Exercise Zone : Pangkat (Eksponen)

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Tingkat Kesulitan :

• Dasar
• SBMPTN
• Olimpiade

• 1
• 2

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No. 1

Hitunglah hasil perpangkatan berikut
\left(\dfrac{27}{125}\right)^{\frac13}\left(\dfrac19\right)^{-2}

$\begin{array}{rl}{\left({\displaystyle \frac{27}{125}}\right)}^{\frac{1}{3}}{\left({\displaystyle \frac{1}{9}}\right)}^{-2}& ={\left({\displaystyle \frac{3^3}{5^3}}\right)}^{\frac{1}{3}}{\left({\displaystyle \frac{1}{3^2}}\right)}^{-2}\\ & ={\displaystyle \frac{{\left(3^3\right)}^{\frac{1}{3}}}{{\left(5^3\right)}^{\frac{1}{3}}}}{\left(3^{-2}\right)}^{-2}\\ & ={\displaystyle \frac{3}{5}}\left(3^4\right)\\ & ={\displaystyle \frac{3}{5}}\left(81\right)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{243}{5}}}}}}\end{array}$

Jadi, \left(\dfrac{27}{125}\right)^{\frac13}\left(\dfrac19\right)^{-2}=\dfrac{243}5.

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No. 2

Diketahui a, b, dan c adalah bilangan real positif, jika \dfrac{\sqrt{bc}}{\sqrt[5]{ab^4}}= ab, maka nilai c^5 adalah

1. a^{11}b^{12}
2. a^{12}b^{11}
3. a^{14}b^{13}

4. a^{13}b^{12}
5. a^{12}b^{13}

$\begin{array}{rl}{\displaystyle \frac{\sqrt{bc}}{\sqrt[5]{a{b}^4}}}& =ab\\ {\left({\displaystyle \frac{\sqrt{bc}}{\sqrt[5]{a{b}^4}}}\right)}^{10}& =(ab{)}^{10}\\ {\displaystyle \frac{(bc{)}^5}{{\left(a{b}^4\right)}^2}}& =a^{10}{b}^{10}\\ {\displaystyle \frac{b^5{c}^5}{a^2{b}^8}}& =a^{10}{b}^{10}\\ c^5& ={\displaystyle \frac{a^2{b}^8\cdot a^{10}{b}^{10}}{b^5}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle a^{12}{b}^{13}}}}}\end{array}$

Jadi, c^5=a^{12}b^{13}.
JAWAB: E

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No. 3

{\dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}=} ....

1. -1
2. 0
3. \dfrac12

4. 1
5. A^{x+y}

Zenius.net

$\begin{array}{rl}{\displaystyle \frac{1}{1+A^{x-y}}}+{\displaystyle \frac{1}{1+A^{y-x}}}& ={\displaystyle \frac{1}{1+{\displaystyle \frac{A^x}{A^y}}}}+{\displaystyle \frac{1}{1+{\displaystyle \frac{A^y}{A^x}}}}\\ & ={\displaystyle \frac{1}{{\displaystyle \frac{A^y+A^x}{A^y}}}}+{\displaystyle \frac{1}{{\displaystyle \frac{A^x+A^y}{A^x}}}}\\ & ={\displaystyle \frac{A^y}{A^y+A^x}}+{\displaystyle \frac{A^x}{A^x+A^y}}\\ & ={\displaystyle \frac{A^y}{A^y+A^x}}+{\displaystyle \frac{A^x}{A^y+A^x}}\\ & ={\displaystyle \frac{A^y+A^x}{A^y+A^x}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 1}}}}\end{array}$

Jadi, \dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}=1.
JAWAB: D

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No. 4

Bentuk sederhana dari \left(\dfrac{p^{-2}q^2r}{pq^{-1}r^3}\right)^2 adalah ....

1. \dfrac{p^6}{q^5r^4}
2. \dfrac{p^6}{q^6r^2}
   
3. \dfrac{p^3}{q^5r^2}
   

4. \dfrac{q^6}{p^6r^4}
5. \dfrac{q^5p^6}{r^2}
   

$\begin{array}{rl}{\left({\displaystyle \frac{p^{-2}{q}^{2}r}{p{q}^{-1}{r}^3}}\right)}^2& ={\left({\displaystyle \frac{q^{2-(-1)}}{p^{1-(-2)}{r}^{3-1}}}\right)}^2\\ & ={\left({\displaystyle \frac{q^3}{p^3{r}^2}}\right)}^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{q^6}{p^6{r}^4}}}}}}\end{array}$

Jadi, \left(\dfrac{p^{-2}q^2r}{pq^{-1}r^3}\right)^2=\dfrac{q^6}{p^6r^4}.
JAWAB: D

