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}

\(\begin{aligned} \left(\dfrac{27}{125}\right)^{\frac13}\left(\dfrac19\right)^{-2}&=\left(\dfrac{3^3}{5^3}\right)^{\frac13}\left(\dfrac1{3^2}\right)^{-2}\\[8pt] &=\dfrac{\left(3^3\right)^{\frac13}}{\left(5^3\right)^{\frac13}}\left(3^{-2}\right)^{-2}\\[8pt] &=\dfrac35\left(3^4\right)\\[8pt] &=\dfrac35\left(81\right)\\ &=\boxed{\boxed{\dfrac{243}5}} \end{aligned}\)

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}

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

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

\(\begin{aligned} \dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}&=\dfrac{1}{1+\dfrac{A^x}{A^y}}+\dfrac{1}{1+\dfrac{A^y}{A^x}}\\[20pt] &=\dfrac{1}{\dfrac{A^y+A^x}{A^y}}+\dfrac{1}{\dfrac{A^x+A^y}{A^x}}\\[20pt] &=\dfrac{A^y}{A^y+A^x}+\dfrac{A^x}{A^x+A^y}\\[8pt] &=\dfrac{A^y}{A^y+A^x}+\dfrac{A^x}{A^y+A^x}\\[8pt] &=\dfrac{A^y+A^x}{A^y+A^x}\\ &=\boxed{\boxed{1}} \end{aligned}\)

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}
   

\(\begin{aligned} \left(\dfrac{p^{-2}q^2r}{pq^{-1}r^3}\right)^2&=\left(\dfrac{q^{2-(-1)}}{p^{1-(-2)}r^{3-1}}\right)^2\\[8pt] &=\left(\dfrac{q^3}{p^3r^2}\right)^2\\[8pt] &=\boxed{\boxed{\dfrac{q^6}{p^6r^4}}} \end{aligned}\)

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

\(\eqalign{ \dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}}&=\dfrac{8^{-\frac35}\left(3^2\right)^{\frac54}}{\left(3^4\right)^{-\frac18}\left(8^2\right)^{\frac15}}\\ &=\dfrac{8^{-\frac35}3^{\frac52}}{3^{-\frac12}8^{\frac25}}\\ &=\dfrac{3^{\frac52+\frac12}}{8^{\frac25+\frac35}}\\ &=\dfrac{3^{\frac62}}{8^{\frac55}}\\ &=\dfrac{3^3}{8^1}\\ &=\boxed{\boxed{\dfrac{27}8}} }\)

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

\(\eqalign{ 3^{2x-4}&=1\\ 3^{2x-4}&=3^0\\ 2x-4&=0\\ 2x&=4\\ x&=\dfrac42\\ &=\boxed{\boxed{2}} }\)

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. \(\eqalign{ 64\cdot2^{-5}:16^3&=2^6\cdot2^{-5}:\left(2^4\right)^3\\ &=2^1:2^{12}\\ &=2^{1-12}\\ &=2^{-11} }\)
2. \(\eqalign{ 81\cdot27^{-4}:243^5&=3^4\cdot\left(3^3\right)^{-4}:\left(3^5\right)^5\\ &=3^4\cdot3^{-12}:3^{25}\\ &=3^{4+(-12)-25}\\ &=3^{-33} }\)
3. \(\eqalign{ 0,004&=4\cdot10^{-3}\\ &=2^2\cdot(2\cdot5)^{-3}\\ &=2^2\cdot2^{-3}\cdot5^{-3}\\ &=2^{-1}\cdot5^{-3} }\)

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

\(\eqalign{ 4^x\cdot2^x&=512\\ \left(2^2\right)^x\cdot2^x&=2^9\\ 2^{2x}\cdot2^x&=2^9\\ 2^{2x+x}&=2^9\\ 2^{3x}&=2^9\\ 3x&=9\\ x&=\boxed{\boxed{3}} }\)

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

\(\eqalign{ \dfrac{x^5\cdot y^7}{x^4\cdot y^8}&=\dfrac{x^{5-4}}{y^{8-7}}\\ &=\boxed{\boxed{\dfrac{x}y}} }\)

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

\(\eqalign{ \left(\dfrac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot2r^2}\right)^2&=\left(\dfrac{3q^{5-4}r^{3-2}}{p^{7-2}}\right)^2\\ &=\left(\dfrac{3qr}{p^5}\right)^2\\ &=\boxed{\boxed{\dfrac{6q^2r^2}{p^{10}}}} }\)

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}}.