BILANGAN BERPANGKAT (EKSPONEN)
Defenisi Bilangan Berpangkat
Untuk $a$ bilangan real dan $n$ bilang bulat positif lebih dari 1,maka $a$ pangkat $n$ ditulis ($a^{n})$ adalah perkalian $n$ buah bilangan $a$. Defenisi ini dapat ditulis:
\begin{align} a^n=\underbrace{a×a×a×...×a}_{n\;kali}
\end{align}
Contoh$2^{3}=2\times2\times2=8$
$\left (\frac{1}{3} \right)^{2}=\frac{1}{3}\times \frac{1}{3}=\frac{1}{9}$
Sifat - Sifat Bilangan Berpangkat
Misalkan $a$ dan $b$ bilangan real $(a∈R)$ serta $m$, $n$ dan $p$ bilangan bulat positif maka berlaku:
- $a^m × a^m=a^{m+n}$
- $\frac{a^{m}}{a^{n}}=a^{m-n}, a≠0$
- $(a^{m})^{n}=(a)^{mn}$
- $(\frac{a^{m}}{b^{n}})^{p}=\frac{a^{mp}}{b^{np}},b≠0$
- $a^{-n}=\frac{1}{a^{n}},a≠0$
- $a^{0}=1$
Dengan memanfaatkan sifat-sifat tersebut kita dapat menyelesaikani soal-soal perpangkatan yang sangat kompleks dengan mudah. Perhatikanlah beberapa contoh soal berikut.
Contoh soal 1
Sederhanakan dan nyatakan dalam pangkat positif dari $\frac{x^{-3}y^{-5}z^{2}}{x^{-1}y^{-2}z^{-3}}$.
Jawab
\begin{align*} \frac{x^{-3}y^{-5}z^{2}}{x^{-1}y^{-2}z^{-3}}&=x^{(-3+1)}y^{(-5+2)}z^{(2+3)}\\ &=x^{-2}y^{-3}z^{5}\\ &=\frac{z^{5}}{x^{2}y^{3}} \end{align*}$
Contoh Soal 2
Sederhanakanlah!
\begin{equation}
\left(\frac{-2a^{3}.b^{-4}}{5a^{-5}.b^{-6}}\right)^{-2}
\end{equation}
Jawab
$\begin{align*} \left(\frac{-2a^{3}b^{-4}}{5a^{-5}b^{-6}}\right)^{-2}&=\frac{(-2)^{-2}a^{-6}b^{8}}{(5)^{-2}a^{10}b^{12}}\\ &=\frac{25}{4}a^{-16}b^{-4}\\ &=\frac{25}{4}.\frac{1}{a^{16}}.\frac{1}{b^{4}}\\ &=\frac{25}{a^{16}b^{4}} \end{align*}$
Contoh soal 3
Jika $a>0$ dan $b>0$ serta $a≠b$, maka $\begin{align*}\frac{(a+b)^{-1}(a^{-2}-b^{-2})}{(a^{-1}+b^{-1})(ab^{-1}-a^{-1}b)}\end{align*}$ = ....
(A) $\begin{align*}\frac{1}{\left(a+b\right)^{2}}\end{align*}$
(B) $(a + b)^{2}$
(C) $\begin{align*}\frac{-ab}{(a+b)^{2}}\end{align*}$
(D) $\begin{align*}\frac{ab}{a+b}\end{align*}$
(E) $\begin{align*}ab\end{align*}$
Jawab
Contoh Soal 4
Jika $a+b≠0$, maka bentuk sederhana dari $\begin{align*}\frac{a^{-1}-b^{-1}}{a^{-2}-b^{-2}} =\end{align*}$....
(A). $\begin{align*}\frac{a+b}{ab}\end{align*}$
(B). $\begin{align*}\frac{ab}{2(a+b)}\end{align*}$
(C). $\begin{align*}\frac{ab}{a+b}\end{align*}$
(D). $\begin{align*}\frac{ab}{a^{2}-b^{2}}\end{align*}$
(E). $\begin{align*}\frac{a - b}{a + b}\end{align*}$
Jawab
$\begin{align*} \frac{a^{-1}-b^{-1}}{a^{-2}-b^{-2}}&=\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a^{2}}-\frac{1}{b^{2}}}\\ &=\frac{\left(\frac{b-a}{ab}\right)}{\left(\frac{b^{2}-a^{2}}{(ab)^{2}} \right )}\\ &=\frac{b-a}{ab}\times \frac{(ab)^{2}}{b^{2}-a^{2}}\\ &=\frac{(b-a)(ab)}{(b+a)(b-a)}\\ &=\frac{ab}{a+b} \end{align*}$
Contoh soal 5
Nyatakan $\begin{align*} \frac{x^{2}(x-y)^{-1}-x}{y^{2}(x-y)^{-1}+y} \end{align*}$, dimana $x\neq y$ dalam bentuk yang paling sederhana!
