Haii adik-adik...
Selamat datang di halaman contoh soal dan pembahasan "Logaritma". Di halaman ini akan membahas tentang contoh soal dan pembahasan lengkap mengenai Logaritma.
Untuk lebih jelasnya mari kita lihat contoh soal dan pembahasan dibawah ini!
1. \(\textrm{Nilai }2^{6}\log16-3^{6}\log4+^{6}\log9=...\)
a) \(3\)
b) \(-3\)
c) \(2\)
d) \(-2\)
e) \(1\)
\(\ 2 \ ^{6}\log16-3 \ ^{6}\log4+^{6}\log9=^{6}\log16^{2}-^{6}\log4^{3}+^{6}\log9=^{6}\log\left ( \frac{16^{2}}{4^{3}}\times 9 \right )=^{6}\log36=2\)
2. \(\textrm{Diketahui }^{2}\log 3=p\textrm{ dan }^{2}\log 5=q\textrm{ maka }^{2}\log 45=....\)
a) \(p^{2}+q\)
b) \(2p+q\)
c) \(2(p+q)\)
d) \(p^{2}+q^{2}\)
e) \(p+2p\)
\(^{2}\log3=p\textrm{ dan }^{2}\log 5=q\\\textrm{Nilai dari }^{2}\log 45=\frac{\log 45}{\log 2}\\=\frac{\log 3^{2}}{\log2}=\frac{\log 3^{2}+\log 5}{\log2}\\=\frac{^{2}\log3^{2}+^{2}\log5}{^{2}\log2}=\frac{p^{2}+q}{1}=p^{2}+q\)
3. \(\textrm{Jika }p=^{6}\log 5;q=^{5}\log 4\textrm{ maka }^{150}\log 4=...\)
a) \(\frac{p\cdot q}{p+q}\)
b) \(\frac{p\cdot q}{2p-1}\)
c) \(\frac{p\cdot q}{2p+1}\)
d) \(\frac{p\cdot q}{2p-q}\)
e) \(\frac{p\cdot q}{2p+q}\)
\(p=^{6}\log 5;q=^{5}\log4\\\textrm{Nilai dari:}^{150}\log4=\frac{\log4}{\log150}=\frac{\log4}{\log5^{2}\cdot 6}=\frac{\log4}{\log5^{2}+\log6}=\frac{^{5}\log 4}{^{5}\log 5^{2}+^{5}\log 6}=\frac{q}{2+\frac{1}{p}}\\=\frac{q}{\frac{2p+1}{p}}=\frac{q}{1}\times \frac{p}{2p+1}=\frac{pq}{2p+1}\)
4. \(\textrm{Nilai x yang memenuhi }b^{2x}+10< 7\cdot b^{x}\textrm{ dengan }b> 1\textrm{ adalah...}\)
a) \(x< ^{b}\log2\)
b) \(x>^{b}\log5\)
c) \(x< ^{b}\log2\textrm{ atau }x>^{b}\log5\)
d) \( ^{b}\log2<x<^{b}\log5\)
e) \(x>^{b}\log2\)
\(\textrm{Diketahui:}\\b^{2x}+10<7\cdot b^{x}\\\Rightarrow b^{2x}-7b^{x}+10<0\\\textrm{Miasal:}b^{x}=p\\\Rightarrow p^{2}-7p+10<0\\\Rightarrow (p-5)(p-2)<0\\\textrm{Sehingga, batas nilai p adalah:}\\p=5\textrm{ dan }p=2\\\bullet p=5\\\Rightarrow b^{x}=5\Rightarrow x=^{b}\log5\\\bullet p=2\\\Rightarrow b^{x}=2\Rightarrow x=^{b}\log2\\\textrm{Misalkan, diambil nilai x=0 maka dari:}b^{2x}+10<7\cdot b^{x}\textrm{ diperoleh 1+10<7.1}\\ \textrm{( bernilai salah )sehingga daerah penyelesaiannya adalah :}\\HP=\left \{ ^{b}\log2<x<^{b} \log5\right \}\)
5. \(\textrm{Himpunanan penyelesaian dari:}\\\log(x^{2}+4x+4)\leq (\log(5x+10)\textrm{ adalah...}\)
a) \(\left \{ x\mid -2<x\leq 3 \right \}\)
b) \(\left \{ x\mid x<3 \right \}\)
c) \(\left \{ x\mid -3<x\leq -2 \right \}\)
d) \(\left \{ x\mid x\leq -2\textrm{ atau }x\geq 3 \right \}\)
e) \(\left \{ x\mid -2\leq x\leq 3 \right \}\)
\(\textrm{Diketahui :}\\\log(x^{2}+4x+4)\leq\log(5x+10)\\\Rightarrow x^{2}+4x+4\leq 5x+10\\\Rightarrow x^{2}-x-6\leq 0\\\Rightarrow (x-3)(x+2)\leq 0\\\textrm{Diperoleh x=3 dan x=-2}\\\textrm{Jadi daerah penyelesaiannya }\left \{ x\mid -2<x\leq 3 \right \}\)
