Pastikan kau cukup percaya pada mimpi-mimpimu..sampai-sampai langit dan bumi tidak bisa tidak membuktikan kepercayaan tersebut..dengan cara mewujudkan mimpi-mimpi itu menjadi nyata.- R. Rusandi , "Inward"
Soal berikut ini dipilih khusus untuk meningkatkan kemampuan sobat dalam menyelesaikan berbagai variasi kesulitan soal pada bab Eksponen .
Eksponen sendiri merupakan perkalian berulang kompleks atau operasi bilangan dengan pangkat. Di sini Soal-soal yang kami sajikan memenuhi kompetensi materi Eksponen yang terdiri oleh empat subbab: (1) Definisi Eksponen, (2) Sifat Eksponen, (3) Persamaan Eksponen, (4) Pertidaksamaan Eksponen.
Oke, berikut ini kami sajikan soal pilihan dan pembahasan...
Nomor 1 $a=9$ , $b=16$ , dan $c=36$ . Nilai $\begin{aligned} \sqrt{\left( a^{- \frac{1}{3}} b^{- \frac{1}{2}} c \right)^{3}} = \cdots \end{aligned}$
1 3 9 12 18 Pembahasan
PENYELESAIAN.
$\begin{aligned} \sqrt{ \left( a^{- \frac{1}{3}} b^{- \frac{1}{2}} c \right)^{3} } & = \sqrt{ \left( 9^{- \frac{1}{3}} \cdot 16^{- \frac{1}{2}} \cdot 36 \right)^{3} } \\ & = \sqrt{ \left( 3^{- \frac{2}{3}} \cdot 2^{- \frac{4}{2}} \cdot 2^{2} \cdot 3^{2} \right)^{3} } \\ & = \sqrt{ \left( 3^{- \frac{2}{3} + 2} \cdot 2^{-2+2} \right)^{3} } \\ & = \sqrt{ \left( 3^{\frac{4}{3}} \right)^{3} } \\ & = \sqrt{3^{4}} \\ & = 9 \end{aligned}$
(Jawaban C)
Nomor 2 $$\sqrt{8^{x^{2} -4x + 3}} = \dfrac{1}{32^{x-1}}$$ $p$ dan $q$ , dengan $p > q$ , maka $p+6q = \cdots$
-17 -1 3 6 19 Pembahasan
PENYELESAIAN.
$\begin{aligned} \Leftrightarrow 2^{\frac{3(x^{3}-4x+3)}{2}} = 2^{-5(x-1)} \\ \Leftrightarrow 3x^{2} - 12x + 9 = -10(x-1) \\ \Leftrightarrow 3x^{2} - 2x - 1 = 0 \\ \Leftrightarrow (3x+1)(x-1) = 0 \\ \Leftrightarrow x = - \dfrac{1}{3} \lor x = 1 \\ \Leftrightarrow q = - \dfrac{1}{3} \land p=1 \end{aligned}$
Jadi,
$\begin{aligned} p + 6q & = 1 + 6 \left( - \dfrac{1}{3} \right) \\ & = 1-2 \\ & = -1 \end{aligned}$
(Jawaban B)
Nomor 3 $$ 2^{2x+1} - 17 \cdot 2^{x} + 8 = 0 $$ $\cdots$
$\{ {-3},{-1} \}$ $\{ {-3},{1} \}$ $\{ {-1},{3} \}$ $\{ {2},{-3} \}$ $\{ {1},{3} \}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \text{Misal} 2^{x} = p ~ \text{, maka} \\ \Leftrightarrow 2^{2x+1} - 17 \cdot 2^{x} + 8 = 0 \\ \Leftrightarrow 2p^{2} - 17p + 8 = 0 \\ \Leftrightarrow (2p-1)(p+8)=0 \\ \Leftrightarrow p = \dfrac{1}{2} \lor p = 8 \\ 2^{x} = 2^{-1} \lor 2^{x} = 2^{3} \\ \Leftrightarrow x=-1 \lor x=3 \end{aligned}$
(Jawaban C)
Nomor 4 $$ 3^{2x^{2}+5x-3} = 27^{2x+3} $$ $\alpha$ dan $\beta$ . Nilai $\alpha \beta = \cdots$
-6 -3 1 3 6 Pembahasan
PENYELESAIAN.
