& =5^1\\ x_1+x_2& =1\end{array}\end{array}$

$\begin{array}{l}\\ 47.& \text{Jumlah akar-akar persamaan}\\ & {2020}^{x^2-7x+7}={2021}^{x^2-7x+7}\text{adalah}....\\ \\ & \begin{array}{l}\\ \text{a}.& -7\\ \text{b}.& -5\\ \text{c}.& -3\\ \text{d}.& 5\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{2020}^{x^2-7x+7}& ={2021}^{x^2-7x+7}\\ \text{Karena basis}& \text{tidak sama},\\ \text{maka harusl}& \text{ah pangkatnya}=0,\\ x^2-7x+7& =0\\ \text{dan jumlah}& \text{akar-akarnya adalah}:\\ x_1+x_2& =-{\displaystyle \frac{b}{a},\text{dari persamaan}}\\ x^2-7x+7& =0\{\begin{array}{ll}a& =1\\ b& =-7\\ c& =7\end{array}\\ \text{maka}{x}_1+x_2& =-{\displaystyle \frac{b}{a}=-\frac{-7}{1}=7}\end{array}\end{array}$

$\begin{array}{l}\\ 48.& \text{Nilai dari}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 5\\ \text{c}.& 10\\ \text{d}.& 20\\ \text{e}.& 40\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{2^{2020}+2^{2018}}{2^{2018}+2^{2016}}}& ={\displaystyle \frac{2^4{.2}^{2016}+2^2{.2}^{2016}}{2^2{.2}^{2018}+2^{2016}}}\\ & ={\displaystyle \frac{2^{2016}(2^4+2^2)}{2^{2016}(2^2+1)}}\\ & ={\displaystyle \frac{16+4}{4+1}}\\ & ={\displaystyle \frac{20}{5}}\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 49.& (\mathbf{\text{UM IPB}})\text{Jika}ab=a^b\text{dan}{\displaystyle \frac{a}{b}=a^{3b}}\\ & \text{maka nilai}a\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 0,5\\ \text{c}.& 1\\ \text{d}.& 0,25\\ \text{e}.& 0,75\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}\text{Diketahui}& \\ ab& =a^b\\ b& ={\displaystyle \frac{a^b}{a}=a^{b-1}.....\mathbf{\text{1}}}\\ \text{maka}& \\ {\displaystyle \frac{a}{b}}& =a^{3b}...............\mathbf{\text{2}}\\ \mathbf{\text{1}}& ke\mathbf{\text{2}}\\ {\displaystyle \frac{a}{a^{b-1}}}& =a^{3b}\\ a^{2-b}& =a^{3b}\\ 2-b& =3b\\ -4b& =-2\\ b& ={\displaystyle \frac{1}{2}................\mathbf{\text{3}}}\\ \mathbf{\text{3}}& ke\mathbf{\text{1}}\\ a\left({\displaystyle \frac{1}{2}}\right)& =a^{\frac{1}{2}}\\ {\displaystyle \frac{1}{4}{a}^2}& =a\\ a^2-4a& =0\\ a(a-4)& =0\\ a=0& \text{atau}a=4\end{array}\end{array}$

$\begin{array}{l}\\ 50.& \text{Jika}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}=4,\text{maka}x=....}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 1,1}\\ \text{b}.& {\displaystyle 1,2}\\ \text{c}.& 1,3\\ \text{d}.& {\displaystyle 1,4}\\ \text{e}.& 1,5\\ \\ & & & (\mathbf{\text{SAT Test Math Level 2}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle 3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}}& =4\\ {\left(3.x^{{}^{{}^{{}^{\frac{3}{2}}}}}\right)}^2& =4^2\\ 3^2.x^3& =4^2\\ x^3& ={\displaystyle \frac{4^2}{3^2}}\\ x^3& ={\displaystyle \frac{4^2}{3^2}\times \frac{3}{3}}\\ x^3& \le {\displaystyle \frac{4^2}{3^2}\times \frac{4}{3}}\\ x^3& \le \left({\displaystyle \frac{4^3}{3^3}}\right)\\ x^3& \le {\left({\displaystyle \frac{4}{3}}\right)}^3\\ x& \le {\displaystyle \frac{4}{3}}\\ x& \le 1,\overline{333}\\ x& \approx 1,3\end{array}\end{array}$

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

Pembahasan:

$\begin{array}{rl}{x}^8& =2207-{\displaystyle \underset{x^8}{\underbrace{{\displaystyle \frac{1}{2207-\frac{1}{2207-\frac{1}{2207-...}}}}}}}\\ x^8& =2207-{\displaystyle \frac{1}{x^8}}\\ x^8+{\displaystyle \frac{1}{x^8}}& =2207\\ {(x^4+{\displaystyle \frac{1}{x^4}})}^2& =2207+2\\ (x^4+{\displaystyle \frac{1}{x^4}})& =\sqrt{2209}=47\end{array}$

$\begin{array}{rl}{x}^4+{\displaystyle \frac{1}{x^4}}& =47\\ {(x^2+{\displaystyle \frac{1}{x^2}})}^2& =47+2\\ x^2+{\displaystyle \frac{1}{x^2}}& =\sqrt{49}=7\\ {(x+{\displaystyle \frac{1}{x}})}^2& =7+2\\ x+{\displaystyle \frac{1}{x}}& =\sqrt{9}=3\\ x^2-3x+1& =0,\\ & \text{persamaan kuadrat dalam x,}\\ & \mathbf{\text{gunakan rumus abc}}\\ x_{1,2}=& {\displaystyle \frac{3\pm \sqrt{5}}{2}={\displaystyle \frac{3\pm 1\sqrt{5}}{2}={\displaystyle \frac{a\pm b\sqrt{c}}{d}}}}\\ & \mathbf{\text{Sehingga}},\{\begin{array}{ll}& a=3\\ & b=1\\ & c=5\\ & d=2\end{array}\end{array}$

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. 

