Latihan Logaritma 02

Jawaban = B

\(^2\log_{}{7} + ^2\log_{}{4} \:-\: ^2\log_{}{14}\)

\(^2\log_{}{\frac{7\times 4}{14}}\)

\(^2\log_{}{\frac{28}{14}}\)

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

\(1\)

Jawaban = C

\(^9\log_{}{4} + ^3\log_{}{18} \:-\: 2 \cdot ^3\log_{}{2}\)

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

\(\frac{2}{2}\cdot ^3\log_{}{2} + ^3\log_{}{18} \:-\: ^3\log_{}{4}\)

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

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

\(^3\log_{}{\frac{36}{4}}\)

\(^3\log_{}{9}\)

\(^{3^1}\log_{}{3^2}\:\:\:\:\:\color{blue}\text{gunakan sifat}\: ^{a^m}\log_{}{b^n}=\frac{n}{m}\cdot ^a\log_{}{b}\)

\(\frac{2}{1}\times ^3\log_{}{3}\:\:\:\:\:\color{blue}\text{ingat}\:^a\log_{}{a}=1\)

\(2\times 1\)

\(2\)

Jawaban = B

\(^3\log_{}{27} + ^7\log_{}{\frac{1}{7^2}} \:-\: \cancel{3}^{^\cancel{3}\log_{}{4}}\)

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

\(^{3^1}\log_{}{3^3} + ^{7^1}\log_{}{7^{-2}}\:-\: 4\)

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

\(\frac{3}{1} \cdot ^3\log_{}{3} + \frac{-2}{1} \cdot ^7\log_{}{7}\:-\: 4\)

\(3 + (-2)\:-\: 4\)

\(-3\)

Jawaban = D

\(\dfrac{(^3\log_{}{45})^2 \:-\: (^3\log_{}{5})^2 }{^3\log_{}{15}}\)

\(\dfrac{(^3\log_{}{45} + ^3\log_{}{5})(^3\log_{}{45} \:-\: ^3\log_{}{5})}{^3\log_{}{15}}\)

\(\color{blue} \text{a² − b²=(a + b)(a − b)}\)

\(\dfrac{(^3\log_{}{(45\times 5)})(^3\log_{}{\frac{45}{15}})}{^3\log_{}{15}}\)

\(\dfrac{^3\log_{}{225}\cdot ^3\log_{}{3}}{^3\log_{}{15}}\)

\(\dfrac{^3\log_{}{15^2}}{^3\log_{}{15}}\)

\(2\cdot \dfrac{\cancel{^3\log_{}{15}}}{\cancel{^3\log_{}{15}}}\)

\(2\)

Jawaban = B

\(^2\log_{}{3^2}\times  ^3\log_{}{\sqrt{5}}\times ^5\log_{}{8}\)

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

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

\(\cancel{2}\cdot \frac{1}{\cancel{2}} \cdot 3 \times ^2\log_{}{\cancel{3}}\cdot ^\cancel{3}\log_{}{\cancel{5}}\cdot ^\cancel{5}\log_{}{2}\)

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

\(3\)

Jawaban = A

Sederhanakan bentuk \(^4\log_{}{27} = 6m\) terlebih dahulu,

\(^{2^2}\log_{}{3^3} = 6m\)

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

\(\frac{3}{2}\cdot ^2\log_{}{3} =6m\)

\(^2\log_{}{3} = \frac{2}{3}\cdot 6m\)

\(^2\log_{}{3} = 4m\)

Selanjutnya,

\(^9\log_{}{16}=^{3^2}\log_{}{2^4}\)

\(\frac{4}{2}\cdot ^3\log_{}{2}\)

\(\frac{4}{2}\cdot \frac{1}{^2\log_{}{3}}\)

\(2\cdot \frac{1}{4m}\)

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

Jawaban = C

Gunakan sifat logaritma \(^a\log_{}{b}=\frac{\log_{}{b}}{\log_{}{a}}\) untuk menyederhanakan \(^{35}\log_{}{40}\)

\(^{35}\log_{}{40}  = \dfrac{^m\log_{}{40}}{^m\log_{}{35}}\)

\(\dfrac{^m\log_{}{(2^3\times 5)}}{^m\log_{}{(5\times 7)}}\)