$\begin{aligned} \Leftrightarrow 3^{2x^{2}+5x-3} = 27^{2x+3} \\ \Leftrightarrow 3^{2x^{2}+5x-3} = 3^{3(2x+3)} \\ \Leftrightarrow 2x^{2} + 5x -3 = 6x + 9 \\ \Leftrightarrow 2x^{2} -x -12 = 0 \\ \Leftrightarrow \alpha + \beta = \dfrac{c}{a} = - \dfrac{12}{2} = -6 \end{aligned}$
(Jawaban A)
Nomor 5 $x$ yang memenuhi pertidaksamaan$$ \sqrt{3^{5x-1}} < \sqrt[3]{27^{x^{2}-4}} $$ $\cdots$
$-1 < x < 3 \frac{1}{2}$ $-3 \frac{1}{2} < x < 1$ $x < -1 ~ \text{atau} ~ x > 3 \frac{1}{2}$ $x < -3 \frac{1}{2} ~ \text{atau} ~ x > 1$ $x < 1 ~ \text{atau} ~ x > 7$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \Leftrightarrow \sqrt{3^{5x-1}} < \sqrt[3]{27^{x^{2}-4}} \\ \Leftrightarrow 3^{\frac{5x-1}{2}} < 3^{\frac{3(x^{2}-4)}{3}} \\ \Leftrightarrow \dfrac{5x-1}{2} < x^{2} - 4 \\ \Leftrightarrow 5x-1 < 2x^{2} - 8 \\ \Leftrightarrow 2x^{2}-5x-7 > 0 \\ \Leftrightarrow (2x-7)(x+1) > 0 \\ \Leftrightarrow x < -1 \lor x > \dfrac{7}{2} \end{aligned}$
(Jawaban C)
Nomor 6 $$ \left( \dfrac{1}{2} \right)^{8+2x-x^{2}} > \left( \dfrac{1}{2} \right)^{x+2} $$ $\cdots$
$\{ x \vert x < -2 ~ \text{atau} ~ x > 5 \}$ $\{ x \vert x < -2 ~ \text{atau} ~ x > 3 \}$ $\{ x \vert x < -2 ~ \text{atau} ~ x > 5 \}$ $\{ x \vert -2 < x < 3 \}$ $\{ x \vert -3 < x < 5 \}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \Leftrightarrow \left( \dfrac{1}{2} \right)^{8+2x-x^{2}} > \left( \dfrac{1}{2} \right)^{x+2} \\ \Leftrightarrow 8 + 2x - x^{2} < x+2 \\ \Leftrightarrow x^{2} - x - 6 > 0 \\ \Leftrightarrow (x+2)(x-3) > 0 \\ \Leftrightarrow x < -2 \lor x > 3 \end{aligned}$
(Jawaban B)
Nomor 7 $$ \dfrac{7x^{3}y^{-4}z^{-6}}{84x^{-7}y^{-1}z^{-4}} = \cdots $$ $\cdots$
$\frac{x^{10} z^{10}}{12y^{3}}$ $\frac{z^{2}}{12x^{4}y^{3}}$ $\frac{x^{10}y^{5}}{12z^{2}}$ $\frac{y^{3}z^{2}}{12x^{4}}$ $\frac{x^{10}}{12y^{3}z^{2}}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \dfrac{7x^{3} y^{-4} z^{-6}}{84x^{-7} y^{-1} z^{-4}} & = \dfrac{7}{84} \cdot \dfrac{x^{3}}{x^{-7}} \cdot \dfrac{y^{-4}}{y^{-1}} \cdot \dfrac{z^{-6}}{z^{-4}} \\ & = \dfrac{1}{12} \cdot x^{10} \cdot x^{10} \cdot y^{-3} \cdot z^{-2} \\ & = \dfrac{x^{10}}{12y^{3} z^{2}} \end{aligned}$
(Jawaban E)
Nomor 8 $$ 2^{2x} - 2^{x+3} + 16 = 0 $$ $\cdots$
1 2 3 4 5 Pembahasan
PENYELESAIAN.