Contoh Soal 9 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 41.& \text{Penyelesaian pertidaksamaan eksponen}\\ & {\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}>\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x>{\displaystyle \frac{6}{5}}\\ \text{b}.& x<-{\displaystyle \frac{6}{5}}\\ \text{c}.& x>{\displaystyle \frac{5}{6}}\\ \text{d}.& x<-2\\ \text{e}.& x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{1}{3}}\right)}^{2x+1}& >\sqrt{{\displaystyle \frac{27}{3^{x-1}}}}\\ 3^{-(2x+1)}& >3^{\frac{1}{2}(3-(x-1))}\\ -(2x+1)& >{\displaystyle \frac{1}{2}(3-(x-1))}\\ -4x-2& >4-x\\ -4x+x& >4+2\\ -3x& >6\\ 3x& <-6\\ x& <-2\end{array}\end{array}$

$\begin{array}{l}\\ 42.& (\mathbf{\text{UMPTN 01}})\text{Nilai}x\text{yang memenuhi}\\ & 4^{x^2-x-2}{.2}^{x^2+3x-10}<{\displaystyle \frac{1}{16}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& x<-5\text{atau}x>-2\\ \text{b}.& x<-2\text{atau}x>{\displaystyle \frac{5}{3}}\\ \text{c}.& -2<x<-1\\ \text{d}.& -2<x<{\displaystyle \frac{5}{3}}\\ \text{e}.& -5<x<2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{4}^{x^2-x-2}{.2}^{x^2+3x-10}& <{\displaystyle \frac{1}{16}}\\ 2^{2(x^2-x-2)+(x^2+3x-10)}& <2^{-4}\\ 2(x^2-x-2)+(x^2+3x-10)& <-4\\ 3x^2-2x+3x-4-10+4& <0\\ 3x^2+x-10+& <0\\ (x+2)(3x-5)& <0\\ \therefore -2<x& <{\displaystyle \frac{5}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 43.& (\mathbf{\text{SPMB 04 Mat IPA}})\text{Himpunan Penyelesaian}\\ & \text{pertidaksamaan eksponen}\\ & 2\sqrt{4^{x^2-3x+2}}<\sqrt[3]{{\left({\displaystyle \frac{1}{2}}\right)}^{3-6x}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{x|x>4\}\\ \text{b}.& \{x|x>2\}\\ \text{c}.& \{x|x<1\}\\ \text{d}.& \{x|1<x<4\}\\ \text{e}.& \{x|2\le x\le 3\}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}2\sqrt{4^{x^2-3x+2}}& <\sqrt[3]{{\left({\displaystyle \frac{1}{2}}\right)}^{3-6x}}\\ 2^{1+\frac{2}{2}(x^2-3x+2)}& <2^{-\frac{3-6x}{3}}\\ 1+(x^2-3x+2)& <-{\displaystyle \frac{3-6x}{3}}\\ x^2-3x+3& <-1+2x\\ x^2-5x+4& <0\\ (x-1)(x-4)& <0\\ 1<x& <4\end{array}\end{array}$

$\begin{array}{l}\\ 44.& (\mathbf{\text{SBMPTN 2015 Mat IPA}})\\ & \text{Nilai}c\text{yang memenuhi}\\ & (0,12{)}^{4x^2+8x+c}<(0,0144{)}^{x^2+4x+4}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& c>0\\ \text{b}.& c>2\\ \text{c}.& c>4\\ \text{d}.& c>6\\ \text{e}.& c>8\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}(0,12{)}^{4x^2+8x+c}& <(0,0144{)}^{x^2+4x+4}\\ (0,12{)}^{4x^2+8x+c}& <(0,12{)}^{2(x^2+4x+4)}\\ 4x^2+8x+c& >2(x^2+4x+4)\\ 2x^2+c-8& >0\text{haruslah definit positif}\\ \text{Syaratnya}& \{\begin{array}{ll}a& =2>0\\ D& =b^2-4ac<0\end{array}\\ \text{Maka}D& =b^2-4ac<0\\ \mathbf{\text{ambil dari}}& 2x^2-c-8=0\{\begin{array}{ll}a& =2\\ b& =0\\ c& =c-8\end{array}\\ & =0^2-4.2(c-8)<0\\ -8c& +64<0\\ -8c& <-64\\ 8c& >64\\ c& >8\end{array}\end{array}$

$\begin{array}{l}\\ 45.& \text{Nilai}x\text{yang memenuhi}\\ & 9^{2x}-{10.9}^x+9>0,x\in \mathbb{R}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& x<1\text{atau}x>0\\ \text{b}.& x<0\text{atau}x>1\\ \text{c}.& x<-1\text{atau}x>2\\ \text{d}.& x<1\text{atau}x>2\\ \text{e}.& x<-1\text{atau}x>1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{9}^{2x}-{10.9}^x+9& >0\\ {\left(9^x\right)}^2-10.\left(9^x\right)+9& >0\\ (9^x-1)(9^x-9)& >0\\ 9^x<1\text{atau}{9}^x& >9\\ 9^x<9^0\text{atau}{9}^x& >9^1\\ x<0\text{atau}x& >1\end{array}\end{array}$

Contoh Soal 8 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 36.& \text{Bentuk sederhana dari}\\ & \sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}=....\\ \\ & (\mathbf{\text{SIMAK UI 2012 Mat IPA}})\\ & \begin{array}{l}\\ \text{a}.& 2-\sqrt{2}\\ \text{b}.& 8-\sqrt{2}\\ \text{c}.& -2+\sqrt{2}\\ \text{d}.& 2+5\sqrt{2}\\ \text{e}.& 8+5\sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{misalkan},\\ & \begin{array}{rl}x& =\sqrt{33+\sqrt{800}}-\sqrt{27-2\sqrt{162}}\\ & =(\sqrt{33+20\sqrt{2}}-\sqrt{27-2.9\sqrt{2}})\\ & =\sqrt{33+20\sqrt{2}}-\sqrt{27-18\sqrt{2}}\\ x^2& =33+20\sqrt{2}+27-18\sqrt{2}-2\sqrt{(33+20\sqrt{2})(27-18\sqrt{2})}\\ & =60+2\sqrt{2}-2\sqrt{33.27-33.18\sqrt{2}+27.20\sqrt{2}-20.18.2}\\ & =60+2\sqrt{2}-2\sqrt{891-720+540\sqrt{2}-594\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-54\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2.27\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2\sqrt{27.27}\sqrt{2}}\\ & =60+2\sqrt{2}-2\sqrt{171-2\sqrt{162.9}}\\ & =60+2\sqrt{2}-2\sqrt{162+9-2\sqrt{162.9}}\\ & =60+2\sqrt{2}-2(\sqrt{162}-\sqrt{9})\\ & =60+2\sqrt{2}-2(9\sqrt{2}-3)\\ x^2& =66-16\sqrt{2}\\ x& =\sqrt{66-2.8\sqrt{2}}\\ & =\sqrt{64+2-2\sqrt{64.2}}\\ & =\sqrt{64}-\sqrt{2}\\ & =8-\sqrt{2}\end{array}\end{array}$