Jawab
$\begin{align*} \frac{x^{2}(x-y)^{-1}-x}{y^{2}(x-y)^{-1}+y}&=\frac{\frac{x^{2}}{x-y}-x}{\frac{y^{2}}{x-y}+y}\\ &=\frac{\left(\frac{x^{2}}{x-y}\right)-\left(\frac{x(x-y)}{x-y}\right)}{\left (\frac{y^{2}}{x-y} \right )+\left (\frac{y(x-y)}{x-y} \right )}\\ &=\frac{\left (\frac{x^{2}-x(x-y)}{x-y} \right )}{\left (\frac{y^{2}+y(x-y)}{x-y} \right )}\\ &=\frac{x^{2}-x^{2}+xy}{y^{2}+xy-y^{2}}\\ &=\frac{xy}{xy}\\ &=1 \end{align*}$
Berikut beberapa soal Ujian Nasional SMA tentang perpangkatan.
Soal UN 2011
Bentuk sederhana dari $\begin{align*} \frac{7x^{3}y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}} \end{align*}$ adalah ....
A. $\begin{align*} \frac{x^{10}z^{10}}{12y^{3}} \end{align*}$
B. $\begin{align*} \frac{z^{2}}{12x^{4}y^{3}} \end{align*}$
C. $\begin{align*} \frac{x^{10}y^{5}}{12z^{2}} \end{align*}$
D. $\begin{align*} \frac{y^{3}z^{2}}{12x^{4}} \end{align*}$
E. $\begin{align*} \frac{x^{10}}{12y^{3}z^{2}} \end{align*}$
Pembahasan
$\begin{align*} \frac{7x^{3}y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}}&=\frac{1}{12}.x^{(3-(-7))}y^{(-4-(-1))}z^{(-6-(-4))}\\ &=\frac{1}{12}.x^{10}y^{-3}z^{-2}\\ &=\frac{1}{12}.\frac{x^{10}}{y^{3}z^{2}}\\ &=\frac{x^{10}}{12y^{3}z^{2}} \end{align*}$
Kunci A
Soal UN 2012
Diketahui $a=4$, $b=2$, dan $c=\frac{1}{2}$. Nilai dari $\begin{align*} \left ( a^{-1} \right )^{2}\times \frac{b^{4}}{c^{-3}} \end{align*}$ adalah ....
A. $\begin{align*} \frac{1}{2} \end{align*}$
B. $\begin{align*} \frac{1}{4} \end{align*}$
C. $\begin{align*} \frac{1}{8} \end{align*}$
D. $\begin{align*} \frac{1}{16} \end{align*}$
E. $\begin{align*} \frac{1}{32} \end{align*}$
Pembahasan
$\begin{align*} \left ( a^{-1} \right )^{2}\times \frac{b^{4}}{c^{-3}}&=a^{-2}\times \frac{b^{4}}{c^{-3}}\\ &=\frac{b^{4}c^{3}}{a^{2}}\\ \end{align*}$
Substitusi nilai $a=4$, $b=2$, dan $c=\frac{1}{2}$ ke bentuk terakhir, maka diperoleh:
$\begin{align*} \frac{b^{4}c^{3}}{a^{2}}&=\frac{({\color{Red} 2^{4}})(\frac{1}{2})^{3}}{({\color{Red} 4^{2}})}\\ &=\left ( \frac{1}{2} \right )^{3}\\ &=\frac{1}{8} \end{align*}$
Kunci: C
Soal UN 2013
Bentuk sederhana dari $\begin{align*} \left (\frac{4a^{-2}b^{2}c}{12a^{-5}b^{4}c^{-1}} \right )^{-1} \end{align*}$ adalah ....