$\begin{aligned} 2^{2x} - 2^{x+3} + 16 & = 0 \\ 2^{2x} - 2^{x} \cdot 2^{3} + 16 & = 0 \end{aligned}$
Dengan memisalkan $2^{x} = p$ , maka persamaan menjadi
$\begin{aligned} p^{2} - 8p + 16p & = 0 \\ (p-4)(p-4) & = 0 \\ p & = 4 \end{aligned}$
$\begin{aligned} \text{Untuk} ~ p=4 \Rightarrow & 2^{x} = 4 \\ & 2^{x} = 2^{2} \\ & x=2 \end{aligned}$
Jadi, $HP = \{ 2 \}$ .
(Jawaban B)
Nomor 9 $$ 7^{x^{2}-5x+6} = 8^{x^{2}-5x+6} $$ $\cdots$
(1,6) (1,-6) (-1,6) (1,7) (-1,7) Pembahasan
PENYELESAIAN.
$\begin{aligned} 7^{x^{2}-5x+6} & = 8^{x^{2}-5x+6} \\ x^{2}-5x+6 & = 0 \\ (x-6)(x+1) & = 0 \\ x=6 & \lor x=-1 \end{aligned}$
Jadi $HP = \{ {-1},{6} \}$ .
(Jawaban C)
Nomor 10 $$ 6^{x-3} = 9^{x-3} $$ $\cdots$
9 6 4 3 2 Pembahasan
PENYELESAIAN.
$\begin{aligned} 6^{x-3} & = 9^{x-3} \\ x-3 & = 0 \\ x & = 3 \end{aligned}$
Jadi $HP = \{ 3 \}$ .
(Jawaban D)
Nomor 11 $$ 9^{x^{2}+x} = 27^{x^{2}-1} $$ $\cdots$
(2,3) (1,3) (-1,3) (4,1) (2,-3) Pembahasan
PENYELESAIAN.
$\begin{aligned} 9^{x^{2}+x} & = 27^{x^{2}-1} \\ 3^{2(x^{2}+x)} & = 3^{3(x^{2}-1)} \\ 2(x^{2}+x) & = 3(x^{2}-1) \\ 2x^{2} + 2x & = 3x^{2} - 3 \\ x^{2}-2x-3 & = 0 \\ (x-3)(x+1) & = 0 \\ x=3 & \lor x=-1 \end{aligned}$
Jadi $HP = \{ {-1},{3} \}$ .
(Jawaban C)
Nomor 12 $$ 25^{x+2} = (0,2)^{1-x} $$ $\cdots$
-1 -2 -3 -4 -5 Pembahasan
PENYELESAIAN.
$\begin{aligned} 25^{x+2} & = (0,2)^{1-x} \\ 5^{2(x+2)} & = 5^{-1(1-x)} \\ 2x+4 & = -1 + x \\ 2x-x & = -1-4 \\ x & = -5 \end{aligned}$
Jadi $HP = \{ -5 \}$ .
(Jawaban E)
Nomor 13 $$ \sqrt[x+2]{8} = \sqrt[x-4]{32} $$ $\cdots$
-8 9 -10 11 -11 Pembahasan
PENYELESAIAN.
$\begin{aligned} \sqrt[x+2]{8} & = \sqrt[x-4]{32} \\ 2^{\frac{3}{x+2}} & = 2^{\frac{5}{x-4}} \\ \dfrac{3}{x+2} & = \dfrac{5}{x-4} \\ 3(x-4) & = 5(x+2) \\ 3x-12 & = 5x+10 \\ -2x & = 22 \\ x & = -11 \end{aligned}$
Jadi $HP = \{ -11 \}$ .
(Jawaban E)
Nomor 14 $$ 5^{2x-1} = 125 $$ $\cdots$
5 1 2 3 4 Pembahasan
PENYELESAIAN.
$\begin{aligned} 5^{2x-1} & = 125 \\ 5^{2x-1} & = 5^{3} \\ 2x-1 & = 3 \\ 2x & = 4 \\ x & = 2 \end{aligned}$
(Jawaban C)
Nomor 15 $$ 2^{2x-7} = \dfrac{1}{32} $$ $\cdots$
1 2 3 4 5 Pembahasan
PENYELESAIAN.
$\begin{aligned} 2^{2x-7} & = \dfrac{1}{32} \\ 2^{2x-7} & = 2^{-5} \\ 2x-7 & = -5 \\ 2x & = 2 \\ x & = 1 \end{aligned}$
(Jawaban A)
Nomor 16 $$ \sqrt{3^{3x-10}} = \dfrac{1}{27} \sqrt{3} $$ $\cdots$
1/3 2/3 4/3 5/3 7/3 Pembahasan
PENYELESAIAN.
