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Exercise Zone : Pangkat (Eksponen)

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Idham Muqoddas
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Berikut ini adalah kumpulan soal mengenai Pangkat (Eksponen). Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

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StandarSNBTHOTS


No.

Hitunglah hasil perpangkatan berikut $$\left(\dfrac{27}{125}\right)^{\frac13}\left(\dfrac19\right)^{-2}$$
Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}\\[4pt] &=\dfrac{\left(3^3\right)^{\frac13}}{\left(5^3\right)^{\frac13}}\left(3^{-2}\right)^{-2}\\[4pt] &=\dfrac35\left(3^4\right)\\[4pt] &=\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.

No.

Diketahui a, b, dan c adalah bilangan real positif, jika {\dfrac{\sqrt{bc}}{\sqrt[5]{ab^4}}= ab}, maka nilai c5 adalah
  1. a11b12
  2. a12b11
  3. a14b13
  1. a13b12
  2. a12b13
SUMBER
ALTERNATIF PENYELESAIAN
\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, c5 = a12b13.
JAWAB: E

No.

{\dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}=....}
  1. −1
  2. 0
  3. \dfrac12
  1. 1
  2. Ax + y
Zenius.net
ALTERNATIF PENYELESAIAN
\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}}\\[4pt] &=\dfrac{1}{\dfrac{A^y+A^x}{A^y}}+\dfrac{1}{\dfrac{A^x+A^y}{A^x}}\\[4pt] &=\dfrac{A^y}{A^y+A^x}+\dfrac{A^x}{A^x+A^y}\\[4pt] &=\dfrac{A^y}{A^y+A^x}+\dfrac{A^x}{A^y+A^x}\\[4pt] &=\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

No.

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}
  1. \dfrac{q^6}{p^6r^4}
  2. \dfrac{q^5p^6}{r^2}
SUMBER
ALTERNATIF PENYELESAIAN
\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\\[4pt] &=\left(\dfrac{q^3}{p^3r^2}\right)^2\\[4pt] &=\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

No.

Hasil dari \dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}} adalah
  1. \dfrac{27}2
  2. \dfrac92
  3. \dfrac{27}8
  1. \dfrac98
  2. \dfrac8{27}
SUMBER
ALTERNATIF PENYELESAIAN
\begin{aligned} \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}} \end{aligned}
Jadi, \dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}}=\dfrac{27}8.
JAWAB: C

No.

25 × 27 = ....
  1. 28
  2. 212
  1. 218
  2. 225
  1. 235
Grup WhatsApp
ALTERNATIF PENYELESAIAN
\begin{aligned} 2^5\times2^7&=2^{5+7}\\ &=\color{blue}\boxed{\boxed{\color{black}2^{12}}} \end{aligned}
Jadi, 25 × 27 = 212.
JAWAB: B

No.

$\dfrac{3^5}{3^2}=$ ....
  1. 1
  2. 3
  1. 9
  2. 27
  1. 81
Grup WhatsApp
ALTERNATIF PENYELESAIAN
\(\begin{aligned} \dfrac{3^5}{3^2}&=3^{5-2}\\ &=3^3\\ &=\color{blue}\boxed{\boxed{\color{black}27}} \end{aligned}\)
Jadi, $\dfrac{3^5}{3^2}=27$.
JAWAB: D

No.

Bentuk sederhana dari $\dfrac{7x^3y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}}=$ ....
  1. $\dfrac{x^{10}z^{10}}{12y^3}$
  2. $\dfrac{z^2}{12x^4y^3}$
  1. $\dfrac{x^{10}y^5}{12z^2}$
  2. $\dfrac{y^3z^2}{12x^4}$
  1. $\dfrac{x^{10}}{12y^3z^2}$
Big Book M3qatematika SMA
ALTERNATIF PENYELESAIAN
\begin{aligned} \dfrac{7x^3y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}}&=\dfrac{x^{3-(-7)}}{12y^{-1-(-4)}z^{-4-(-6)}}\\ &=\color{blue}\boxed{\boxed{\color{black}\dfrac{x^{10}}{12y^3z^2}}} \end{aligned}
Jadi, $\dfrac{7x^3y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}}=\dfrac{x^{10}}{12y^3z^2}$.
JAWAB: E

No.

Bentuk sederhana dari \dfrac{x^5\cdot y^7}{x^4\cdot y^8}
  1. \dfrac{x}y
  2. x2⋅ y
  1. x⋅ y
  2. x⋅ y2
Grup Telegram
ALTERNATIF PENYELESAIAN
\begin{aligned} \dfrac{x^5\cdot y^7}{x^4\cdot y^8}&=\dfrac{x^{5-4}}{y^{8-7}}\\ &=\boxed{\boxed{\dfrac{x}y}} \end{aligned}
Jadi, \dfrac{x^5\cdot y^7}{x^4\cdot y^8}=\dfrac{x}y.
JAWAB: A

No.

Sederhanakan bentuk dari \left(\dfrac{6p^2\cdot q^5\cdot r^3}{p^7\cdot q^4\cdot2r^2}\right)^2
Grup WhatsApp
ALTERNATIF PENYELESAIAN
\begin{aligned} \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}}}} \end{aligned}
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}}.


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