A. $\begin{align*} \frac{3b^{6}}{a^{3}c} \end{align*}$
B. $\begin{align*} \frac{3b^{6}}{a^{7}c^{2}} \end{align*}$
C. $\begin{align*} \frac{3b^{2}}{a^{3}c^{2}} \end{align*}$
D. $\begin{align*} \frac{a^{3}c^{2}}{3b^{2}} \end{align*}$
E. $\begin{align*} \frac{a^{7}c^{2}}{3b^{6}} \end{align*}$
Pembahasan
$\begin{align*} \left (\frac{4a^{-2}b^{2}c}{12a^{-5}b^{4}c^{-1}} \right )^{-1}&=\left ( \frac{1}{3}a^{3}b^{-2}c^{2} \right )^{-1}\\ &=3a^{-3}b^{2}c^{-2}\\ &=\frac{3b^{2}}{a^{3}c^{2}} \end{align*}$
Kunci C
Soal UN 2015
Bentuk Sederhana dari $\begin{align*} \left (\frac{4x^{-\frac{2}{3}}y^{\frac{4}{5}}z^{-\frac{5}{2}}}{5z^{-\frac{3}{2}}x^{-\frac{2}{3}}y^{-\frac{1}{5}}} \right )^{2} \end{align*}$ adalah ....
A. $\begin{align*} \frac{16xy}{25z} \end{align*}$
B. $\begin{align*} \frac{4xy}{5z} \end{align*}$
C. $\begin{align*} \frac{4x^{2}y^{2}}{5z^{2}} \end{align*}$
D. $\begin{align*} \frac{16x^{2}y^{2}}{25z^{2}} \end{align*}$
E. $\begin{align*} \frac{16xy}{25z^{2}} \end{align*}$
Pembahasan
$\begin{align*} \left (\frac{4x^{-\frac{2}{3}}y^{\frac{4}{5}}z^{-\frac{5}{2}}}{5z^{-\frac{3}{2}}x^{-\frac{5}{3}}y^{-\frac{1}{5}}} \right )^{2}&=\left (\frac{4}{5}x^{(-\frac{2}{3}+\frac{5}{3})}y^{(\frac{4}{5}+\frac{1}{5})}z^{(-\frac{5}{2}+\frac{3}{2})} \right )^{2}\\ &=\left (\frac{4}{5}x^{1}y^{1}z^{-1} \right )^{2}\\ &=\left ( \frac{4xy}{5z} \right )^{2}\\ &=\frac{16x^{2}y^{2}}{25z^{2}} \end{align*}$
Kunci D
Soal UN 2016
Nilai dari $\begin{align*} \frac{(125)^{\frac{2}{3}}-(25)^{\frac{1}{2}}}{(81)^{\frac{1}{4}}+(27)^{\frac{1}{3}}} \end{align*}$ adalah ....
A. $\begin{align*} \frac{8}{3} \end{align*}$
B. $\begin{align*} \frac{10}{3} \end{align*}$
C. $\begin{align*} \frac{14}{3} \end{align*}$
D. $\begin{align*} \frac{16}{3} \end{align*}$
E. $\begin{align*} \frac{20}{3} \end{align*}$
Pembahasan
$\begin{align*} \frac{(125)^{\frac{2}{3}}-(25)^{\frac{1}{2}}}{(81)^{\frac{1}{4}}+(27)^{\frac{1}{3}}}&=\frac{\left(5^{3}\right)^{\frac{2}{3}}-\left(5^{2}\right)^{\frac{1}{2}}}{\left(3^{4}\right)^{\frac{1}{4}}+\left(3^{3}\right)^{\frac{1}{3}}}\\ &=\frac{5^{2}-5}{3+3}\\ &=\frac{20}{6}\\ &=\frac{10}{3} \end{align*}$
Contoh soal 1
Sederhanakan dan nyatakan dalam pangkat positif dari $\frac{x^{-3}y^{-5}z^{2}}{x^{-1}y^{-2}z^{-3}}$.
