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Eksponen 3. Pangkat Pecahan

Eksponen-Pangkat Pecahan

A. Pangkat Pecahan yang Berbentuk $a^{\frac{m}{n}}$

Jika $a$ bilangan real, $m$ bilangan bulat, $n$ bilangan asli dan $n\ge 2$, maka ${{a}^{\frac{m}{n}}}=\sqrt[n]{{{a}^{m}}}={{\left( \sqrt[n]{a} \right)}^{m}}$.

B. Sifat-sifat Bilangan Berpangkat dengan Pangkat Rasional

Jika $a,b\in R$, $a\ne 0$, $b\ne 0$ serta $m$ dan $n$ bilangan rasional, maka berlaku:
  1. ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$
  2. ${{a}^{m}}:{{a}^{n}}={{a}^{m-n}}$
  3. ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$
  4. ${{\left( ab \right)}^{n}}={{a}^{n}}\times {{b}^{n}}$
  5. ${{\left( \frac{a}{b} \right)}^{m}}=\frac{{{a}^{m}}}{{{b}^{m}}}$
  6. ${{\left( {{a}^{m}}.{{b}^{m}} \right)}^{n}}={{a}^{mn}}.{{b}^{mn}}$
  7. ${{\left( \frac{{{a}^{m}}}{{{b}^{m}}} \right)}^{n}}=\frac{{{a}^{mn}}}{{{b}^{mn}}}$
  8. ${{a}^{-\frac{m}{n}}}=\frac{1}{{{a}^{\frac{m}{n}}}}$
  9. ${{a}^{\frac{m}{n}}}=\frac{1}{{{a}^{-\frac{m}{n}}}}$
Contoh 1.
Tentukan nilai dari $\sqrt[5]{{{1024}^{2}}}$.
Penyelesaian:
$\begin{align}\sqrt[5]{{{1024}^{2}}} &= {{\left( 1024 \right)}^{\frac{2}{5}}} \\ &= {{\left( {{2}^{10}} \right)}^{\frac{2}{5}}} \\ &= {{2}^{10\times \frac{2}{5}}} \\ &= 2^4 \\ \sqrt[5]{{{1024}^{2}}} &= 16 \end{align}$

Contoh 2.
Tentukan nilai dari $\sqrt[4]{0,0081}$.
Penyelesaian:
$\begin{align}\sqrt[4]{0,0081} &={{\left( 0,0081 \right)}^{\frac{1}{4}}} \\ &={{\left( \frac{81}{10000} \right)}^{\frac{1}{4}}} \\ &={{\left( \frac{{{3}^{4}}}{{{10}^{4}}} \right)}^{\frac{1}{4}}} \\ &=\frac{{{3}^{4\times \frac{1}{4}}}}{{{10}^{4\times \frac{1}{4}}}} \\ &=\frac{3}{10} \\ \sqrt[4]{0,0081} &=0,3 \end{align}$

Contoh 3.
Tentukan nilai dari ${{\left( -243 \right)}^{\frac{2}{5}}}$.
Penyelesaian:
$\begin{align}{{\left( -243 \right)}^{\frac{2}{5}}} &={{\left( -{{3}^{5}} \right)}^{\frac{2}{5}}} \\ &={{\left( {{(-3)}^{5}} \right)}^{\frac{2}{5}}} \\ &={{(-3)}^{5\times \frac{2}{5}}} \\ &={{(-3)}^{2}} \\ {{\left( -243 \right)}^{\frac{2}{5}}} &=9 \end{align}$

Contoh 4.
Jika $p=25$ dan $q=64$, tentukan nilai $\frac{{{p}^{-\frac{3}{2}}}.{{q}^{\frac{2}{3}}}}{{{q}^{\frac{1}{3}}}-{{p}^{\frac{1}{2}}}}$.
Penyelesaian:
$p=25$ dan $q=64$ maka:
$\begin{align}\frac{{{p}^{-\frac{3}{2}}}.{{q}^{\frac{2}{3}}}}{{{q}^{\frac{1}{3}}}-{{p}^{\frac{1}{2}}}} &= \frac{{{25}^{-\frac{3}{2}}}{{.64}^{\frac{2}{3}}}}{{{64}^{\frac{1}{3}}}-{{25}^{\frac{1}{2}}}} \\ &= \frac{{{({{5}^{2}})}^{-\frac{3}{2}}}.{{({{4}^{3}})}^{\frac{2}{3}}}}{{{({{4}^{3}})}^{\frac{1}{3}}}-{{({{5}^{2}})}^{\frac{1}{2}}}} \\ &= \frac{{{5}^{2\times \left( -\frac{3}{2} \right)}}{{.4}^{3\times \frac{2}{3}}}}{{{4}^{3\times \frac{1}{3}}}-{{5}^{2\times \frac{1}{2}}}} \\ &=\frac{{{5}^{-3}}{{.4}^{2}}}{4-5} \\ &=\frac{\frac{1}{{{5}^{3}}}.16}{-1} \\ \frac{{{p}^{-\frac{3}{2}}}.{{q}^{\frac{2}{3}}}}{{{q}^{\frac{1}{3}}}-{{p}^{\frac{1}{2}}}} &=-\frac{16}{125} \end{align}$

