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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!
51. \(\textrm{Nilai x yang memenuhi }\log x=4\log\left ( a+b \right )+2\left ( a-b \right )-3\log\left ( a^{2}-b^{2} \right )-\log\frac{a+b}{a-b}\textrm{ adalah...}\)
a) \(a+b\)
b) \(a-b\)
c) \(\left ( a+b \right )^{2}\)
d) \(10\)
e) \(1\)
\(\log x=4\log\left ( a+b \right )+2\log\left ( a-b \right )-3\log\left ( a^{2} -b^{2}\right )-\log\frac{a+b}{a-b}\\\Rightarrow \log x=\log\left ( a+b \right )^{4}+\log\left ( a-b \right )^{2}-\log\left ( a^{2}-b^{2} \right )^{3}-\log\frac{a+b}{a-b}\\\Rightarrow \log x=\log\frac{\left ( a+b \right )^{4}.\left ( a-b \right )^{2}}{\left ( a^{2-b^{2}} \right )^{3}\left ( \frac{a+b}{a-b} \right )}\\\Rightarrow x=\frac{\left ( a+b \right )^{4}.\left ( a-b \right )^{2}}{\left ( a^{2-b^{2}} \right )^{3}\left ( \frac{a+b}{a-b} \right )}\\\Rightarrow \frac{\left ( a+b \right )^{4}.\left ( a-b \right )^{2}}{\left ( \left ( a+b \right )\left ( a-b \right ) \right )^{3}\left ( \frac{a+b}{a-b} \right )}\\\Rightarrow x=\frac{\left ( a+b \right )^{4}.\left ( a-b \right )^{2}}{\left ( a+b \right )^{3}\left ( a-b \right )^{3}\left ( \frac{a+b}{a-b} \right )}\\x=\frac{\left ( a+b \right )^{4}.\left ( a-b \right )^{2}}{\left ( a+b \right )^{4}\left ( a-b \right )^{2}}=1\)
52. \(\textrm{Bilangan }^{y}\log\left ( x-1 \right ),^{y}\log\left ( x+1 \right ),^{y}\log\left ( 3x-1 \right )\\ \textrm{ merupakan tiga suku berurutan dari deret aritmatika.}\\ \textrm{ jika jumlah tiga bilangan itu adalah 6 maka x+y=...}\)
a) \(2\)
b) \(3\)
c) \(4\)
d) \(5\)
e) \(6\)
\(\textrm{Bilangan }^{y}\log\left ( x-1 \right ),^{y}\log\left ( x+1 \right ),^{y}\log\left ( 3x-1 \right )\\ \textrm{ merupakan tiga suku berurutan dari deret aritmatika.}\\^{y}\log\left ( x+1 \right )-^{y}\log\left ( x-1 \right )=^{y}\log\left ( 3x-1 \right )-^{y}\log\left ( x-1 \right )\\\Rightarrow ^{y}\log\left ( \frac{x+1}{x-1} \right )=^{y}\log\left ( \frac{3x-1}{x+1} \right )\\\Rightarrow \frac{x+1}{x-1}=\frac{3x-1}{x+1}\\\Rightarrow \left ( x+1 \right )\left ( x+1 \right )=\left ( 3x-1 \right )\left ( x-1 \right )\\x^{2}+2x+1=3x^{2}-4x+1\\\Rightarrow 2x^{2}-6x=0\\\Rightarrow x\left ( 2x-6 \right )=0\\\Rightarrow x=0\textrm{(tidak mungkin)}\textrm{ atau }x=3\\\textrm{Maka deret aritmatika tersebut adalah }^{y}\log 2,^{y}\log 4,^{y}\log 8\\\textrm{Jumlah tiga bilangan itu adalah 6, artinya:}\\^{y}\log2+^{y}\log 4+^{y}\log 8=6\Rightarrow ^{y}\log64=6\\y^{6}=64\Rightarrow y=2\textrm{ maka }x+y=5\)