$\begin{array}{l}\\ 37.& \text{Jika}{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}={\displaystyle \frac{a\sqrt{2}+b\sqrt{3}+c\sqrt{5}}{12},}}\\ \\ & \text{maka}a+b+c=....\\ & (\mathbf{\text{UM UGM 2016 Mat Das}})\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 1\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 4\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}}& ={\displaystyle \frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\times {\displaystyle \frac{(\sqrt{2}+\sqrt{3}-\sqrt{5})}{(\sqrt{2}+\sqrt{3}-\sqrt{5})}}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{{(\sqrt{2}+\sqrt{3})}^2-{\left(\sqrt{5}\right)}^2}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+3+2\sqrt{2.3}-5}}\\ & ={\displaystyle \frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2\sqrt{6}}\times {\displaystyle \frac{\sqrt{6}}{\sqrt{6}}}}\\ & ={\displaystyle \frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}}\\ & ={\displaystyle \frac{3\sqrt{2}+2\sqrt{3}-\sqrt{30}}{12}}\\ & \{\begin{array}{ll}a& =3\\ b& =2\\ c& =-1\end{array}\\ a+b+c& =3+2+(-1)\\ & =4\end{array}\end{array}$

$\begin{array}{l}\\ 38.& \text{Tunjukkan bahwa}\\ & \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots }}}}}=3\\ \\ & \text{Bukti}\\ & \begin{array}{rl}{x}^2& =x^2\\ x^2& =1+(x^2-1)\\ & =1+(x-1)(x+1)\\ & =1+(x-1)\sqrt{{(x+1)}^2}\\ & =1+(x-1)\sqrt{1+{(x+1)}^2-1}\\ & =1+(x-1)\sqrt{1+(x+1-1)(x+1+1)}\\ & =1+(x-1)\sqrt{1+x(x+2)}\\ & =1+(x-1)\sqrt{1+x\sqrt{{(x+2)}^2}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+{(x+2)}^2-1}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+2-1)(x+2+1)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)(x+3)}}\\ & =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{{(x+3)}^2}}}\\ x^2& =1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}\\ x& =\sqrt{1+(x-1)\sqrt{1+x\sqrt{1+(x+1)\sqrt{...}}}}\end{array}\end{array}$

$\begin{array}{l}\\ 39.& \text{Jika bentuk}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}\\ & \text{dinyatakan dalam pangkat positif}=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a^2}{a-b}}& & & \\ \\ \text{b}.& {\displaystyle \frac{a^2}{a-1}}\\ \\ \text{c}.& {\displaystyle \frac{b-a}{ab}}\\ \\ \text{d}.& {\displaystyle \frac{a^2}{b-a}}\\ \\ \text{e}.& {\displaystyle \frac{1}{a-b}}\\ \\ & & & & (\mathbf{\text{textit{SAT Test Math Level 2}}})\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}}& ={\displaystyle \frac{a{b}^{-1}}{a^{-1}-b^{-1}}\times \frac{b}{b}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}}\\ & ={\displaystyle \frac{a}{a^{-1}b-1}\times \frac{a}{a}}\\ & ={\displaystyle \frac{a^2}{b-a}}\end{array}\end{array}$

$\begin{array}{l}\\ 40.& \text{Jika terdapat hubungan berikut}\\ & \text{a}.2^p=3^q=6^r,\text{tunjukkan bahwa}pr+qr-pq=0\\ & \text{b}.2^x=3^{2y}=6^z,\text{tunjukkan bahwa }2xy-2yz-xz=0\\ & \text{c}.3^{15a}=5^{5b}={15}^{3c},\text{tunjukkan bahwa }5ab-bc-3ac=0\end{array}$

$\mathbf{\text{bukti}}$

Yang akan ditunjukkan adalah no. 40 yang poin c, yaitu:

$\begin{array}{rl}{3}^{15a}& =5^{5b}={15}^{3c}\{\begin{array}{ll}3=5^{\frac{5b}{15a}}& \\ 3^{\frac{15a}{5b}}=b& (a^b=c^d\to a=c^{\frac{d}{b}}\text{atau}{a}^{\frac{b}{d}}=c)\end{array}\\ 3^{15a}& ={15}^{3c}\\ 3^{15a}& =(3\times 5{)}^{3c}\\ 3^{15a}& =(3\times 3^{\frac{15a}{5b}}{)}^{3c}\\ 3^{15a}& =3^{3c+\frac{9c}{b}}\\ a^{f(x)}& =a^{g(x)}\\ f(x)& =g(x)\\ 15a& =3c+\frac{9ac}{b}\\ 15ab& =3bc+9ac\\ 5ab& =bc+3ac\\ 5ab-bc-3ac& =0◼\end{array}$

Contoh Soal 7 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 31.& \text{Nilai}x\text{yang memenuhi}\\ & \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}=3\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 6\\ \text{c}.& 7\\ \text{d}.& 8\\ \text{e}.& 9\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misalkan}A& =\sqrt{+\sqrt{x+\sqrt{+\cdots }}}\\ \sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =3\\ \text{dikuadratkan}& \\ x+\sqrt{x+\sqrt{x+\sqrt{x+\cdots }}}& =9\\ x+3& =9\\ x& =9-3\\ x& =6\end{array}\end{array}$

$\begin{array}{l}\\ 32.& \text{Nilai}x\text{yang memenuhi}\\ & x=\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 7\sqrt[7]{7}\\ \text{b}.& 7\\ \text{c}.& 14\\ \text{d}.& 49\\ \text{e}.& \sqrt[3]{81}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}x& =\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49\sqrt[3]{49\sqrt[3]{49\sqrt[3]{49\cdots }}}\\ x^3& =49x\\ x^2& =49\\ x& =\sqrt{49}\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 33.& \text{Nilai}x\text{yang memenuhi}\\ & x^{x^{x^{x^{x^{\cdots }}}}}=2020\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{2020}\\ \text{b}.& \sqrt[2020]{2020}\\ \text{c}.& {2020}^{\sqrt{2020}}\\ \text{d}.& {\sqrt{2020}}^{\sqrt{2020}}\\ \text{e}.& \sqrt{2020\sqrt{2020}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{x}^{x^{x^{x^{x^{\cdots }}}}}& =2020\\ x^{2020}& =2020\\ x& =\sqrt[2020]{2020}\end{array}\end{array}$

$\begin{array}{l}\\ 34.& \text{Nilai dari}\\ & {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& \sqrt[3]{2}+1\\ \text{c}.& \sqrt[3]{2}-1\\ \text{d}.& \sqrt[3]{4}+1\\ \text{e}.& \sqrt[3]{4}-1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& {\displaystyle \frac{1+\sqrt[3]{2}}{1+\sqrt[3]{2}+\sqrt[3]{4}}\times \frac{\sqrt[3]{2}-1}{\sqrt[3]{2}-1}}\\ & ={\displaystyle \frac{{\left(\sqrt[3]{2}\right)}^2-1}{\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{8}-1-\sqrt[3]{2}-\sqrt[3]{4}}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{\sqrt[3]{8}-1}}\\ & ={\displaystyle \frac{\sqrt[3]{4}-1}{2-1}}\\ & =\sqrt[3]{4}-1\end{array}\end{array}$