$\begin{aligned} \sqrt{3^{3x-10}} & = \dfrac{1}{27} \sqrt{3} \\ 3^{\frac{3x-10}{2}} & = 3^{-3} \cdot 3^{\frac{1}{2}} \\ 3^{\frac{3x-10}{2}} & = 3^{-\frac{5}{2}} \\ \dfrac{3x-10}{2} & = - \dfrac{5}{2} \\ 3x-10 & = -5 \\ 3x & = 5 \\ x & = \dfrac{5}{3} \end{aligned}$
(Jawaban D)
Nomor 17 $$ 3^{5x-10} = 1 $$ $\cdots$
2 4 6 8 10 Pembahasan
PENYELESAIAN.
$\begin{aligned} 3^{5x-10} & = 1 \\ 3^{5x-10} & = 3^{0} \\ 5x-10 & = 0 \\ 5x & = 10 \\ x & = 2 \end{aligned}$
(Jawaban A)
Nomor 18 $$ 2^{2x^{2}+3x-5}=1 $$ $\cdots$
(-3/2 , 1) (3/2 , 1) (3,1) (4,1) (-5/2 , 1) Pembahasan
PENYELESAIAN.
$\begin{aligned} 2^{2x^{2}+3x-5} & = 1 \\ 2^{2x^{2}+3x-5} & = 2^{0} \\ 2^{2x} + 3x -5 & = 0 \\ (2x+5)(x-1) & = 0 \\ 2x+5=0 & \lor x-1=0 \\ x = - \dfrac{5}{2} & \lor x = 1 \end{aligned}$
(Jawaban E)
Nomor 19 $$ 9^{2x-1} = 27 $$ $\cdots$
2/3 4/7 5/4 7/4 1/3 Pembahasan
PENYELESAIAN.
$\begin{aligned} 9^{2x-1} & = 27 \\ 3^{2(2x-1)} & = 3^{3} \\ 3^{4x-2} & = 3^{3} \\ 4x-2 & = 3 \\ 4x & = 5 \\ x & = \dfrac{5}{4} \end{aligned}$
(Jawaban C)
Nomor 20 $$ 2^{3x-1} = \sqrt{\left( \frac{1}{32} \right)^{x-4}} $$ $\cdots$
1 2 3 4 5 Pembahasan
PENYELESAIAN.
$\begin{aligned} 2^{3x-1} & = \sqrt{\left( \frac{1}{32} \right)^{x-4}} \\ 2^{3x-1} & = \sqrt{\left( 2^{-5} \right)^{x-4}} \\ 2^{3x-1} & = \sqrt{2^{20-5x}} \\ 2^{3x-1} & = 2^{\frac{1}{2}(20-5x)} \\ 2^{3x-1} & = 2^{10-\frac{5}{2}x} \\ 3x -1 & = 10 - \dfrac{5}{2} x \\ 6x-2 & = 20 - 5x \\ 11x & = 22 \\ x & = 2 \end{aligned}$
(Jawaban B)
Nomor 21 $$ \left( a^{8} : a^{6} \right)^{3} $$ $\cdots$
$a^{2}$ $a^{3}$ $a^{4}$ $a^{5}$ $a^{6}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \left( a^{8} : a^{6} \right)^{3} & = \left( a^{8-6} \right)^{3} \\ & = a^{2 \cdot 3} \\ & = a^{6} \end{aligned}$
(Jawaban E)
Nomor 22 $$ \left( \frac{p^{2}}{q^{-3}} \right)^{3} \cdot \left( \frac{2q}{p^{3}} \right)^{2} $$ $\cdots$
$3q^{11}$ $4q^{11}$ $5q^{11}$ $6q^{11}$ $7q^{11}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \left( \frac{p^{2}}{q^{-3}} \right)^{3} \cdot \left( \frac{2q}{p^{3}} \right)^{2} & = \left( \frac{p^{6}}{q^{-9}} \right) \cdot \left( \frac{4q^{2}}{p^{6}} \right) \\ & = \left( p^{6} \cdot q^{9} \right) \left( 4q^{2} \cdot p^{-6} \right) \\ & = 4p^{6+(-6)} \cdot q^{9+2} \\ & = 4p^{0} q^{11} \\ & = 4q^{11} \end{aligned}$