Contoh 5.
Tuliskan bentuk sederhana dari $\sqrt[3]{{{x}^{2}}\sqrt{x\sqrt[5]{{{x}^{3}}}}}$.
Penyelesaian:
$\begin{align}\sqrt[3]{{{x}^{2}}\sqrt{x\sqrt[5]{{{x}^{3}}}}} &= \sqrt[3]{{{x}^{2}}\sqrt{x.{{x}^{\frac{3}{5}}}}} \\ &=\sqrt[3]{{{x}^{2}}\sqrt{{{x}^{1+\frac{3}{5}}}}} \\ &=\sqrt[3]{{{x}^{2}}\sqrt{{{x}^{\frac{8}{5}}}}} \\ &=\sqrt[3]{{{x}^{2}}.{{x}^{\frac{8}{5}\times \frac{1}{2}}}} \\ &=\sqrt[3]{{{x}^{2}}.{{x}^{\frac{4}{5}}}} \\ &=\sqrt[3]{{{x}^{2+\frac{4}{5}}}} \\ &=\sqrt[3]{{{x}^{\frac{14}{5}}}} \\ &={{x}^{\frac{14}{5}\times \frac{1}{3}}} \\ \sqrt[3]{{{x}^{2}}\sqrt{x\sqrt[5]{{{x}^{3}}}}} &={{x}^{\frac{14}{15}}} \end{align}$

Contoh 6.
Jika $\sqrt{{{2}^{x}}}+\sqrt{{{2}^{-x}}}=\sqrt{7}$ maka nilai ${{2}^{2x}}+{{2}^{-2x}}$ adalah …
Penyelesaian:
$\begin{align}\sqrt{{{2}^{x}}}+\sqrt{{{2}^{-x}}} &=\sqrt{7} \\ {{2}^{\frac{x}{2}}}+{{2}^{-\frac{x}{2}}} &=\sqrt{7} \\ {{\left( {{2}^{\frac{x}{2}}}+{{2}^{-\frac{x}{2}}} \right)}^{2}} &= {{\left( \sqrt{7} \right)}^{2}} \\ \left( {{2}^{\frac{x}{2}}}+{{2}^{-\frac{x}{2}}} \right).\left( {{2}^{\frac{x}{2}}}+{{2}^{-\frac{x}{2}}} \right) &= 7 \\ {{2}^{\frac{x}{2}+\frac{x}{2}}}+{{2}^{\frac{x}{2}-\frac{x}{2}}}+{{2}^{-\frac{x}{2}+\frac{x}{2}}}+{{2}^{-\frac{x}{2}-\frac{x}{2}}} &= 7 \\ {{2}^{x}}+{{2}^{0}}+{{2}^{0}}+{{2}^{-x}} &= 7 \\ {{2}^{x}}+1+1+{{2}^{-x}} &= 7 \\ {{2}^{x}}+{{2}^{-x}} &= 5 \end{align}$

$\begin{align}{{2}^{x}}+{{2}^{-x}} &=5 \\ {{\left( {{2}^{x}}+{{2}^{-x}} \right)}^{2}} &={{5}^{2}} \\ \left( {{2}^{x}}+{{2}^{-x}} \right)\left( {{2}^{x}}+{{2}^{-x}} \right) &=25 \\ {{2}^{x+x}}+{{2}^{x-x}}+{{2}^{-x+x}}+{{2}^{-x-x}} &=25 \\ {{2}^{2x}}+{{2}^{0}}+{{2}^{0}}+{{2}^{-2x}} &=25 \\ {{2}^{2x}}+1+1+{{2}^{-2x}} &=25 \\ {{2}^{2x}}+{{2}^{-2x}} &=23 \end{align}$

C. Soal Latihan

  1. Tuliskan bentuk sederhana dari $\left( \frac{{{a}^{-\frac{2}{3}}}}{{{b}^{-\frac{1}{3}}}} \right)\times {{\left( {{a}^{\frac{3}{2}}}.{{b}^{\frac{1}{2}}} \right)}^{2}}:\left( \frac{{{a}^{\frac{1}{2}}}}{{{b}^{\frac{1}{3}}}} \right)$.
  2. Tuliskan bentuk sederhana dari $\frac{\sqrt[3]{{{a}^{4}}\sqrt[3]{a\sqrt{a}}}}{\sqrt{a\sqrt[3]{a}}}$.
  3. Tuliskan bentuk sederhana dari $\frac{{{(-2a)}^{3}}{{(2a)}^{-\frac{2}{3}}}}{{{(16{{a}^{4}})}^{\frac{1}{3}}}}$.
  4. Diketahui nilai $a=9$, $b=16$ dan $c=36$. Tentukan nilai $\sqrt{{{\left( {{a}^{-\frac{1}{3}}}{{b}^{-\frac{1}{2}}}c \right)}^{3}}}$.
  5. Diketahui $a=\frac{1}{8}$, $b=16$ dan $c=4$. Tentukan nilai ${{a}^{-1\frac{1}{3}}}{{b}^{\frac{1}{4}}}{{c}^{-1\frac{1}{2}}}$.
By: Catatan Matematika
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