$\begin{array}{l}\\ 35.& \text{Nilai dari}\\ & {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & \text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\sqrt{2}-1\\ \text{c}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \text{d}.& \sqrt{{\displaystyle \frac{5}{3}}}\\ \text{e}.& \sqrt{{\displaystyle \frac{2}{5}}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}& {\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\times \frac{\sqrt{\sqrt{5}+1}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}}}\\ & ={\displaystyle \frac{\sqrt{7+3\sqrt{5}}+\sqrt{3-\sqrt{5}}}{\sqrt{5}+1}-(\sqrt{2}-1)}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{3+\sqrt{5}}{\sqrt{2}}}\right)+\left({\displaystyle \frac{\sqrt{5}-1}{\sqrt{2}}}\right)}{\sqrt{5}+1}+1-\sqrt{2}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2+2\sqrt{5}}{\sqrt{2}}}}{1+\sqrt{5}}+1-\sqrt{2}}\\ & ={\displaystyle \frac{2}{\sqrt{2}}+1-\sqrt{2}}\\ & =\sqrt{2}+1-\sqrt{2}\\ & =1\end{array}\end{array}$

Contoh Soal 6 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 26.& (\mathbf{\text{UM UGM 05}})\text{Hasil dari}\\ & \sqrt{0,3+\sqrt{0,08}}=\sqrt{a}+\sqrt{b},\text{maka}{\displaystyle \frac{1}{a}+\frac{1}{b}=....}\\ & \begin{array}{l}\\ \text{a}.& 25\\ \text{b}.& 20\\ \text{c}.& 15\\ \text{d}.& 10\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\sqrt{0,3+\sqrt{0,08}}& =\sqrt{0,2+0,1+\sqrt{4\times 0,2\times 0,1}}\\ & =\sqrt{0,2+0,1+2\sqrt{\times 0,2\times 0,1}}\\ & =\sqrt{0,2}+\sqrt{0,1}\\ \text{maka},a=0,2& ,b=0,1\\ \text{sehingga}{\displaystyle \frac{1}{a}+\frac{1}{b}}& ={\displaystyle \frac{1}{0,2}+\frac{1}{0,1}=5+10=15}\end{array}\end{array}$

$\begin{array}{l}\\ 27.& (\mathbf{\text{SPMB 06}})\text{Jika bilangan bulat}a\text{dan}b\text{memenuhi}\\ & {\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}=a+b\sqrt{30},\text{maka}ab=....}\\ & \begin{array}{l}\\ \text{a}.& -22\\ \text{b}.& -11\\ \text{c}.& -9\\ \text{d}.& 2\\ \text{e}.& 13\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}}& ={\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}+\sqrt{6}}\times {\displaystyle \frac{\sqrt{5}-\sqrt{6}}{\sqrt{5}-\sqrt{6}}}}\\ & ={\displaystyle \frac{5-2\sqrt{30}+6}{5-6}}\\ & ={\displaystyle \frac{11-2\sqrt{30}}{-1}}\\ & =-11+2\sqrt{30}\\ \text{sehingga}& a=-11,b=2,\text{maka}\\ ab& =(-11)\times 2\\ & =-22\end{array}\end{array}$

$\begin{array}{l}\\ 28.& (\mathbf{\text{OSK 2013}})\text{Misal}a\text{dan}b\text{bilangan asli}\\ & \text{dengan}a>b.\text{Jika}\sqrt{94+2\sqrt{2013}}=\sqrt{a}+\sqrt{b}\\ & \text{maka nilai}a-b\text{adalah... .}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\sqrt{94+2\sqrt{2013}}& =\sqrt{61+33+2\sqrt{61\times 33}}\\ & =\sqrt{61}+\sqrt{33}\\ & =\sqrt{a}+\sqrt{b}\\ \text{Sehingga}a& =61,b=33,\text{maka}\\ a-b& =61-33\\ & =28\end{array}\end{array}$

$\begin{array}{l}\\ 29.& \text{Daerah hasil dari fungsi eksponen}y=x^{-\frac{2}{3}}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& y<0\\ \text{b}.& y>0\\ \text{c}.& y\ge 0\\ \text{d}.& y\le 0\\ \text{e}.& \text{Semua bilangan real}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \text{Perhatikanlah gambar berikut}\end{array}$

$.\begin{array}{rl}\text{diketahui}& \\ y& =x^{-{\displaystyle \frac{2}{3}}}\\ y^3& =x^{-2}\\ y^3& ={\displaystyle \frac{1}{x^2},\text{atau}}\\ y^3\times x^2& =1,\\ \text{sehingga}& y\text{tidak mungkin berharga}0\end{array}$

$\begin{array}{l}\\ 30.& \text{Jika}f(x)=b^x,\text{di mana konstan positif},\\ \\ & {\displaystyle \frac{f(x^2+x)}{f(x+1)}=....}\\ & \begin{array}{l}\\ \text{a}.& f\left(x^2\right)& & & \\ \text{b}.& f(x+1)f(x-1)\\ \text{c}.& f(x+1)+f(x-1)\\ \text{d}.& f(x+1)-f(x-1)\\ \text{e}.& f(x^2-1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{f(x^2+x)}{f(x+1)}}& =\frac{b^{x^2+x}}{b^{x+1}}\\ & =b^{x^2+x-(x+1)}\\ & =b^{x^2-1}\\ & =f(x^2-1)\end{array}\end{array}$

Contoh Soal 5 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{l}\\ 21.& \text{Nilai}p-q^{p-q}\text{untuk}p=2\text{dan}q=-2\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -18\\ \text{b}.& -14\\ \text{c}.& 1\\ \text{d}.& 18\\ \text{e}.& 256\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}p-q^{p-q}& =(2-(-2){)}^{2-(-2)}\\ & =4^4\\ & =256\end{array}\end{array}$

$\begin{array}{l}\\ 22.& \text{Bentuk sederhana dari}{\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.& 49\\ \text{b}.& 9\\ \text{c}.& 7\\ \text{d}.& 7^{2x+2}\\ \text{e}.& 3^{2x-1}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{3^{x+8}{(7^{3x+1}\times 3^{4x-1})}^2}{(7^{2x}\times 3^{3x+2}{)}^3}}& ={\displaystyle \frac{3^{x+8+2(4x-1)}\times 7^{2(3x+1)}}{7^{3.2x}\times 3^{3(3x+2)}}}\\ & ={\displaystyle \frac{3^{x+8x+8-2}\times 7^{6x+2}}{3^{9x+6}\times 7^{6x}}}\\ & =3^0\times 7^2\\ & =1\times 49\\ & =49\end{array}\end{array}$