53. \(\textrm{Nilai x yang memnuhi: }\left ( ^{4}\log x \right )^{2}-^{2}\log\sqrt{x}-\frac{3}{4}=0\textrm{ adalah ...}\)
a) \(16\textrm{ atau }4\)
b) \(16\textrm{ atau }\frac{1}{4}\)
c) \(8\textrm{ atau }2\)
d) \(8\textrm{ atau }\frac{1}{2}\)
e) \(8\textrm{ atau }4\)
\(\left ( ^{4}\log x \right )^{2}-^{2}\log\sqrt{x}-\frac{3}{4}=0\\\Rightarrow \left ( ^{4}\log x \right )^{2}-^{2^{2}}\log\left ( \sqrt{x} \right )^{2}-\frac{3}{4}=0\\\Rightarrow \left ( ^{4} \log x\right )^{2}-^{4}\log x-\frac{3}{4}=0\\\textrm{Misal:}^{4}\log x=p\\\Rightarrow p^{2}-p-\frac{3}{4}=0\\\Rightarrow 4p^{2}-4p-3=0\\\Rightarrow \left ( 2p-3 \right )\left ( 2p+1 \right )=0\\\Rightarrow p_{1}=\frac{3}{4}\textrm{ atau }p_{2}=-\frac{1}{2}\\\bullet p_{1}=\frac{3}{4}\Rightarrow ^{4}\log x=\frac{3}{4}\\\Rightarrow x=4^{\frac{3}{2}}=\sqrt{4^{3}}=8\\\bullet p_{2}=-\frac{1}{2}\Rightarrow ^{4}\log x=-\frac{1}{2}\\\Rightarrow x=4^{-\frac{1}{2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}\\\textrm{Jadi, nilai x yang memenuhi adalah 8 atau }\frac{1}{2}\)
54. \(\textrm{Penyelesaian persamaan }^{3}\log\left ( x^{2}-8x+20 \right )=^{3}\log8 \textrm{ adalah }x_{1}\textrm{ dan }x_{2}\textrm{ dengan }x_{2}< x_{1}.\textrm{ Nilai dari }^{2}\log\left ( x_{1}-x _{2}\right )\textrm{ adalah ...}\)
a) \(1\)
b) \(2\)
c) \(3\)
d) \(4\)
e) \(5\)
\(^{3}\log \left ( x^{2} -8x+20\right )=^{3}\log 8\\\Rightarrow x^{2}-8x+20=8\\\Rightarrow x^{2}-8x+20=0\\\Rightarrow \left ( x-6 \right )\left ( x-2 \right )=0\\x_{1}=6\textrm{ dan }x_{2}=2\\\textrm{Sehingga, nilai dari: }\\^{2}\log \left ( x_{1}-x_{2}\right )=^{2}\log\left ( 6-2 \right )=^{2}\log 4=2\)
55. \(\textrm{Nilai maksimum fungsi: }f(x)=^{2}\log\left ( x+5 \right )+^{2}\log\left ( 3-1 \right )\textrm{ adalah...}\)
a) \(4\)
b) \(8\)
c) \(12\)
d) \(15\)
e) \(16\)
\(\textrm{Tiketahui :}\\f(x)=^{2}\log(x+5)+^{2}\log(3-x)=^{2}\log(x+5)(3-x)=^{2}\log(-x^{2}+2x+15)\\\textrm{Misalkan :}g(x)=-x^{2}+2x+15\textrm{ maka }\\f(x)^{2}\log g(x)\\\textrm{f(x) akan mencapai nilai maksimum jika g(x) mencapai nilai maksimum.}\\g(x)\textrm{ akan maksimum jika:}\\x=\frac{-b}{2a}=\frac{-2}{2(-1)}=1\\\textrm{Sehingga, f(x) akan maksimum ketika x=1}\\f(1)=^{2}log -(1)^{2}+2\cdot 1+15=^{2}\log 16=4\)
56. \(\textrm{Jika }^{4}\log 32=\frac{p}{p+3}\textrm{ maka }\frac{6p+20}{p}=...\)
a) \(-2\)
b) \(-1\)
c) \(0\)
d) \(1\)
e) \(2\)
\(^{4}\log 32=\frac{p}{p+3}\Rightarrow ^{2^{2}}\log 2^{5}=\frac{p}{p+3}\\\Rightarrow \frac{5}{2}^{2}\log 2=\frac{p}{p+3}\Rightarrow \frac{5}{2}=\frac{p}{p+3}\\\Rightarrow 5\left ( p+3 \right )=2p\\\Rightarrow 5p+15=2p\\\Rightarrow 3p=-15\Rightarrow p=-5\\\textrm{Jadi, nilai dari:}\\\frac{6p+20}{p}=\frac{6\left ( -5 \right )+20}{-5}=\frac{-10}{-5}=2\)
