tolong bantu dong ak sudah surender di kumpulin besok
Logaritma.
e. 2⁶ = 64 ⇔ ²log 64 = 6
f. 2¹ = 2 ⇔ ²log 2 = 1
g. 2⁰ = 1 ⇔ ²log 1 = 0
h. 2⁻¹ = [tex]\frac{1}{2^{1}}[/tex] = [tex]\frac{1}{2}[/tex] ⇔ ²log [tex]\frac{1}{2}[/tex] = –1
i. 2⁻² = [tex]\frac{1}{2^{2}}[/tex] = [tex]\frac{1}{4}[/tex] ⇔ ²log [tex]\frac{1}{4}[/tex] = –2
j. 2⁻³ = [tex]\frac{1}{2^{3}}[/tex] = [tex]\frac{1}{8}[/tex] ⇔ ²log [tex]\frac{1}{8}[/tex] = –3
k. 2⁻⁴ = [tex]\frac{1}{2^{4}}[/tex] = [tex]\frac{1}{16}[/tex] ⇔ ²log [tex]\frac{1}{16}[/tex] = –4
l. 2⁻⁵ = [tex]\frac{1}{2^{5}}[/tex] = [tex]\frac{1}{32}[/tex] ⇔ ²log [tex]\frac{1}{32}[/tex] = –5
m. 2⁻⁶ = [tex]\frac{1}{2^{6}}[/tex] = [tex]\frac{1}{64}[/tex] ⇔ ²log [tex]\frac{1}{64}[/tex] = –6
Untuk soal berikutnya dapat disimak di penjelasan dengan langkah-langkah.
Penjelasan dengan langkah-langkah
Logaritma merupakan salah satu invers dari perpangkatan. Definisinya:
- ᵃlog b = n artinya aⁿ = b
dengan syarat a > 0, b > 0, a ≠ 1
Diketahui
[tex]r. \: 3^{-1} = \frac{. . .}{. . .} = \frac{. . .}{. . .}\: \Leftrightarrow \: ^{3}log \: \frac{. . .}{. . .} = . . .[/tex]
[tex]s. \: 3^{-2} = \frac{. . .}{. . .} = \frac{. . .}{. . .} \:\Leftrightarrow \: ^{3}log \: \frac{. . .}{. . .} = . . .[/tex]
[tex]t. \: 3^{-3} = \frac{. . .}{. . .} = \frac{. . .}{. . .} \:\Leftrightarrow \: ^{3}log \: \frac{. . .}{. . .} = . . .[/tex]
[tex]u. \: 4^{\frac{1}{2}} = \sqrt{4} = 2 \:\Leftrightarrow \: ^{4}log \: 2 = \frac{1}{2}[/tex]
[tex]v. \: 16^{\frac{1}{2}} = \sqrt{. . .} = . . .\: \Leftrightarrow \: ^{16}log \: . . . = \frac{. . .}{. . .}[/tex]
[tex]w. \: 8^{\frac{1}{3}} = \sqrt[3]{8} = 2 \:\Leftrightarrow \: ^{8}log \: 2 = \frac{1}{3}[/tex]
[tex]x. \: 16^{\frac{1}{4}} = \sqrt[. . .]{. . .} = . . . \:\Leftrightarrow \: ^{16}log \: . . . = \frac{. . .}{. . .}[/tex]
[tex]y. \: 27^{\frac{1}{3}} = \sqrt[. . .]{. . .} = . . . \: \Leftrightarrow \:^{27}log \: . . . = \frac{. . .}{. . .} [/tex]
Ditanyakan
Lengkapi isian pada titik-titik di atas!
Jawab
Bagian r
[tex]3^{-1} = \frac{1}{3^{1}} = \frac{1}{3}\: \Leftrightarrow \: ^{3}log \: \frac{1}{3} = -1[/tex]
Bagian s
[tex]3^{-2} = \frac{1}{3^{2}} = \frac{1}{9} \:\Leftrightarrow \: ^{3}log \: \frac{1}{9} = -2[/tex]
Bagian t
[tex]3^{-3} = \frac{1}{3^{3}} = \frac{1}{27} \:\Leftrightarrow \: ^{3}log \: \frac{1}{27} = -3[/tex]
Bagian u
[tex]4^{\frac{1}{2}} = \sqrt{4} = 2 \:\Leftrightarrow \: ^{4}log \: 2 = \frac{1}{2}[/tex]
Bagian v
[tex]16^{\frac{1}{2}} = \sqrt{16} = 4\: \Leftrightarrow \: ^{16}log \: 4 = \frac{1}{2}[/tex]
Bagian w
[tex]8^{\frac{1}{3}} = \sqrt[3]{8} = 2 \:\Leftrightarrow \: ^{8}log \: 2 = \frac{1}{3}[/tex]
Bagian x
[tex]16^{\frac{1}{4}} = \sqrt[4]{16} = 2 \:\Leftrightarrow \: ^{16}log \: 2 = \frac{1}{4}[/tex]
Bagian y
[tex]27^{\frac{1}{3}} = \sqrt[3]{27} = 3 \: \Leftrightarrow \:^{27}log \: 3 = \frac{1}{3} [/tex]
Pelajari lebih lanjut
Materi tentang logaritma https://brainly.co.id/tugas/51937796
#BelajarBersamaBrainly #SPJ1
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