$\begin{array}{l}\\ 23.& \text{Jika}{\displaystyle \frac{(0,24{)}^3(0,243{)}^5}{(3,6{)}^7}=\frac{2^p{3}^q}{5^r},}\\ & \text{maka nilai}p+q+r\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 7\\ \text{b}.& 8\\ \text{c}.& 9\\ \text{d}.& 10\\ \text{e}.& 11\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{(0,24{)}^3(0,243{)}^5}{(3,6{)}^7}}& ={\displaystyle \frac{{\left(\frac{24}{100}\right)}^3\times {\left(\frac{243}{1000}\right)}^5}{{\left(\frac{36}{10}\right)}^7}}\\ & ={\displaystyle \frac{(8\times 3{)}^3\times {\left(3^5\right)}^5}{(2^2\times 3^2{)}^7}\times {\displaystyle \frac{{10}^7}{{100}^3\times {1000}^5}}}\\ & ={\displaystyle \frac{(2^3\times 3{)}^3\times 3^{25}}{(2^{14}\times 3^{14})}\times {\displaystyle \frac{{10}^7}{{\left({10}^2\right)}^3\times {\left({10}^3\right)}^5}}}\\ & =2^{9-14}{.3}^{3+25-14}{.10}^{7-6-15}\\ & =2^{-5}{.3}^{14}{.10}^{-14}=2^{-5}{.3}^{14}.(2.5{)}^{-14}\\ & =2^{-5-14}{.3}^{21}{.5}^{-14}\\ & ={\displaystyle \frac{2^{-19}{.3}^{14}}{5^{14}}}\\ \text{Sehingga}& p+q+r=-19+14+14=9\end{array}\end{array}$

$\begin{array}{l}\\ 24.& \text{Nilai eksak dari}\\ & {\displaystyle \frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+{\displaystyle \frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}}}\\ & \begin{array}{l}\\ \text{a}.& 2020& & & \\ \text{b}.& 2020,5\\ \text{c}.& 2021\\ \text{d}.& 2021,5\\ \text{e}.& 2022\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{Misal},x& =\frac{1}{{10}^{-2020}+1}+\frac{1}{{10}^{-2019}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ \text{maka},x& =\frac{1}{\frac{1}{{10}^{2020}}+1}+\frac{1}{\frac{1}{{10}^{2019}}+1}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{{10}^{2019}}{1+{10}^{2019}}+...+\frac{1}{{10}^{2019}+1}+\frac{1}{{10}^{2020}+1}\\ & =\frac{{10}^{2020}}{1+{10}^{2020}}+\frac{1}{{10}^{2020}+1}+\frac{{10}^{2019}}{1+{10}^{2019}}+\frac{1}{{10}^{2019}+1}+...+\frac{1}{1^0+1}\\ & =\frac{{10}^{2020}+1}{{10}^{2020}+1}+\frac{{10}^{2019}+1}{{10}^{2019}+1}+\frac{{10}^{2018}+1}{{10}^{2018}+1}+...+\frac{{10}^1+1}{{10}^1+1}+\frac{1}{1^0+1}\\ & =\underset{\text{sebanyak}2020}{\underbrace{1+1+1+1+1+...+1}}+\frac{1}{2}\\ & =2020,5\end{array}\end{array}$

$\begin{array}{l}\\ 25.& \text{Nilai dari}\\ & \sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 10\\ \text{b}.& 11\\ \text{c}.& 12\\ \text{d}.& 5\sqrt{6}\\ \text{e}.& 6\sqrt{6}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}\text{misal diketah}& \text{ui}\\ x& =\sqrt{54+14\sqrt{5}}+\sqrt{12-2\sqrt{35}}+\sqrt{32-10\sqrt{7}}\\ \text{untuk}& \\ \sqrt{54+14\sqrt{5}}& =\sqrt{49+5+2.7\sqrt{5}}=7+\sqrt{5}\\ \sqrt{12-2\sqrt{35}}& =\sqrt{7+5-2\sqrt{7.5}}=\sqrt{7}-\sqrt{5}\\ \sqrt{32-10\sqrt{7}}& =\sqrt{25+7-2.5\sqrt{7}}=5-\sqrt{7}+\\ & ---------------\\ & =7+5\\ & =12\end{array}\end{array}$

Contoh Soal 3 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 11.& (\mathbf{\text{SPMB 04}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{27}{3^{2x-1}}={81}^{-0,125}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -1{\displaystyle \frac{3}{4}}\\ \text{b}.& -{\displaystyle \frac{3}{4}}\\ \text{c}.& {\displaystyle \frac{3}{4}}\\ \text{d}.& 1{\displaystyle \frac{1}{4}}\\ \text{e}.& 2{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{27}{3^{2x-1}}}& ={81}^{-0,125}\\ 3^{3-(2x-1)}& =3^{4(\frac{1}{8})}\\ 3-2x+1& =-{\displaystyle \frac{1}{2}}\\ -2x+4& =-{\displaystyle \frac{1}{2}}\\ -x+2& =-{\displaystyle \frac{1}{4}}\\ -x& =-2-{\displaystyle \frac{1}{4}}\\ -x& =-2{\displaystyle \frac{1}{4}}\\ x& =2{\displaystyle \frac{1}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 12.& (\mathbf{\text{UMPTN 98}})\text{Bentuk}{\left({\displaystyle \frac{x^{\frac{2}{3}}.y^{\frac{-4}{3}}}{y^{\frac{2}{3}}.x^2}}\right)}^{-\frac{3}{4}}\\ & \text{dapat diserdernakan menjadi}\\ & \begin{array}{l}\\ \text{a}.& \sqrt{x{y}^2}\\ \text{b}.& x\sqrt{y}\\ \text{c}.& \sqrt{x^{2}y}\\ \text{d}.& xy\sqrt{y}\\ \text{e}.& xy\sqrt{x}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{x^{\frac{2}{3}}.y^{\frac{-4}{3}}}{y^{\frac{2}{3}}.x^2}}\right)}^{-\frac{3}{4}}& ={(x^{\frac{2}{3}-2}.y^{-\frac{3}{4}-\frac{2}{3}})}^{-\frac{3}{4}}\\ & =x^{-\frac{3}{4}(\frac{2}{3}-2)}.y^{-\frac{3}{4}(-\frac{3}{4}-\frac{2}{3})}\\ & =x^{-\frac{1}{2}+\frac{3}{2}}.y^{1+\frac{1}{2}}\\ & =x^1.y^{1\frac{1}{2}}\\ & =xy\sqrt{y}\end{array}\end{array}$