57. \(\textrm{Jika }\frac{1}{2}\log \left ( 2x^{2}-x-2 \right )=\log\left ( x+2 \right )\textrm{ maka nilai maksimum }f(y)=-y^{2}+4xy+5x^{2}\textrm{ adalah...}\)
a) \(302\)
b) \(306\)
c) \(212\)
d) \(318\)
e) \(324\)
\(\frac{1}{2}\log\left ( 2x^{2}-x-2 \right )=\log\left ( x+2 \right )\\\Rightarrow \log\left ( 2x^{2}-x-2 \right )=2.\log\left ( x+2 \right )\\\Rightarrow \log\left ( 2x^{2}-x-2 \right )=\log\left ( x+2 \right )^{2}\\\Rightarrow \left ( 2x^{2}-x-2 \right )=\left ( x+2 \right )^{2}\\\Rightarrow 2x^{2}-x-2=x^{2}+4x+4\\\Rightarrow x^{2}-5x-6=0\\\Rightarrow \left ( x-6 \right )\left ( x+1 \right )=0\\\textrm{Diperoleh:}\\x=6\textrm{ dan }x=-1\textrm{(tidak memenuhi syarat logaritma )}\\\textrm{Sehingga:}\\f(y)=-y^{2}+4xy+5x^{2}=-y^{2}+4\left ( 6 \right )y+5\left ( 6 \right )^{2}=-y^{2}+24y+180\\\textrm{Nilai maksimum dai f(y) diperoleh ketika }x=\frac{-b}{2a}=12,\textrm{ yaitu:}f(12)=-144+(24)(12)+180=324\)
58. \(\log b+\log ab^{2}+\log a^{2}b^{3}+...+\log a^{9}b^{10}=...\)
a) \(\log 45a+\log55b\)
b) \(\left ( \log a \right )^{45}+\left ( \log a \right )^{55}\)
c) \(45\log a+55\log b\)
d) \(\left ( 91 \right )\log a+\left ( 101 \right )\log b\)
e) \(45\log\left ( ab \right )\)
\(\log b+\log ab^{2}+\log a^{2}b^{3}+...+\log a^{9}b^{10}\\=\log\left ( ab^{2} \right )\left ( a^{2}b^{3} \right )...\left ( a^{9}b^{10} \right )\\=\log a^{1+2+3+...+10}.b^{1+2+3+...10}\\=\log a^{45}.b^{55}\\=\log a^{45}+\log b^{55}=45\log a+55\log b\)
59. \(\textrm{Jika }^{a}\log\left ( 1-^{2}\log\frac{1}{32} \right )=3\textrm{ maka nilai a adalah...}\)
a) \(\sqrt[3]{2}\)
b) \(\sqrt[3]{4}\)
c) \(\sqrt[3]{5}\)
d) \(\sqrt[3]{6}\)
e) \(\sqrt[3]{7}\)
\(^{a}\log \left ( 1-^{2}\log \frac{1}{32} \right )=3\\\Rightarrow ^{a}\log\left ( 1-^{2}\log 32^{-1} \right )=3\\\Rightarrow ^{a}\log\left ( 1-^{2}\log 2^{-5} \right )=3\\\Rightarrow ^{a}\log\left ( 1-\left ( -5 \right ) \right )=3\\\Rightarrow ^{a}\log 6=3\\\Rightarrow a^{3}=6\Rightarrow a=\sqrt[3]{6}\)
60. \(\textrm{Jika }\log x+\log 2x+\log 4x+\log 8x+...+\log1024x=22\textrm{ maka x}=...\)
a) \(5,5\)
b) \(3,125\)
c) \(2,75\)
d) \(1,375\)
e) \(0,625\)
\(^{a}\log \left ( 1-^{2}\log \frac{1}{32} \right )=3\\\Rightarrow ^{a}\log\left ( 1-^{2}\log 32^{-1} \right )=3\\\Rightarrow ^{a}\log\left ( 1-^{2}\log 2^{-5} \right )=3\\\Rightarrow ^{a}\log\left ( 1-\left ( -5 \right ) \right )=3\\\Rightarrow ^{a}\log 6=3\\\Rightarrow a^{3}=6\Rightarrow a=\sqrt[3]{6}\)
