Exercise Zone : Pangkat (Eksponen)

Berikut ini adalah kumpulan soal mengenai Pangkat (Eksponen) tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.

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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

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

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

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