$\begin{array}{ll}\\ 85.&\textrm{Perhatikan pernyataan berikut}\\ &(1).\quad ^{2}\log 7+\, ^{2}\log 2=\, ^{2}\log 14\\ &(2).\quad ^{2}\log 12-\, ^{2}\log 4=\, ^{2}\log 8\\ &(3).\quad ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=3\\ &(4).\quad ^{2}\log \displaystyle \frac{1}{2}\times \, ^{2}\log 16=3\\ &\textrm{Pernyataan di atas yang benar adalah}\\ &\textrm{a}.\quad (1)\: \: \textrm{dan}\: \: (2)\\ &\textrm{b}.\quad (1)\: \: \textrm{dan}\: \: (3)\\ &\textrm{c}.\quad (2)\: \: \textrm{dan}\: \: (3)\\ &\textrm{d}.\quad (2)\: \: \textrm{dan}\: \: (4)\\ &\textrm{e}.\quad (3)\: \: \textrm{dan}\: \: (4)\\\\ &\begin{aligned}&\textbf{Jawab}:\quad \textbf{b}\\ &\textrm{Perhatikan pernyataan (1)}\\ &\color{red}^{2}\log 7+\, ^{2}\log 2=\, ^{2}\log 14\\ &\textrm{benar karena},\\ &^{a}\log b+\, ^{a}\log c=\, ^{a}\log bc\\ &\textrm{Perhatikan pernyataan (2)}\\ &\color{red}^{2}\log 12-\, ^{2}\log 4=\, ^{2}\log 8\\ &\textrm{adalah salah, seharusnya}\\ &^{2}\log 12-\, ^{2}\log 4=\, \color{blue}^{2}\log \frac{12}{4}=\, ^{2}\log 3\\ &\textrm{ingat sifat berikut}\\ &^{a}\log b-\, ^{a}\log c=\, ^{a}\log \displaystyle \frac{b}{c}\\ &\textrm{Perhatikan pernyataan (3)}\\ & ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=3\\ &\textrm{Benar, karena sama dengan sifat no (1) di atas}\\ &\textrm{yaitu}:\\ &^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=\, ^{2}\log \displaystyle \frac{1}{2}.16\\ &\qquad =\, ^{2}\log 8=\, ^{2}\log 2^{3}=\color{red}3 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 86.&\textrm{Perhatikan pernyataan berikut}\\ &(1).\quad ^{2}\log 10+\, ^{2}\log 3=\, ^{2}\log 13\\ &(2).\quad ^{2}\log 20-\, ^{2}\log 4=\, ^{2}\log 5\\ &(3).\quad ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=-4\\ &(4).\quad ^{2}\log \displaystyle \frac{1}{2}\times \, ^{2}\log 16=-4\\ &\textrm{Pernyataan di atas yang benar adalah}\\ &\textrm{a}.\quad (1)\: \: \textrm{dan}\: \: (2)\\ &\textrm{b}.\quad (1)\: \: \textrm{dan}\: \: (3)\\ &\textrm{c}.\quad (2)\: \: \textrm{dan}\: \: (3)\\ &\textrm{d}.\quad (2)\: \: \textrm{dan}\: \: (4)\\ &\textrm{e}.\quad (3)\: \: \textrm{dan}\: \: (4)\\\\ &\textbf{Jawab}:\quad \textbf{d}\\ &\begin{aligned} &\textrm{Perhatikan pernyataan (1)}\\ &^{2}\log 10+\, ^{2}\log 3=\, ^{2}\log 13\\ &\textrm{adalah salah karena},\\ &^{a}\log b+\, ^{a}\log c=\, ^{a}\log bc\\ &\color{red}\textrm{Perhatikan pernyataan (2)}\\ &^{2}\log 20-\, ^{2}\log 4=\, ^{2}\log 5\\ &\textrm{benar, karena}\\ &^{2}\log 20-\, ^{2}\log 4=\, ^{2}\log \frac{20}{4}=\, ^{2}\log 5\\ &\textrm{ingat sifat berikut}\\ &^{a}\log b-\, ^{a}\log c=\, ^{a}\log \displaystyle \frac{b}{c}\\ &\color{red}\textrm{Perhatikan pernyataan (3)}\\ & ^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=-4\\ &\textrm{salah, karena sama dengan sifat no (1) di atas}\\ &\textrm{yaitu}:\\ &^{2}\log \displaystyle \frac{1}{2}+\, ^{2}\log 16=\, ^{2}\log \displaystyle \frac{1}{2}.16\\ &\qquad =\, ^{2}\log 8=\, ^{2}\log 2^{3}=3\\ &^{2}\log \displaystyle \frac{1}{2}\times \, ^{2}\log 16=^{2}\log 2^{-1}\times \, ^{2}\log 16\\ &\qquad =\, ^{2}\log \displaystyle 2^{-1}\times \, ^{2}\log 2^{4}=(-1).(4)=\color{red}-4 \end{aligned} \end{array}$.