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No. 5

Hasil dari \dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}} adalah

1. \dfrac{27}2
2. \dfrac92
3. \dfrac{27}8

4. \dfrac98
5. \dfrac8{27}

SUMBER

$\begin{array}{rl}{\displaystyle \frac{8^{-\frac{3}{5}}{9}^{\frac{5}{4}}}{{81}^{-\frac{1}{8}}{64}^{\frac{1}{5}}}}& ={\displaystyle \frac{8^{-\frac{3}{5}}{\left(3^2\right)}^{\frac{5}{4}}}{{\left(3^4\right)}^{-\frac{1}{8}}{\left(8^2\right)}^{\frac{1}{5}}}}\\ & ={\displaystyle \frac{8^{-\frac{3}{5}}{3}^{\frac{5}{2}}}{3^{-\frac{1}{2}}{8}^{\frac{2}{5}}}}\\ & ={\displaystyle \frac{3^{\frac{5}{2}+\frac{1}{2}}}{8^{\frac{2}{5}+\frac{3}{5}}}}\\ & ={\displaystyle \frac{3^{\frac{6}{2}}}{8^{\frac{5}{5}}}}\\ & ={\displaystyle \frac{3^3}{8^1}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{27}{8}}}}}}\end{array}$

Jadi, \dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}}=\dfrac{27}8.
JAWAB: C

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No. 6

Nilai x yang memenuhi persamaan {3^{2x-4}=1} adalah ....

1. -3
2. -2

3. 2
4. 3

Grup WhatsApp

$\begin{array}{rl}{3}^{2x-4}& =1\\ 3^{2x-4}& =3^0\\ 2x-4& =0\\ 2x& =4\\ x& ={\displaystyle \frac{4}{2}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$

Jadi, x=2.
JAWAB: C

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No. 7

Nyatakan bilangan berikut dalam bentuk perpangkatan

1. 64\cdot2^{-5}:16^3 (dalam basis 2)
2. 81\cdot27^{-4}:243^5 (dalam basis 3)
3. 0,004 (dalam basis 2 dan 5)

Grup Telegram

1. $\begin{array}{rl}64\cdot 2^{-5}:{16}^3& =2^6\cdot 2^{-5}:{\left(2^4\right)}^3\\ & =2^1:2^{12}\\ & =2^{1-12}\\ & =2^{-11}\end{array}$
2. $\begin{array}{rl}81\cdot {27}^{-4}:{243}^5& =3^4\cdot {\left(3^3\right)}^{-4}:{\left(3^5\right)}^5\\ & =3^4\cdot 3^{-12}:3^{25}\\ & =3^{4+(-12)-25}\\ & =3^{-33}\end{array}$
3. $\begin{array}{rl}0,004& =4\cdot {10}^{-3}\\ & =2^2\cdot (2\cdot 5{)}^{-3}\\ & =2^2\cdot 2^{-3}\cdot 5^{-3}\\ & =2^{-1}\cdot 5^{-3}\end{array}$

Jadi,

1. 64\cdot2^{-5}:16^3=2^{-11}
2. 81\cdot27^{-4}:243^5=3^{-33}
3. 0,004=2^{-1}\cdot5^{-3}

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No. 8

Tentukan x jika {4^x\cdot2^x=512}

Grup Telegram

$\begin{array}{rl}{4}^x\cdot 2^x& =512\\ {\left(2^2\right)}^x\cdot 2^x& =2^9\\ 2^{2x}\cdot 2^x& =2^9\\ 2^{2x+x}& =2^9\\ 2^{3x}& =2^9\\ 3x& =9\\ x& =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$

Jadi, x=3

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No. 9

Bentuk sederhana dari \dfrac{x^5\cdot y^7}{x^4\cdot y^8}

1. \dfrac{x}y
2. x^2\cdot y

3. x\cdot y
4. x\cdot y^2

Grup Telegram

$\begin{array}{rl}{\displaystyle \frac{x^5\cdot y^7}{x^4\cdot y^8}}& ={\displaystyle \frac{x^{5-4}}{y^{8-7}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{x}{y}}}}}}\end{array}$

Jadi, \dfrac{x^5\cdot y^7}{x^4\cdot y^8}=\dfrac{x}y.
JAWAB: A

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No. 10

Sederhanakan bentuk dari \left(\dfrac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot2r^2}\right)^2

Grup WhatsApp

$\begin{array}{rl}{\left({\displaystyle \frac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot 2r^2}}\right)}^2& ={\left({\displaystyle \frac{3q^{5-4}{r}^{3-2}}{p^{7-2}}}\right)}^2\\ & ={\left({\displaystyle \frac{3qr}{p^5}}\right)}^2\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{6q^2{r}^2}{p^{10}}}}}}}\end{array}$

Jadi, \left(\dfrac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot2r^2}\right)^2=\dfrac{6q^2r^2}{p^{10}}.