Jawab
\begin{align*} \frac{x^{-3}y^{-5}z^{2}}{x^{-1}y^{-2}z^{-3}}&=x^{(-3+1)}y^{(-5+2)}z^{(2+3)}\\ &=x^{-2}y^{-3}z^{5}\\ &=\frac{z^{5}}{x^{2}y^{3}} \end{align*}$
Contoh Soal 2
Sederhanakanlah!
\begin{equation}
\left(\frac{-2a^{3}.b^{-4}}{5a^{-5}.b^{-6}}\right)^{-2}
\end{equation}
Jawab
$\begin{align*} \left(\frac{-2a^{3}b^{-4}}{5a^{-5}b^{-6}}\right)^{-2}&=\frac{(-2)^{-2}a^{-6}b^{8}}{(5)^{-2}a^{10}b^{12}}\\ &=\frac{25}{4}a^{-16}b^{-4}\\ &=\frac{25}{4}.\frac{1}{a^{16}}.\frac{1}{b^{4}}\\ &=\frac{25}{a^{16}b^{4}} \end{align*}$
Contoh soal 3
Jika $a>0$ dan $b>0$ serta $a≠b$, maka $\begin{align*}\frac{(a+b)^{-1}(a^{-2}-b^{-2})}{(a^{-1}+b^{-1})(ab^{-1}-a^{-1}b)}\end{align*}$ = ....
(A) $\begin{align*}\frac{1}{\left(a+b\right)^{2}}\end{align*}$
(B) $(a + b)^{2}$
(C) $\begin{align*}\frac{-ab}{(a+b)^{2}}\end{align*}$
(D) $\begin{align*}\frac{ab}{a+b}\end{align*}$
(E) $\begin{align*}ab\end{align*}$
Jawab
Contoh Soal 4
Jika $a+b≠0$, maka bentuk sederhana dari $\begin{align*}\frac{a^{-1}-b^{-1}}{a^{-2}-b^{-2}} =\end{align*}$....
(A). $\begin{align*}\frac{a+b}{ab}\end{align*}$
(B). $\begin{align*}\frac{ab}{2(a+b)}\end{align*}$
(C). $\begin{align*}\frac{ab}{a+b}\end{align*}$
(D). $\begin{align*}\frac{ab}{a^{2}-b^{2}}\end{align*}$
(E). $\begin{align*}\frac{a - b}{a + b}\end{align*}$
Jawab
$\begin{align*} \frac{a^{-1}-b^{-1}}{a^{-2}-b^{-2}}&=\frac{\frac{1}{a}-\frac{1}{b}}{\frac{1}{a^{2}}-\frac{1}{b^{2}}}\\ &=\frac{\left(\frac{b-a}{ab}\right)}{\left(\frac{b^{2}-a^{2}}{(ab)^{2}} \right )}\\ &=\frac{b-a}{ab}\times \frac{(ab)^{2}}{b^{2}-a^{2}}\\ &=\frac{(b-a)(ab)}{(b+a)(b-a)}\\ &=\frac{ab}{a+b} \end{align*}$
Contoh soal 5
Nyatakan $\begin{align*} \frac{x^{2}(x-y)^{-1}-x}{y^{2}(x-y)^{-1}+y} \end{align*}$, dimana $x\neq y$ dalam bentuk yang paling sederhana!
Jawab
$\begin{align*} \frac{x^{2}(x-y)^{-1}-x}{y^{2}(x-y)^{-1}+y}&=\frac{\frac{x^{2}}{x-y}-x}{\frac{y^{2}}{x-y}+y}\\ &=\frac{\left(\frac{x^{2}}{x-y}\right)-\left(\frac{x(x-y)}{x-y}\right)}{\left (\frac{y^{2}}{x-y} \right )+\left (\frac{y(x-y)}{x-y} \right )}\\ &=\frac{\left (\frac{x^{2}-x(x-y)}{x-y} \right )}{\left (\frac{y^{2}+y(x-y)}{x-y} \right )}\\ &=\frac{x^{2}-x^{2}+xy}{y^{2}+xy-y^{2}}\\ &=\frac{xy}{xy}\\ &=1 \end{align*}$
Berikut beberapa soal Ujian Nasional SMA tentang perpangkatan.
Soal UN 2011
Bentuk sederhana dari $\begin{align*} \frac{7x^{3}y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}} \end{align*}$ adalah ....
