___________________________

» Pendahuluan

Bilangan bereksponen (berpangkat) dinyatakan dengan:

a × a × a × ....... a × a = aⁿ

Notasi: an dibaca “a pangkat n”

a disebut bilangan pokok (basis)

n disebut bilangan pangkat

Berlaku sifat:

1. pᵃ x pᵇ = pᵃ ⁺ ᵇ,

2. pᵃ : pᵇ = pᵃ ⁻ ᵇ,

3. (pᵃ)ᵇ = pᵃ ˣ ᵇ,

4. (p x q)ᵇ = pᵇ x qᵇ,

5. (p : q)ᵇ = pᵇ : qᵇ,

6. p⁰ = 1,

Contoh soal Eksponen (Berulang-ulang)

= 10⁵

= (10 . 10 . 10 . 10 . 10)

= 100.000

Contoh soal Eksponen (Sederhanakan)

= 14¹² ÷ 14¹¹

= 14(¹²-¹¹)

= 14¹

= 14

Contoh Eksponen (pangkat 2 dan 3)

1² = 1 × 1 = 1

2² = 2 × 2 = 4

3² = 3 × 3 = 9

4² = 4 × 4 = 16

5² = 5 × 5 = 25

6² = 6 × 6 = 36

7² = 7 × 7 = 49

8² = 8 × 8 = 64

9² = 9 × 9 = 81

10² = 10 × 10 = 100

1³ = 1 × 1 × 1 = 1

2³ = 2 × 2 × 2 = 8

3³ = 3 × 3 × 3 = 27

4³ = 4 × 4 × 4 = 64

5³ = 5 × 5 × 5 = 125

6³ = 6 × 6 × 6 = 216

7³ = 7 × 7 × 7 = 34

8³ = 8 × 8 × 8 = 512

9³ = 9 × 9 × 9 = 729

10³ = 10 × 10 × 10 = 1000

___________________________

» Penyelesaian

pᵃ ÷ pᵇ = pᵃ ⁻ ᵇ

= Q.¹²³ ÷ Q.⁴⁵⁶

= Q.¹²³ ⁻ ⁴⁵⁶

= Q.⁻³³³ ☑

___________________________

» Pelajari Lebih Lanjut

Contoh Soal

• yomemimo.com/tugas/47210028

• yomemimo.com/tugas/47203868

• yomemimo.com/tugas/47202217

• yomemimo.com/tugas/47201917

___________________________

» Detail Jawaban

Mapel : Matematika

Kelas : VI

Materi : Operasi Hitung Bilangan Bulat

Bab : 1 - Bilangan Bulat

Kode Soal : 2

Kode Kategorisasi : 6.2.1

Kata Kunci : Bilangan bulat, Bilangan positif, Bilangan negatif, Garis bilangan

___________________________