\(\dfrac{^m\log_{}{2^3}+^m\log_{}{5}}{^m\log_{}{5}+^m\log_{}{7}}\)

\(\dfrac{3\cdot^m\log_{}{2}+^m\log_{}{5}}{^m\log_{}{5}+^m\log_{}{7}}\)

Selanjutnya pilih basis \(m\) yang cocok,

\(\dfrac{3\cdot^5\log_{}{2}+^5\log_{}{5}}{^5\log_{}{5}+^5\log_{}{7}}\:\:\:\:\:\color{blue}\text{pilih m = 5}\)

\(\dfrac{3\cdot \frac{1}{^2\log_{}{5}}+1}{1+^5\log_{}{7}}\)

Selanjutnya ganti \(^2\log_{}{5}\) dengan \(a\) dan \(^5\log_{}{7}\) dengan \(b\)

\(\dfrac{3\cdot \frac{1}{a}+1}{1+b}\)

\(\color{blue}\text{kalikan pembilang dan penyebut dengan } a\)

\(\dfrac{3+a}{a+ab}\)

Jawaban = C

\(\log_{}{120}=\log_{}{(2^3\times 3 \times 5})\)

\(\log_{}{120}=3\log_{}{2} + \log_{}{3} + \log_{}{5}\)

\(\log_{}{120}=3(0,301) + 0,477 + \log_{}{5}\)

Jadi,

\(\log_{}{120}=3(0,301) + 0,477 + \log_{}{5}\)

\(\log_{}{120}=0,903+ 0,477 + 0,699=2,079\)

Jawaban = D

\(2\cdot ^2\log_{}{2\sqrt{2}} + ^2\log_{}{0,125}\:-\: ^{0,5}\log_{}{2}\)

\(2\cdot ^2\log_{}{2^{\frac{3}{2}}} + ^2\log_{}{\frac{1}{8}}\:-\: ^{2^{-1}}\log_{}{2}\)

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

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

\(1\)

Jawaban = D

\(\dfrac{2+\log_{}{(\log_{}{5})}}{2\log_{}{(\log_{}{5^{100}})}}\)

\(\color{blue}\text{ubah 2 menjadi }\log_{}{100}\)

\(\dfrac{\log_{}{100}+\log_{}{(\log_{}{5})}}{2\log_{}{(\log_{}{5^{100}})}}\)

\(\dfrac{\log_{}{(100\cdot\log_{}{5})}}{2\log_{}{(\log_{}{5^{100}})}}\)

\(\dfrac{\cancel{(\log_{}{\log_{}{5^{100}}})}}{2\cancel{\log_{}{(\log_{}{5^{100}}})}}\)

\(\dfrac{1}{2}\)
\(0,5\)

Jawaban = C

\(\text{Misal } ^{\text{4p}}\log_{}{100} = a\)

\((\text{4p})^a = 100\)

\(4^a\cdot \text{p}^a = 100\)

\(\text{p}^a = \dfrac{100}{4^a}\dotso\dotso (1)\)

\(^{\text{p}}\log_{}{25} = a\)

\(\text{p}^a = 25\dotso\dotso (2)\)

Substitusikan persamaan (1) ke persamaan (2)

\(\dfrac{100}{4^a} = 25\)

\(100 = 25\cdot 4^a\)

\(\dfrac{100}{25} = 4^a\)

\(4 = 4^a\)

\(a = 1\)

Untuk mencari nilai \(p\) substitusikan \(a = 1\) ke persamaan (1)

\(\text{p}^1 = \dfrac{100}{4^1}\)

\(\text{p}= 25\)

\(\text{2p}^2 = 2\cdot(25)^2\)

\(\text{2p}^2 = 1250\)

Jawaban = A

\(3^x = 2\:-\:\sqrt{3} \color{red} \times \dfrac{2 + \sqrt{3}}{2 + \sqrt{3}}\)

\(3^x = \dfrac{1}{2 + \sqrt{3}}\)

\(2 + \sqrt{3} = 3^{-x}\)

\(^{2 + \sqrt{3}}\log 27^x = ^{3^{-x}}\log 3^{3x}\)

\(^{2 + \sqrt{3}}\log 27^x = \dfrac{3x}{-x}\cdot ^3 \log 3\)

\(^{2 + \sqrt{3}}\log 27^x = -3\)