A. $\begin{align*} \frac{x^{10}z^{10}}{12y^{3}} \end{align*}$
B. $\begin{align*} \frac{z^{2}}{12x^{4}y^{3}} \end{align*}$
C. $\begin{align*} \frac{x^{10}y^{5}}{12z^{2}} \end{align*}$
D. $\begin{align*} \frac{y^{3}z^{2}}{12x^{4}} \end{align*}$
E. $\begin{align*} \frac{x^{10}}{12y^{3}z^{2}} \end{align*}$
Pembahasan
$\begin{align*} \frac{7x^{3}y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}}&=\frac{1}{12}.x^{(3-(-7))}y^{(-4-(-1))}z^{(-6-(-4))}\\ &=\frac{1}{12}.x^{10}y^{-3}z^{-2}\\ &=\frac{1}{12}.\frac{x^{10}}{y^{3}z^{2}}\\ &=\frac{x^{10}}{12y^{3}z^{2}} \end{align*}$
Kunci A
Soal UN 2012
Diketahui $a=4$, $b=2$, dan $c=\frac{1}{2}$. Nilai dari $\begin{align*} \left ( a^{-1} \right )^{2}\times \frac{b^{4}}{c^{-3}} \end{align*}$ adalah ....
A. $\begin{align*} \frac{1}{2} \end{align*}$
B. $\begin{align*} \frac{1}{4} \end{align*}$
C. $\begin{align*} \frac{1}{8} \end{align*}$
D. $\begin{align*} \frac{1}{16} \end{align*}$
E. $\begin{align*} \frac{1}{32} \end{align*}$
Pembahasan
$\begin{align*} \left ( a^{-1} \right )^{2}\times \frac{b^{4}}{c^{-3}}&=a^{-2}\times \frac{b^{4}}{c^{-3}}\\ &=\frac{b^{4}c^{3}}{a^{2}}\\ \end{align*}$
Substitusi nilai $a=4$, $b=2$, dan $c=\frac{1}{2}$ ke bentuk terakhir, maka diperoleh:
$\begin{align*} \frac{b^{4}c^{3}}{a^{2}}&=\frac{({\color{Red} 2^{4}})(\frac{1}{2})^{3}}{({\color{Red} 4^{2}})}\\ &=\left ( \frac{1}{2} \right )^{3}\\ &=\frac{1}{8} \end{align*}$
Kunci: C
Soal UN 2013
Bentuk sederhana dari $\begin{align*} \left (\frac{4a^{-2}b^{2}c}{12a^{-5}b^{4}c^{-1}} \right )^{-1} \end{align*}$ adalah ....
A. $\begin{align*} \frac{3b^{6}}{a^{3}c} \end{align*}$
B. $\begin{align*} \frac{3b^{6}}{a^{7}c^{2}} \end{align*}$
C. $\begin{align*} \frac{3b^{2}}{a^{3}c^{2}} \end{align*}$
D. $\begin{align*} \frac{a^{3}c^{2}}{3b^{2}} \end{align*}$
E. $\begin{align*} \frac{a^{7}c^{2}}{3b^{6}} \end{align*}$
Pembahasan
$\begin{align*} \left (\frac{4a^{-2}b^{2}c}{12a^{-5}b^{4}c^{-1}} \right )^{-1}&=\left ( \frac{1}{3}a^{3}b^{-2}c^{2} \right )^{-1}\\ &=3a^{-3}b^{2}c^{-2}\\ &=\frac{3b^{2}}{a^{3}c^{2}} \end{align*}$
Kunci C
Soal UN 2015
Bentuk Sederhana dari $\begin{align*} \left (\frac{4x^{-\frac{2}{3}}y^{\frac{4}{5}}z^{-\frac{5}{2}}}{5z^{-\frac{3}{2}}x^{-\frac{2}{3}}y^{-\frac{1}{5}}} \right )^{2} \end{align*}$ adalah ....
A. $\begin{align*} \frac{16xy}{25z} \end{align*}$
B. $\begin{align*} \frac{4xy}{5z} \end{align*}$
C. $\begin{align*} \frac{4x^{2}y^{2}}{5z^{2}} \end{align*}$
D. $\begin{align*} \frac{16x^{2}y^{2}}{25z^{2}} \end{align*}$
E. $\begin{align*} \frac{16xy}{25z^{2}} \end{align*}$
Pembahasan
$\begin{align*} \left (\frac{4x^{-\frac{2}{3}}y^{\frac{4}{5}}z^{-\frac{5}{2}}}{5z^{-\frac{3}{2}}x^{-\frac{5}{3}}y^{-\frac{1}{5}}} \right )^{2}&=\left (\frac{4}{5}x^{(-\frac{2}{3}+\frac{5}{3})}y^{(\frac{4}{5}+\frac{1}{5})}z^{(-\frac{5}{2}+\frac{3}{2})} \right )^{2}\\ &=\left (\frac{4}{5}x^{1}y^{1}z^{-1} \right )^{2}\\ &=\left ( \frac{4xy}{5z} \right )^{2}\\ &=\frac{16x^{2}y^{2}}{25z^{2}} \end{align*}$
Kunci D
Soal UN 2016
Nilai dari $\begin{align*} \frac{(125)^{\frac{2}{3}}-(25)^{\frac{1}{2}}}{(81)^{\frac{1}{4}}+(27)^{\frac{1}{3}}} \end{align*}$ adalah ....