$\begin{array}{l}\\ 13.& (\mathbf{\text{UMPTN 00}})\\ & \text{Bentuk}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ & \text{Jika}{x}_0\text{memenuhi persamaan, maka nilai}\\ & 1-{\displaystyle \frac{3}{4}{x}_0=....}\\ & \begin{array}{l}\\ \text{a}.& 1\frac{3}{16}\\ \text{b}.& 1\frac{1}{4}\\ \text{c}.& 1\frac{3}{4}\\ \text{d}.& 2\frac{1}{3}\\ \text{e}.& 2\frac{3}{4}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\left(\sqrt[3]{{\displaystyle \frac{1}{243}}}\right)}^{3x}& ={\left({\displaystyle \frac{3}{3^{x-2}}}\right)}^2\sqrt[3]{{\displaystyle \frac{1}{9}}}\\ 3^{-5x}& =3^{2(1-(x-2))}{.3}^{-\frac{2}{3}}\\ -5x& =2(1-(x-2))+(-{\displaystyle \frac{2}{3}}),\text{dikali}3\\ -15x& =6(3-x)+(-2)\\ -15x& =18-6x-2\\ 6x-15x& =16\\ -9x& =16\\ x& ={\displaystyle \frac{16}{-9}}\\ x_0& =-{\displaystyle \frac{16}{9},\text{selanjutnya}}\\ 1-{\displaystyle \frac{3}{4}{x}_0}& =1-{\displaystyle \frac{3}{4}\times (-\frac{16}{9})}\\ & =1+\frac{4}{3}\\ & =1+1{\displaystyle \frac{1}{3}}\\ & =2{\displaystyle \frac{1}{3}}\end{array}\end{array}$

$\begin{array}{l}\\ 14.& \text{Diketahui}{x}^{\frac{1}{2}}+x^{-\frac{1}{2}}=3\\ & \text{Nilai}x+x^{-1}=....\\ & \begin{array}{l}\\ \text{a}.& 7\\ \text{b}.& 8\\ \text{c}.& 8\\ \text{d}.& 10\\ \text{e}.& 11\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{x}^{\frac{1}{2}}+x^{-\frac{1}{2}}& =3\\ \text{dikadrat}& \text{kan}\\ {(x^{\frac{1}{2}}+x^{-\frac{1}{2}})}^2& =3^2\\ x+2+x^{-1}& =9\\ x+x^{-1}& =9-2\\ & =7\end{array}\end{array}$

$\begin{array}{l}\\ 15.& \text{Diketahui}{2}^{2x}+2^{-2x}=2\\ & \text{Nilai}{2}^x+2^{-x}=....\\ & \begin{array}{l}\\ \text{a}.& 1\\ \text{b}.& 2\\ \text{c}.& \sqrt{2}\\ \text{d}.& 3\\ \text{e}.& \sqrt{3}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{2}^{2x}+2^{-2x}& =2\\ \text{jika soal}& \text{dikuadratkan}\\ {(2^x+2^{-x})}^2& =2^{2x}+2+2^{-2x}\\ & =2^{2x}+2^{-2x}+2\\ {(2^x+2^{-x})}^2& =2+2=4\\ 2^x+2^{-x}& =\sqrt{4}\\ & =2\end{array}\end{array}$

Contoh Soal 2 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 6.& \text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle 5^2{\left({\left({\displaystyle \frac{1}{25}}\right)}^{2x+6}\right)}^{\frac{1}{6}}={\displaystyle \frac{1}{25}\text{adalah... .}}}\\ & \begin{array}{l}\\ \text{a}.& -5\\ \text{b}.& -4\\ \text{c}.& -3\\ \text{d}.& 1\\ \text{e}.& 3\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle 5^2{\left({\left({\displaystyle \frac{1}{25}}\right)}^{2x+6}\right)}^{\frac{1}{6}}}& ={\displaystyle \frac{1}{25}}\\ 25.{\displaystyle 5^2{\left({\left({\displaystyle \frac{1}{25}}\right)}^{2x+6}\right)}^{\frac{1}{6}}}& =1\\ 5^2{.5}^2{.5}^{-\frac{4x+12}{6}}& =5^0\\ 5^{2+2-\frac{2}{3}x-2}& =5^0\\ {\displaystyle 2+2-\frac{2}{3}x-2}& =0\\ 2-{\displaystyle \frac{2}{3}x}& =0\\ -{\displaystyle \frac{2}{3}x}& =-2\\ x& =3\end{array}\end{array}$

$\begin{array}{l}\\ 7.& (\mathbf{\text{UM-UGM 03}})\text{Nilai}x\text{yang memenuhi}\\ & {\left({\displaystyle \frac{1}{25}}\right)}^{x-\frac{5}{2}}=\sqrt{{\displaystyle \frac{625}{5^{2-x}}}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{3}{5}}\\ \text{b}.& {\displaystyle \frac{8}{5}}\\ \text{c}.& 2\\ \text{d}.& 3\\ \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{1}{25}}\right)}^{x-\frac{5}{2}}& =\sqrt{{\displaystyle \frac{625}{5^{2-x}}}}\\ 5^{-2x+5}& =5^{\frac{1}{2}(4-(2-x))}\\ -2x+5& ={\displaystyle \frac{1}{2}(4-2+x)}\\ -4x+10& =2+x\\ -5x& =2-10\\ x& ={\displaystyle \frac{-8}{-5}}\\ & =\frac{8}{5}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Nilai}x\text{yang memenuhi}\\ & 2^{\frac{x}{3}-1}=16\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& 5\\ \text{b}.& 10\\ \text{c}.& 15\\ \text{d}.& 20\\ \text{e}.& 25\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{2}^{\frac{x}{3}-1}& =16\\ 2^{\frac{x}{3}-1}& =2^4\\ {\displaystyle \frac{x}{3}-1}& =4\\ {\displaystyle \frac{x}{3}}& =5\\ x& =15\end{array}\end{array}$

$\begin{array}{l}\\ 9.& (\mathbf{\text{SPMB 05}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{\sqrt[3]{(0,08{)}^{7-2x}}}{(0,2{)}^{-4x+5}}=1\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -3\\ \text{b}.& -2\\ \text{c}.& -1\\ \text{d}.& 0\\ \text{e}.& 1\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{\sqrt[3]{(0,08{)}^{7-2x}}}{(0,2{)}^{-4x+5}}}& =1\\ \sqrt[3]{(0,08{)}^{7-2x}}& =(0,2{)}^{-4x+5}\\ (0,2{)}^{\frac{3(7-2x)}{3}}& =(0,2{)}^{-4x+5}\\ {\displaystyle 7-2x}& =-4x+5\\ 4x-2x& =5-7\\ 2x& =-2\\ x& =-1\end{array}\end{array}$

