Berikut ini adalah kumpulan soal mengenai Pangkat (Eksponen) tingkat dasar. Jika ada jawaban yang salah, mohon dikoreksi melalui komentar. Terima kasih.
No. 1
Hitunglah hasil perpangkatan berikut
\left(\dfrac{27}{125}\right)^{\frac13}\left(\dfrac19\right)^{-2}
\(\begin{aligned}
\left(\dfrac{27}{125}\right)^{\frac13}\left(\dfrac19\right)^{-2}&=\left(\dfrac{3^3}{5^3}\right)^{\frac13}\left(\dfrac1{3^2}\right)^{-2}\\[8pt]
&=\dfrac{\left(3^3\right)^{\frac13}}{\left(5^3\right)^{\frac13}}\left(3^{-2}\right)^{-2}\\[8pt]
&=\dfrac35\left(3^4\right)\\[8pt]
&=\dfrac35\left(81\right)\\
&=\boxed{\boxed{\dfrac{243}5}}
\end{aligned}\)
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
- a^{11}b^{12}
- a^{12}b^{11}
- a^{14}b^{13}
- a^{13}b^{12}
- a^{12}b^{13}
\(\begin{aligned}
\dfrac{\sqrt{bc}}{\sqrt[5]{ab^4}}&= ab\\
\left(\dfrac{\sqrt{bc}}{\sqrt[5]{ab^4}}\right)^{10}&=(ab)^{10}\\
\dfrac{(bc)^5}{\left(ab^4\right)^2}&=a^{10}b^{10}\\
\dfrac{b^5c^5}{a^2b^8}&=a^{10}b^{10}\\
c^5&=\dfrac{a^2b^8\cdot a^{10}b^{10}}{b^5}\\
&=\boxed{\boxed{a^{12}b^{13}}}
\end{aligned}\)
No. 3
{\dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}=} ....
\(\begin{aligned}
\dfrac{1}{1+A^{x-y}}+\dfrac{1}{1+A^{y-x}}&=\dfrac{1}{1+\dfrac{A^x}{A^y}}+\dfrac{1}{1+\dfrac{A^y}{A^x}}\\[20pt]
&=\dfrac{1}{\dfrac{A^y+A^x}{A^y}}+\dfrac{1}{\dfrac{A^x+A^y}{A^x}}\\[20pt]
&=\dfrac{A^y}{A^y+A^x}+\dfrac{A^x}{A^x+A^y}\\[8pt]
&=\dfrac{A^y}{A^y+A^x}+\dfrac{A^x}{A^y+A^x}\\[8pt]
&=\dfrac{A^y+A^x}{A^y+A^x}\\
&=\boxed{\boxed{1}}
\end{aligned}\)
No. 4
Bentuk sederhana dari
\left(\dfrac{p^{-2}q^2r}{pq^{-1}r^3}\right)^2 adalah ....
- \dfrac{p^6}{q^5r^4}
- \dfrac{p^6}{q^6r^2}
- \dfrac{p^3}{q^5r^2}
- \dfrac{q^6}{p^6r^4}
- \dfrac{q^5p^6}{r^2}
\(\begin{aligned}
\left(\dfrac{p^{-2}q^2r}{pq^{-1}r^3}\right)^2&=\left(\dfrac{q^{2-(-1)}}{p^{1-(-2)}r^{3-1}}\right)^2\\[8pt]
&=\left(\dfrac{q^3}{p^3r^2}\right)^2\\[8pt]
&=\boxed{\boxed{\dfrac{q^6}{p^6r^4}}}
\end{aligned}\)
No. 5
Hasil dari
\dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}} adalah
- \dfrac{27}2
- \dfrac92
- \dfrac{27}8
\(\eqalign{
\dfrac{8^{-\frac35}9^{\frac54}}{81^{-\frac18}64^{\frac15}}&=\dfrac{8^{-\frac35}\left(3^2\right)^{\frac54}}{\left(3^4\right)^{-\frac18}\left(8^2\right)^{\frac15}}\\
&=\dfrac{8^{-\frac35}3^{\frac52}}{3^{-\frac12}8^{\frac25}}\\
&=\dfrac{3^{\frac52+\frac12}}{8^{\frac25+\frac35}}\\
&=\dfrac{3^{\frac62}}{8^{\frac55}}\\
&=\dfrac{3^3}{8^1}\\
&=\boxed{\boxed{\dfrac{27}8}}
}\)
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