Exercise Zone : Pangkat (Eksponen) [2]

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

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StandarSNBTHOTS

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No. 11

Sederhanakan dan tentukan nilai dari ${64}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{4}}\right)}^2\cdot {32}^{-\frac{3}{5}}$

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ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{64}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{4}}\right)}^2\cdot {32}^{-\frac{3}{5}}& ={\left(2^6\right)}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{2^2}}\right)}^2\cdot {\left(2^5\right)}^{-\frac{3}{5}}\\ & =2^2\cdot {\left(2^{-2}\right)}^2\cdot 2^{-3}\\ & =2^2\cdot 2^{-4}\cdot 2^{-3}\\ & =2^{2-4-3}\\ & =2^{-5}\\ & ={\displaystyle \frac{1}{2^5}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{1}{32}}}}}}\end{array}$$

Jadi, ${64}^{\frac{2}{6}}\cdot {\left({\displaystyle \frac{1}{4}}\right)}^2\cdot {32}^{-\frac{3}{5}}={\displaystyle \frac{1}{32}}$.

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No. 12

Sederhanakanlah operasi perpangkatan di bawah ini
5⁵ × 5⁻⁴

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ALTERNATIF PENYELESAIAN

5⁵ × 5⁻⁴ = 5^{5 + (−4)} = 5¹ = 5

Jadi, 5⁵ × 5⁻⁴ = 5.

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No. 13

Jika K^2x = 3, maka ${\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}=$

1. ${\displaystyle \frac{36}{122}}$
2. ${\displaystyle \frac{39}{122}}$
3. ${\displaystyle \frac{81}{244}}$

4. ${\displaystyle \frac{122}{39}}$
5. ${\displaystyle \frac{244}{81}}$

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}& ={\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}\text{color}red\cdot {\displaystyle \frac{K^{5x}}{K^{5x}}}\\ & ={\displaystyle \frac{K^{8x}-K^{2x}}{K^{10x}+K^0}}\\ & ={\displaystyle \frac{{\left(K^{2x}\right)}^4-K^{2x}}{{\left(K^{2x}\right)}^5+1}}\\ & ={\displaystyle \frac{3^4-3}{3^5+1}}\\ & ={\displaystyle \frac{81-3}{243+1}}\\ & ={\displaystyle \frac{78}{244}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{39}{122}}}}}}\end{array}$$

Jadi, ${\displaystyle \frac{K^{3x}-K^{-3x}}{K^{5x}+K^{-5x}}}={\displaystyle \frac{39}{122}}$.
JAWAB: B

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No. 14

Hasil dari 6⁸ : 6⁵ adalah ....

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{6}^8:6^5& =6^{8-5}\\ & =6^3\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 216}}}}\end{array}$$

Jadi, 6⁸ : 6⁵ = 216.

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No. 15

Bentuk sederhana dari ${\displaystyle \frac{5a^2{b}^{-7}{c}^3}{10a^5{b}^3{c}^{-2}}}$ adalah ....

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{5a^2{b}^{-7}{c}^3}{10a^5{b}^3{c}^{-2}}}& ={\displaystyle \frac{c^{3-(-2)}}{2a^{5-2}{b}^{3-(-7)}}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{c^5}{2a^3{b}^{10}}}}}}}\end{array}$$

Jadi, bentuk sederhana dari ${\displaystyle \frac{5a^2{b}^{-7}{c}^3}{10a^5{b}^3{c}^{-2}}}$ adalah ${\displaystyle \frac{c^5}{2a^3{b}^{10}}}$.

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No. 16

Bila x = 9 dan y = 64 maka nilai ${\displaystyle \frac{x^{-\frac{3}{2}}\sqrt[3]{y^2}}{y^{\frac{1}{3}}-x^{\frac{1}{2}}}}$ adalah ....

Latihan USBN SMK

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}\frac{x^{-\frac{3}{2}}\sqrt[3]{y^2}}{y^{\frac{1}{3}}-x^{\frac{1}{2}}}& =\frac{9^{-\frac{3}{2}}\sqrt[3]{{64}^2}}{{64}^{\frac{1}{3}}-9^{\frac{1}{2}}}\\ & =\frac{{\left(3^2\right)}^{-\frac{3}{2}}{\left(4^3\right)}^{\frac{2}{3}}}{{\left(4^3\right)}^{\frac{1}{3}}-{\left(3^2\right)}^{\frac{1}{2}}}\\ & =\frac{3^{-3}{4}^2}{4-3}\\ & =\frac{\frac{1}{3^3}\cdot 16}{1}\\ & =\frac{1}{27}\cdot 16\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \frac{16}{27}}}}}\end{array}$$

Jadi, nilai $\dfrac{x^{-\frac32}\sqrt[3]{y^2}}{y^{\frac13}-x^{\frac12}}$ adalah ${\displaystyle \frac{16}{27}}$.

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No. 17

Bentuk sederhana dari (2x² ⋅ y⁻⁵)(−2x⁻⁸ ⋅ y⁹) adalah ....

Latihan USBN SMK

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}(2x^2\cdot y^{-5})(-2x^{-8}\cdot y^9)& =(2)(-2)x^{2+(-8)}\cdot y^{-5+9}\\ & =-4x^{-6}{y}^4\\ & =-\frac{4y^4}{x^6}\end{array}$$

Jadi, bentuk sederhana dari (2x² ⋅ y⁻⁵)(−2x⁻⁸ ⋅ y⁹) adalah $-\frac{4y^4}{x^6}$.

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No. 18

Jika a = 2, b = 3, dan c = 4, nilai dari ${\displaystyle \frac{a^2{b}^3}{c^4}}$ adalah …

Latihan Pra UKK Semester Genap Kelas X

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{a^2{b}^3}{c^4}}& =\frac{2^2\cdot 3^3}{4^4}\\ & =\frac{4\cdot 27}{256}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle \frac{27}{64}}}}}\end{array}$$

Jadi, ${\displaystyle \frac{a^2{b}^3}{c^4}}={\displaystyle \frac{27}{64}}$.

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No. 19

Hasil dari (3²)² ⋅ (3³)⁻³ adalah …

SUMBER

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\left(3^2\right)}^2\cdot {\left(3^3\right)}^{-3}& =3^4\cdot 3^{-9}\\ & =3^{4+(-9)}\\ & =3^{-5}\\ & ={\displaystyle \frac{1}{3^5}}\\ & ={\displaystyle \frac{1}{243}}\end{array}$$

Jadi, hasil dari (3²)² ⋅ (3³)⁻³ adalah ${\displaystyle \frac{1}{243}}$.

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No. 20

Jika a = 4, b = 3, dan c = 2, nilai dari ${\displaystyle \frac{a^2\cdot b^3}{c^4}}$ adalah ....

1. 3
2. 9
3. 18

4. 27
5. 36

UKK 2022 Kelas X

ALTERNATIF PENYELESAIAN

$$\begin{array}{rl}{\displaystyle \frac{a^2\cdot b^3}{c^4}}& ={\displaystyle \frac{4^2\cdot 3^3}{2^4}}\\ & ={\displaystyle \frac{16\cdot 27}{16}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 27}}}}\end{array}$$

Jadi, nilai dari ${\displaystyle \frac{a^2\cdot b^3}{c^4}}$ adalah 27.

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JAWAB: D

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