Latihan Logaritma 01

Jawaban = D

3² = 9 

angka 3 menjadi basis (bilangan pokok),

angka 9 menjadi numerus,

angka 2 menjadi hasil logaritma

\(3^2 = 9 \Leftrightarrow ^3\log_{}{9} = 2\)

Jawaban = B

Gunakan sifat \(^a\log_{}{b} = c \Leftrightarrow a^c = b\)

\(^6\log_{}{216} = 3 \Leftrightarrow 216 = 6^3\)

Jawaban = B

Gunakan sifat \(^a\log_{}{b}+^a\log_{}{c} = ^a\log_{}{bc} \)

\(^6\log_{}{12} + ^6\log_{}{3}\)

\(^6\log_{}{(12\times 3)}\)

\(^6\log_{}{36}\)

\(^{6^1}\log_{}{6^2}\:\:\:\:\:\color{blue}\text{gunakan sifat}\: ^{a^m}\log_{}{b^n}=\frac{n}{m}\cdot ^a\log_{}{b}\)

\(\frac{2}{1}\cdot ^6\log_{}{6}\)

\(2\)

Jawaban = A

Gunakan sifat \(^a\log_{}{b}\:-\:^a\log_{}{c} = ^a\log_{}{\frac{b}{c}} \)

\(^5\log_{}{15} \:-\: ^5\log_{}{3}\)

\(^5\log_{}{\frac{15}{3}}\)

\(^5\log_{}{5}\)

\(1\)

Jawaban = D

Ubahlah bentuk desimal dan bentuk akar menjadi bilangan berpangkat, yaitu:

\(0.5 = \frac{5}{10} = \frac{1}{2} = 2^{-1}\)

\(0.25 = \frac{25}{100} = \frac{1}{4} = \frac{1}{2^2} = 2^{-2}\)

\(\sqrt{27} = \sqrt{3^3} = (3^3)^{\frac{1}{2}} = 3^{\frac{3}{2}}\)

Bentuk soalnya akan lebih mudah diselesaikan

\(^{0.5}\log_{}{0.25} + ^3\log_{}{\sqrt{27}}\)

\(^{2^{-1}}\log_{}{2^{-2}} + ^3\log_{}{3^{\frac{3}{2}}}\)

\(\frac{-2}{-1}\cdot ^2\log_{}{2} + \frac{3}{2}\cdot ^3\log_{}{3}\)

\(^2\log_{}{2}\) dan \(^3\log_{}{3}\) masing-masing bernilai 1

\(2 + \frac{3}{2}\)

\(3.5\)

Jawaban = A

Sederhanakan dahulu \(^{25}\log_{}{9}\) menjadi bentuk perpangkatan paling sederhana

\(^{5^2}\log_{}{3^2}\:\:\:\:\:\color{blue}\text{gunakan sifat}\: ^{a^m}\log_{}{b^n}=\frac{n}{m}\cdot ^a\log_{}{b}\)

\(= \frac{2}{2}\cdot ^5\log_{}{3}\)

\(\frac{2}{2}\cdot \dfrac{1}{^3\log 5}\:\:\:\:\:\color{blue} ^a\log_{}{b} = \frac{1}{^b\log_{}{a}}\)

\(\frac{1}{a}\)

Jawaban = A

Sederhanakan dahulu bentuk \(^{125}\log_{}{\sqrt{3}} = m\)

\(^{5^3}\log_{}{3^\frac{1}{2}} = m\)

\(\color{blue}\text{gunakan sifat}\: ^{a^m}\log_{}{b^n}=\frac{n}{m}\cdot ^a\log_{}{b}\)

\(^\frac{\frac{1}{2}}{3}\cdot ^{5}\log_{}{3} = m\)

\(\frac{1}{6}\cdot ^{5}\log_{}{3} = m\)

\(\color{blue}\text{kalikan kedua ruas dengan 6}\)

\(^{5}\log_{}{3} = 6m\)

Bentuk \(^5\log_{}{\frac{1}{3}} = ^5\log_{}{3^{-1}}\)

\(- ^5\log_{}{3}\)

\(- 6m\)

Jawaban = C

Ubah dahulu bentuk desimal menjadi bentuk perpangkatan

\(0,25 = \dfrac{25}{100} = \frac{1}{4} = \frac{1}{2^2} = 2^{-2}\)

\(0,125 = \dfrac{125}{1000} = \frac{1}{8} = \frac{1}{2^3} = 2^{-3}\)

Sehingga,

\(^{0,25}\log_{}{0,125} + 3^{^3\log_{}{7}} \:-\: ^5\log{}{1}\)

\(^{2^{-2}}\log_{}{2^{-3}} + 3^{^3\log_{}{7}} \:-\: ^5\log{}{1}\)

\(\frac{-3}{-2}\cdot ^2\log_{}{2} + \cancel{3}^{^\cancel{3}\log_{}{7}} \:-\: 0\)

\(\frac{3}{2}+ 7\)

\(8,5\)

Jawaban = D

Gunakan rumus pemfaktoran \(\color{blue} a^2 \:-\: b^2 = (a + b)(a \:-\: b)\)

\(\dfrac{(^2\log_{}{6})^2 \:-\: (^2\log_{}{3})^2}{^4\log_{}{18}}\)

\(\dfrac{(^2\log_{}{6}+^2\log_{}{3})(^2\log_{}{6}\:-\:^2\log_{}{3})}{^4\log_{}{18}}\)

\(\dfrac{(^2\log_{}{(6\times 3)})(^2\log_{}{\frac{6}{3}})}{^4\log_{}{18}}\)

\(\dfrac{^2\log_{}{18}\cdot ^2\log_{}{2}}{^4\log_{}{18}}\)

\(\dfrac{^2\log_{}{18}}{^{2^2}\log_{}{18^1}}\)

\(\dfrac{\cancel{^2\log_{}{18}}}{\frac{1}{2}\cdot \cancel{^2\log_{}{18}}}\)

\(\dfrac{1}{\frac{1}{2}}\)

\(2\)

Jawaban = B

\(^2\log_{}{7^2}\times ^7\log_{}{4}\times ^4\log_{}{8}\)

\(2\cdot ^2\log_{}{\cancel{7}}\times ^{\cancel{7}}\log_{}{\cancel{4}}\times ^{\cancel{4}}\log_{}{2^3}\)

\(2 \cdot^2\log_{}{2^3}\)

\(2\times 3 \times ^2\log_{}{2}\)

\(6\)