6. \(\textrm{Jika}\log2=p\ \textrm{dan}\ \log3=q,\textrm{maka}\ \log\left ( \frac{9}{4} \right )=\)
a) \(2(p-q)\)
b) \(2(q-p)\)
c) \(2(p+q)\)
d) \(2pq\)
e) \(2(q+p)\)
\(\log2=p\rightarrow \log3=q\\ \log\left ( \frac{9}{4} \right )=\log9=\log4=2\log3-2\log2=2\left ( \log3-\log2 \right )=2\left ( q-p \right )\)
7. \(\textrm{Jika }^{3}\log 4=a\textrm{ dan }^{3}\log 5=b\textrm{ maka }^{8}\log20=...\)
a) \(\frac{a+b}{2a}\)
b) \(\frac{a+b}{3a}\)
c) \(\frac{2a+2b}{3a}\)
d) \(\frac{3a+3b}{2a}\)
e) \(\frac{a+2b}{3a}\)
\(^{3}\log 4=a\textrm{ dan }^{3}\log 5=b,\textrm{ perhatikan sifat logaritma:}\\^{8}\log 20=\frac{^{3}\log 20}{^{3}\log8}=\frac{^{3}\log4\cdot 5}{^{3}\log(4)^{\frac{3}{2}}}=\frac{a+b}{\frac{3}{2}a}=\frac{2a+2b}{3a}\)
8. \(\textrm{Jika }^{9}\log 8=3m,\textrm{ nilai }^{4}\log 3=....\)
a) \(\frac{1}{4m}\)
b) \(\frac{3}{4m}\)
c) \(\frac{3}{2m}\)
d) \(\frac{m}{4}\)
e) \(\frac{4m}{4}\)
\(^{9}\log 8=3m\\^{3^{2}}\log 2^{3}=\frac{3}{2}^{3}\log2=3m\rightarrow ^{3}\log 2=2m\textrm{ berarti }^{2}\log 3=\frac{1}{2m}\textrm{ dan }^{4}\log 3=\frac{1}{2}^{2}\log 3=\frac{1}{4m}\)
9. \(\textrm{Jika }x_{1}\textrm{ dan }x_{2}\textrm{ adalah akar-akar }4(\log x)(1+\log x)=3\textrm{ maka }x_{1}\cdot x_{1}=....\)
a) \(\frac{1}{10}\)
b) \(\frac{1}{4}\)
c) \(\frac{3}{4}\)
d) \(3\)
e) \(10\)
\(4\log x(1+\log x)=3\\4(\log x)^{2}+4\log x-3=0\\\textrm{Misal :y}=\log x,\textrm{ maka :}4y^{2}+4y-3=0\rightarrow (2y-1)(2y+3)=0\\\square 2y=1\rightarrow y=\frac{1}{2}=\log x=\frac{1}{2}\rightarrow x_{1}=\sqrt{10}\\\square 2y+3=0\rightarrow y=-\frac{3}{2}=\log x=-\frac{3}{2}\rightarrow x_{2}=10^{-\frac{3}{2}}=\frac{1}{10\sqrt{10}}\\\textrm{Berarti }x_{1}\cdot x_{2}=\sqrt{10}\cdot \frac{1}{10\sqrt{10}}=\frac{1}{10}\)
10. \(\textrm{Nilai x yang memenuhi }\log x=4\log\left ( a+b \right )+2\log\left ( a-b \right )-3\log\left ( a^{2} -b^{2}\right )-\log\frac{a+b}{a-b}\ \textrm{adalah...}\)
a) \(a+b\)
b) \(a-b\)
c) \(\left ( a+b \right )^{2}\)
d) \(10\)
e) \(1\)
\(\log x=4\log\left ( a+b \right )+2\log\left ( a-b \right )-3\log\left ( a^{2} -b^{2}\right )-\log\frac{a+b}{a-b}\\ \log x= \log \left ( a+b \right )^{4}+\log\left ( a-b \right )^{2}-\log\left ( a^{2}-b^{2} \right )^{3}-\log\frac{a+b}{a-b}=\log\frac{\left ( a+b \right )^{4}\times \left ( a-b \right )^{2}\times \left ( a-b \right )}{\left ( a^{2}-b^{2} \right )^{3}\times \left ( a+b \right )}=\log\frac{\left ( a+b \right )^{4}\times \left ( a-b \right )^{3}}{\left ( a-b \right )^{3}\times \left ( a+b \right )^{3}\times \left ( a+b \right )^{1}}\\ \log x=\log 1\rightarrow x=1\)