$\begin{array}{l}\\ 10.& (\mathbf{\text{SPMB 05}})\text{Nilai}x\text{yang memenuhi}\\ & {\displaystyle \frac{\sqrt[3]{{\displaystyle \frac{1}{9^{2-x}}}}}{27}=3^{x+1}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& -16\\ \text{b}.& -7\\ \text{c}.& 4\\ \text{d}.& 5\\ \text{e}.& 6\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{a}}\\ & \begin{array}{rl}{\displaystyle \frac{\sqrt[3]{{\displaystyle \frac{1}{9^{2-x}}}}}{27}}& =3^{x+1}\\ \sqrt[3]{{\displaystyle \frac{1}{9^{2-x}}}}& =27\times 3^{x+1}\\ \sqrt[3]{3^{-2(2-x)}}& =3^3{.3}^{x+1}\\ 3^{\frac{-4+2x}{3}}& =3^{3+(x+1)}\\ {\displaystyle \frac{-4+2x}{3}}& =4+x\\ -4+2x& =12+3x\\ 2x-3x& =12+4\\ -x& =16\\ x& =-16\end{array}\end{array}$

Contoh Soal 4 Fungsi Eksponen (Matematika Peminatan Kelas X)

 $\begin{array}{l}\\ 16.& (\mathbf{\text{textrm{UM UNDIP 2012 Math IPA}}})\\ & \text{Jika}f(x)=x^3-3x^2-3x-1,\\ & \text{maka nilai dari}f(1+\sqrt[3]{2}+\sqrt[3]{4})=....\\ & \begin{array}{l}\\ \text{a}.& -\sqrt{2}\\ \text{b}.& -1\\ \text{c}.& 0\\ \text{d}.& 1\\ \text{e}.& \sqrt{2}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}f(x)& =x^3-3x^2-3x-1\\ & ={(x-1)}^3-6x\\ f(1+\sqrt[3]{2}+\sqrt[3]{4})& ={(1+\sqrt[3]{2}+\sqrt[3]{4}-1)}^3-6(1+\sqrt[3]{2}+\sqrt[3]{4})\\ & ={(\sqrt[3]{2}+\sqrt[3]{4})}^3-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & ={\left(\sqrt[3]{2}(\sqrt[3]{2}+1)\right)}^3-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =2(1+3\sqrt[3]{2}+3\sqrt[3]{4}+2)-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =(6+6\sqrt[3]{2}+6\sqrt[3]{4})-6-6\sqrt[3]{2}-6\sqrt[3]{4}\\ & =0\end{array}\end{array}$

$\begin{array}{l}\\ 17.& \text{Jika}\mathbf{\text{a}}\text{dan}\mathbf{\text{b}}\text{adalah bilangan bulat positif}\\ & \text{yang memenuhi persamaan}{\mathbf{\text{a}}}^{\mathbf{\text{b}}}=2^{20}-2^{19},\\ & \text{maka nilai dari}\mathbf{\text{a}}+\mathbf{\text{b}}=....\\ & \begin{array}{l}\\ \text{a}.& 3\\ \text{b}.& 7\\ \text{c}.& 19\\ \text{d}.& 21\\ \text{e}.& 23\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}{\mathbf{\text{a}}}^{\mathbf{\text{b}}}& =2^{20}-2^{19}\\ & =2^{{}^{{}^{(19+1)}}}-2^{19}\\ & =2^{19}{.2}^1-2^{19}\\ & =2^{19}(2-1)\\ & =2^{19}\\ \text{Se}& \text{hingga nilai}\mathbf{\text{a}}+\mathbf{\text{b}}=2+19=21\end{array}\end{array}$

$\begin{array}{l}\\ 18.& \text{Perhatikan gambar berikut}\end{array}$

$\begin{array}{l}\\ .& \text{Persamaan grafik fungsi seperti gambar di atas adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=2^{x-2}\\ \text{b}.& f(x)=2^x-2\\ \text{c}.& f(x)=2^x-1\\ \text{d}.& f(x)={}^2\log (x-1)\\ \text{e}.& f(x)={}^2\log (x+1)\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}\text{a}& \Rightarrow (1,0)\Rightarrow f(1)=2^{1-2}=2^{-1}=\frac{1}{2}\ne 0(\text{salah})\\ \text{b}& \Rightarrow (1,0)\Rightarrow f(1)=2^1-2=2^1-2=0=0(\mathbf{\text{benar}})\\ \text{c}& \Rightarrow (1,0)\Rightarrow f(1)=2^1-1=2^1-1=1\ne 0(\text{salah})\\ \text{d}& \Rightarrow (1,0)\Rightarrow f(1)={}^2\log (1-1)={}^2\log 0=\\ & \mathbf{\text{tidak mungkin}}\ne 0(\text{salah})\\ \text{e}& \Rightarrow (1,0)\Rightarrow f(1)={}^2\log (1+1)={}^2\log 2=1\ne 0(\text{salah})\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Solusi untuk persamaan}{3}^{2x+1}={81}^{x-2}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0\\ \text{b}.& 2\\ \text{c}.& 4\\ \text{d}.& 4,5\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}{3}^{2x+1}& ={81}^{x-2}\\ \left(3^{2x}\right){.3}^1& ={\left(3^4\right)}^{x-2}\\ 3.\left(3^{2x}\right)& =3^{4x}{.3}^{-8}\\ 3.\left(3^{2x}\right)& ={\displaystyle \frac{{\left(3^{2x}\right)}^2}{3^8}}\\ 0& ={\displaystyle \frac{{\left(3^{2x}\right)}^2}{3^8}-3\left(3^{2x}\right)}\\ 0& ={\left(3^{2x}\right)}^2-3^9.\left(3^{2x}\right)\\ 0& =\left(3^{2x}\right)(\left(3^{2x}\right)-3^9)\end{array}& \begin{array}{rl}\mathbf{\text{atau}}& \\ {\left(3^{2x}\right)}^2-27.\left(3^{2x}\right)& =0\\ \left(3^{2x}\right)(\left(3^{2x}\right)-3^9)& =0\\ 3^{2x}=0\text{atau}{3}^{2x}& =3^9\\ (\text{tm})\text{atau}{3}^{2x}& =3^9\\ 2^x& =9\\ x& ={\displaystyle \frac{9}{2}}\\ \text{atau}x& =4{\displaystyle \frac{1}{2}}\end{array}\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 20.& \text{Jika}\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}=\sqrt{a}+\sqrt{b}+\sqrt{c},\\ & \text{maka nilai dari}{(a+b-c)}^{abc}=....\\ & \begin{array}{l}\\ \text{a}.& 1000\\ \text{b}.& 1\\ \text{c}.& 0\\ \text{d}.& -1\\ \text{e}.& -1000\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{r}\begin{array}{rl}{(\sqrt{2}+\sqrt{3}+\sqrt{5})}^2& =(\sqrt{2}+\sqrt{3}+\sqrt{5})\times (\sqrt{2}+\sqrt{3}+\sqrt{5})\\ & =\sqrt{2.2}+\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.2}+\sqrt{3.3}\\ & +\sqrt{3.5}+\sqrt{5.2}+\sqrt{5.3}+\sqrt{5.5}\\ & =2+2\sqrt{6}+2\sqrt{10}+3+2\sqrt{15}+5\\ & =2+3+5+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}\\ & =10+\sqrt{2^2.6}+\sqrt{2^2.10}+\sqrt{2^2.15}\\ & =10+\sqrt{24}+\sqrt{40}+\sqrt{60}\\ \sqrt{2}+\sqrt{3}+\sqrt{5}& =\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\\ & \{\begin{array}{ll}a& =2\\ b& =3\\ c& =5\end{array}.\text{sesuai dengan urutannya}\\ \text{Sehingga nilai}& \\ {(a+b-c)}^{abc}& ={(2+3-5)}^{2.3.5}\\ & =0^{30}\\ & =0\end{array}\end{array}\end{array}$