A. $\begin{align*} \frac{8}{3} \end{align*}$
B. $\begin{align*} \frac{10}{3} \end{align*}$
C. $\begin{align*} \frac{14}{3} \end{align*}$
D. $\begin{align*} \frac{16}{3} \end{align*}$
E. $\begin{align*} \frac{20}{3} \end{align*}$
Pembahasan
$\begin{align*} \frac{(125)^{\frac{2}{3}}-(25)^{\frac{1}{2}}}{(81)^{\frac{1}{4}}+(27)^{\frac{1}{3}}}&=\frac{\left(5^{3}\right)^{\frac{2}{3}}-\left(5^{2}\right)^{\frac{1}{2}}}{\left(3^{4}\right)^{\frac{1}{4}}+\left(3^{3}\right)^{\frac{1}{3}}}\\ &=\frac{5^{2}-5}{3+3}\\ &=\frac{20}{6}\\ &=\frac{10}{3} \end{align*}$
Kunci B
Soal UN 2017
Hasil dari $\begin{align*} \frac{\left (8^{-\frac{3}{5}}.9^{\frac{5}{4}} \right )}{81^{-\frac{1}{8}}.64^{\frac{1}{5}}} \end{align*}$ adalah ....
A. $\begin{align*} \frac{27}{2} \end{align*}$
B. $\begin{align*} \frac{9}{2} \end{align*}$
C. $\begin{align*} \frac{27}{8} \end{align*}$
D. $\begin{align*} \frac{9}{8} \end{align*}$
E. $\begin{align*} \frac{8}{27} \end{align*}$
Pembahasan
$\begin{align*} \frac{\left (8^{-\frac{3}{5}}.9^{\frac{5}{4}} \right )}{\left(81^{-\frac{1}{8}}.64^{\frac{1}{5}} \right )} &=\frac{\left ((2^{3})^{-\frac{3}{5}}.(3^{2})^{\frac{5}{4}} \right )}{(3^{4})^{-\frac{1}{8}}.(2^{6})^{\frac{1}{5}}}\\ &=\frac{2^{-\frac{9}{5}}.3^{\frac{5}{2}}}{3^{-\frac{1}{2}}.2^{\frac{6}{5}}}\\ &=2^{-\frac{15}{5}}.3^{\frac{6}{2}}\\ &=2^{-3}.3^{3}\\ &=\frac{27}{8} \end{align*}$
Kunci C
Soal UN 2017
Hasil dari $\begin{align*} \frac{\left (3^{\frac{3}{5}}.8^{-\frac{3}{2}} \right )}{\left(32^{\frac{1}{10}}.81^{-\frac{3}{5}} \right )} \end{align*}$ adalah ....
A. $\begin{align*} \frac{27}{2} \end{align*}$
B. $\begin{align*} \frac{32}{9} \end{align*}$
C. $\begin{align*} \frac{27}{16} \end{align*}$
D. $\begin{align*} \frac{32}{27} \end{align*}$
E. $\begin{align*} \frac{27}{32} \end{align*}$
Pembahasan
$\begin{align*} \frac{\left (3^{\frac{3}{5}}.8^{-\frac{3}{2}} \right )}{\left(32^{\frac{1}{10}}.81^{-\frac{3}{5}} \right )}&=\frac{\left (3^{\frac{3}{5}}.(2^{3})^{-\frac{3}{2}} \right )}{\left((2^{5})^{\frac{1}{10}}.(3^{4})^{-\frac{3}{5}} \right )}\\ &=\frac{\left (3^{\frac{3}{5}}.2^{-\frac{9}{2}} \right )}{\left(2^{\frac{1}{2}}3^{-\frac{12}{5}} \right )}\\ &=3^{\frac{15}{5}}.2^{-\frac{10}{2}}\\ &=3^{3}.2^{-5}\\ &=\frac{27}{32} \end{align*}$
Kunci E
Posting Komentar