(Jawaban B)
Nomor 23 $p=16$ dan $w=8$ , maka nilai dari$$ \left( \frac{32p^{3} w^{3}}{16^{2} p^{2} w^{4}} \right)^{2} $$ $\cdots$
$\frac{1}{16}$ $\frac{1}{8}$ $\frac{1}{4}$ $\frac{1}{2}$ $\frac{3}{4}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \left( \frac{32p^{3} w^{3}}{16^{2} p^{2} w^{4}} \right)^{2} & = \left( \frac{2^{5} p^{3} w^{3}}{(2^{4})^{2} p^{2} w^{4}} \right)^{2} \\ & = \dfrac{2^{5 \cdot 2} p^{3 \cdot 2} w^{3 \cdot 2}}{2^{8 \cdot 2} p^{2 \cdot 2} w^{4 \cdot 2}} \\ & = \dfrac{2^{10} p^{6} w^{6}}{2^{16} p^{4} w^{8}} \\ & = 2^{10-16} \cdot p^{6-4} \cdot w^{6-8} \\ & = \dfrac{p^{2}}{2^{6} w^{2}} \\ & = \dfrac{p^{2}}{64w^{2}} \end{aligned}$
$\begin{aligned} \dfrac{p^{2}}{64w^{2}} & = \dfrac{(16)^{2}}{64(8)^{2}} \\ & = \dfrac{1}{16} \end{aligned}$
(Jawaban A)
Nomor 24 $$ \left( a^{3} b^{6} c^{4} \right)^{2} $$ $\cdots$
$a^{6} b^{12} c^{8}$ $a^{5} b^{12} c^{8}$ $a^{6} b^{10} c^{8}$ $a^{6} b^{12} c^{6}$ $a^{6} b^{4} c^{8}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \left( a^{3} b^{6} c^{4} \right)^{2} & = a^{3 \cdot 2} b^{6 \cdot 2} c^{4 \cdot 2} \\ & = a^{6} b^{10} c^{8} \end{aligned}$
(Jawaban A)
Nomor 25 $$ \frac{2x^{3} + 4x^{6}}{x^{-2}} $$ $\cdots$
$x^{5} + 4x^{8}$ $2x^{5} + x^{8}$ $2x^{5} - 4x^{8}$ $-2x^{5} + 4x^{8}$ $2x^{5} + 4x^{8}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} \frac{2x^{3} + 4x^{6}}{x^{-2}} & \Leftrightarrow \dfrac{2x^{3}}{x^{-2}} + \dfrac{4x^{6}}{x^{-2}} \\ & \Leftrightarrow 2x^{3} x^{2} + 4x^{6} x^{2} \\ & \Leftrightarrow 2x^{3+2} + 4x^{6+2} \\ & \Leftrightarrow 2x^{5} + 4x^{8} \end{aligned}$
(Jawaban E)
Nomor 26 $$ 2^{x+2} > 16^{x-2} $$ $\cdots$
$x < - \frac{10}{3}$ $x < \frac{8}{3}$ $x < - \frac{9}{3}$ $x < \frac{9}{3}$ $x < \frac{10}{3}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} 2^{x+2} & > 16^{x-2} \\ 2^{x+2} & > 2^{4(x-2)} \\ x+2 & > 4(x-2) \quad \text{...(a > 1, fungsi naik)} \\ x+2 & > 4x-8 \\ 3x & < 10 \\ x & < \dfrac{10}{3} \end{aligned}$
Jadi, himpunan penyelesaian adalah $HP = \{ x \vert x < \frac{10}{3} , x \in R \}$ .