Contoh Soal 1 Fungsi Eksponen (Matematika Peminatan Kelas X)

$\begin{array}{l}\\ 1.& \text{Hasil dari}{\displaystyle \frac{3^4{.5}^3{.7}^2}{27.125.49}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& 2\\ \text{b}.& 3\\ \text{c}.& 4\\ \text{d}.& 9\\ \text{e}.& 16\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\displaystyle \frac{3^4{.5}^3{.7}^2}{27.125.49}}& ={\displaystyle \frac{3^4{.5}^3{.7}^2}{3^3{.5}^3{.7}^2}}\\ & =3^{4-3}{.5}^{3-3}{.7}^{2-2}\\ & =3^1{.5}^0{.7}^0\\ & =3.1.1\\ & =3\end{array}\end{array}$

$\begin{array}{l}\\ 2.& \text{Bentuk sederhana dari}{\displaystyle \frac{5a^4{b}^2}{a^2{b}^{-3}}\text{adalah... .}}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle a{b}^2}\\ \\ \text{b}.& {\displaystyle a^2{b}^2}\\ \\ \text{c}.& {\displaystyle 5a^2{b}^5}\\ \\ \text{d}.& 5a^4{b}^5\\ \\ \text{e}.& 5a^5{a}^2\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{c}}\\ & \begin{array}{rl}{\displaystyle \frac{5a^4{b}^2}{a^2{b}^{-3}}}& =5a^{4-2}{b}^{2-(-3)}\\ & =5a^2{b}^5\end{array}\end{array}$

$\begin{array}{l}\\ 3.& \text{Bentuk sederhana dari}{\left({\displaystyle \frac{a{b}^2}{a^2{b}^3}}\right)}^4\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{b}}\\ \\ \text{b}.& {\displaystyle \frac{1}{ab}}\\ \\ \text{c}.& {\displaystyle \frac{1}{a^2{b}^2}}\\ \\ \text{d}.& {\displaystyle \frac{1}{a{b}^2}}\\ \\ \text{e}.& {\displaystyle \frac{1}{a^4{b}^4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{a{b}^2}{a^2{b}^3}}\right)}^4& ={\left(a^{1-2}{b}^{2-3}\right)}^4\\ & ={\left(a^{-1}{b}^{-1}\right)}^4\\ & ={\left({\displaystyle \frac{1}{ab}}\right)}^4\\ & ={\displaystyle \frac{1}{a^4{b}^4}}\end{array}\end{array}$

$\begin{array}{l}\\ 4.& \text{Bentuk sederhana dari}{\displaystyle \frac{3^{\frac{5}{6}}\times {12}^{\frac{7}{12}}}{2^{-\frac{1}{4}}\times 6^{\frac{2}{3}}}\text{adalah}....}\\ & \begin{array}{l}\\ \text{a}.6^{\frac{1}{4}}& & \text{d}.{\left({\displaystyle \frac{2}{3}}\right)}^{\frac{3}{4}}\\ \text{b}.6^{\frac{3}{4}}& \text{c}.6^{\frac{2}{3}}& \text{e}.{\left({\displaystyle \frac{3}{2}}\right)}^{\frac{3}{4}}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{b}}\\ & \begin{array}{rl}{\displaystyle \frac{3^{\frac{5}{6}}\times {12}^{\frac{7}{12}}}{6^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}& ={\displaystyle \frac{3^{\frac{5}{6}}\times (3\times 4{)}^{\frac{7}{12}}}{(2\times 3{)}^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 4^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times {\left(2^2\right)}^{\frac{7}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{6}}\times 3^{\frac{7}{12}}\times 2^{\frac{14}{12}}}{2^{\frac{2}{3}}\times 3^{\frac{2}{3}}\times 2^{-\frac{1}{4}}}}\\ & =2^{(\frac{14}{12}-\frac{2}{3}+\frac{1}{4})}\times 3^{(\frac{5}{6}+\frac{7}{12}-\frac{2}{3})}\\ & =2^{(\frac{14-8+3}{12})}\times 3^{(\frac{10+7-8}{12})}\\ & =2^{\frac{9}{12}}\times 3^{\frac{9}{12}}\\ & =(2\times 3{)}^{\frac{9}{12}}\\ & =6^{\frac{9}{12}}\\ & =6^{\frac{3}{4}}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Bentuk sederhana dari}{\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{-3}}{a^{-1}{b}^{-\frac{3}{2}}}}\right)}^{\frac{3}{2}}\text{adalah... .}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{a}{b}}\\ \\ \text{b}.& {\displaystyle \frac{b}{a}}\\ \\ \text{c}.& {\displaystyle ab}\\ \\ \text{d}.& \sqrt[a]{b}\\ \\ \text{e}.& \sqrt[4]{{\left({\displaystyle \frac{a}{b}}\right)}^9}\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}{\left({\displaystyle \frac{a^{\frac{1}{2}}{b}^{-3}}{a^{-1}{b}^{-\frac{3}{2}}}}\right)}^{\frac{3}{2}}& ={\displaystyle \frac{a^{\frac{3}{4}}{b}^{\frac{-9}{2}}}{a^{-\frac{3}{2}}{b}^{-\frac{9}{4}}}}\\ & =a^{\frac{3}{4}+\frac{3}{2}}{b}^{\frac{-9}{2}+\frac{9}{4}}\\ & =a^{\frac{3+6}{4}}{b}^{\frac{-18+9}{4}}\\ & =a^{\frac{9}{4}}{b}^{-\frac{9}{4}}\\ & ={\left({\displaystyle \frac{a}{b}}\right)}^{\frac{9}{4}}\\ & =\sqrt[4]{{\left({\displaystyle \frac{a}{b}}\right)}^9}\end{array}\end{array}$

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 $\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}$.