(Jawaban E)
Nomor 27 $$ (3x-10)^{x^{2}} = (3x-10)^{2x} $$ $\cdots$
$\{ {0} , {2} , {\frac{10}{3}} , {\frac{11}{3}} \}$ $\{ {2} , {\frac{10}{3}} , {\frac{11}{3}} \}$ $\{ {\frac{10}{3}} , {\frac{11}{3}} \}$ $\{ {\frac{11}{3}} \}$ $\{ {0} , {-2} , {\frac{10}{3}} , {\frac{11}{3}} \}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} (3x-10)^{x^{2}} = (3x-10)^{2x} \end{aligned}$
$\begin{aligned} \text{i)..} \quad x^{2} - 2x & = 0 \\ x(x-2) & 0 \\ x=0 & \lor x=2 \end{aligned}$
$\begin{aligned} \text{ii)..} \quad 3x-10 & = 0 \\ 3x & = 10 \\ x & = \dfrac{10}{3} \end{aligned}$
$\begin{aligned} \text{iii)..} \quad 3x-10 & = 1 \\ 3x & = 11 \\ x & = \dfrac{11}{3} \end{aligned}$
Sekarang periksa apakah untuk $x = \frac{10}{3}$ , $g(x)$ , dan $h(x)$ keduanya positif?
$\begin{aligned} g \left( \frac{10}{3} \right) & = \left( \frac{10}{3} \right)^{2} \\ & = \dfrac{100}{9} > 0 \end{aligned}$
$\begin{aligned} h \left( \frac{10}{3} \right) & = 2 \cdot \left( \frac{10}{3} \right) \\ & = \dfrac{20}{3} > 0 \end{aligned}$
Jadi, untuk $x = \frac{10}{3}$ , $g(x)$ , dan $h(x)$ keduanya positif, sehingga $x = \frac{10}{3}$ merupakan penyelesaian.
$\begin{aligned} 3x-10 & = -1 \\ 3x & = 9 \\ x & = 3 \end{aligned}$
Sekarang periksa apakah untuk $x=3$ , $g(x)$ , dan $h(x)$ keduanya genap atau keduanya ganjil?
$\begin{aligned} g(3) & = 3^{2} \\ & = 9 \end{aligned}$
$\begin{aligned} h(3) & = 2 \cdot (3) \\ & = 6 \end{aligned}$
Perhatikan bahwa untuk $x=3$ , $g(x)$ ganjil dan $h(x)$ genap sehingga $x=3$ bukan penyelesaian.
Dengan demikian, himpunan penyelesaian $(3x-10)^{x^{2}} = (3x-10)^{2x}$ adalah $\{ {0} , {2} , {\frac{10}{3}} , {\frac{11}{3}} \}$ .
(Jawaban A)
Nomor 28 $$ 25^{x+3} = 5^{x-1} $$ $\cdots$
-6 -7 -8 -9 -10 Pembahasan
PENYELESAIAN.
$\begin{aligned} 25^{x+3} & = 5^{x-1} \\ 5^{2(x+3)} & = 5^{x-1} \\ 2(x+3) & = x-1 \\ 2x+6 & = x-1 \\ x & = -7 \end{aligned}$
Jadi, penyelesaian $25^{x+3} = 5^{x-1}$ adalah $x = -7$ .
(Jawaban B)
Nomor 29 $$ 3 = 27^{1-x} $$ $\cdots$
2 3 2/3 1/3 4/3 Pembahasan
PENYELESAIAN.
$\begin{aligned} 3 & = 27^{1-x} \\ 3^{1} & = 3^{3(1-x)} \\ 3(1-x) & = 1 \\ 1-x & = \dfrac{1}{3} \\ x & = \dfrac{2}{3} \end{aligned}$
Jadi, penyelesaian $3 = 27^{1-x}$ adalah $x = \dfrac{2}{3}$ .
(Jawaban C)
Nomor 30 $$ (3x^{3} \cdot y^{-5})(-3x^{-8} \cdot y^{9}) $$ $\cdots$
$\frac{9y^{4}}{x^{6}}$ $\frac{8y^{4}}{x^{6}}$ $\frac{7y^{4}}{x^{6}}$ $\frac{6}{x^{6}}$ $\frac{5y^{4}}{x^{6}}$ Pembahasan
PENYELESAIAN.
$\begin{aligned} (3x^{3} \cdot y^{-5})(-3x^{-8} \cdot y^{9}) & = (3x^{2}) (-3x^{-8}) (y^{-5}) (y^{9}) \\ & = (3) (-3)x^{2} \cdot x^{-8} \cdot y^{-5} \cdot y^{9} \\ & = -9 x^{-6} \cdot x^{2-8} \cdot y^{-5+9} \\ & = -9x^{-6} \cdot y^{4} \\ & = - \dfrac{9y^{4}}{x^{6}} \end{aligned}$
(Jawaban A)