$\begin{array}{ll} 87.&\textrm{Jika}\: \: \log 2=0,301\: \: \log 3=0,477\\ &\textrm{maka nilai}\: \: \log \sqrt[3]{225}\: \: \textrm{adalah}\: ....\\ &\begin{array}{lllllll} \textrm{a}.&\displaystyle 0,714&&&\textrm{d}.&0,778\\ \textrm{b}.&\displaystyle 0,734\quad&\textrm{c}.&0,756\quad&\textrm{e}.&\color{red}0,784 \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}\log \sqrt[3]{225}&=\log 225^{.^{\frac{1}{3}}}\\ &=\log (15^{2})^{.^{\frac{1}{3}}}\\ &=\log 15^{.^{\frac{2}{3}}}\\ &=\displaystyle \frac{2}{3}\log 15\\ &=\displaystyle \frac{2}{3}\log 3\times 5\\ &=\displaystyle \frac{2}{3}\left ( \log 3+\log 5 \right )\\ &=\displaystyle \frac{2}{3}\left ( \log 3+\log \displaystyle \frac{10}{2} \right )\\ &=\displaystyle \frac{2}{3}\left ( \log 3+\log 10-\log 2 \right )\\ &=\displaystyle \frac{2}{3}\left ( \log 3+1-\log 2 \right )\\ &=\displaystyle \frac{2}{3}\left ( 0,477+1,000+0,301 \right )\\ &=\displaystyle \frac{2}{3}\left ( 1,176 \right )\\ &=\displaystyle \frac{2,352}{3}\\ &=\color{red}0,784 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 88.&\textrm{Fungsi invers dari}\: \: f(x)=5^{x}\\ &\begin{array}{lllllll} \textrm{a}.&f^{-1}(x)=5^{-x}&\\ \textrm{b}.&f^{-1}(x)=\left ( \displaystyle \frac{1}{5} \right )^{x}\\ \textrm{c}.&f^{-1}(x)=\left ( \displaystyle \frac{1}{5} \right )^{-x}&\\ \textrm{d}.&f^{-1}(x)=\, ^{5}\log x\\ \textrm{e}.&f^{-1}(x)=\, ^{x}\log 5\\ \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{d}\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: f(x)=5^{x},\: \textrm{maka inversnya}\\ &\textrm{adalah}:\\ &\textrm{Langkah mula-mula dilogkan}\\ &\textrm{masing-masing ruas untuk mencari nilai}\\ &x,\: \: \textrm{yaitu}:\\ &\log f(x)=\log 5^{x}\Leftrightarrow \log f(x)=x\log 5\\ &\Leftrightarrow x=\displaystyle \frac{\log f(x)}{\log 5}=\, ^{5}\log f(x)\\ &\textrm{Selanjutnya kita ganti}\: \: x\: \: \textrm{dengan}\: \: f^{-1}(x),\: \: \textrm{dan}\\ & f(x)\: \: \textrm{dengan}\: \: x,\: \: \textrm{sehingga menjadi bentuk}\\ &\Leftrightarrow f^{-1}(x)=\, \color{red}^{5}\log x \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 89.&\textrm{Fungsi invers dari}\: \: y=\left ( \displaystyle \frac{1}{3} \right )^{x}\\ &\begin{array}{lllllll} \textrm{a}.&y^{-1}=\, ^{.^{\frac{1}{3}}}\log x&\quad&\textrm{d}.&y^{-1}=\left (\displaystyle \frac{1}{3} \right )^{-x}\\ \textrm{b}.&y^{-1}=\, ^{-3}\log x&&\textrm{e}.&y^{-1}=x^{.^{\frac{1}{3}}}\\ \textrm{c}.&y^{-1}=3^{-x}&&& \end{array}\\\\ &\textbf{Jawab}:\quad \color{red}\textbf{a}\\ &\begin{aligned}&\textrm{Diketahui bahwa}:\: \: y=\left ( \displaystyle \frac{1}{3} \right )^{x},\: \textrm{maka inversnya}\\ &\textrm{sama seperti pada nomor sebelumnya, yaitu}:\\ &\textrm{dilogkan masing-masing ruas untuk mencari nilai}\\ &x,\: \: \textrm{yaitu}:\\ &\log y=\log \left ( \displaystyle \frac{1}{3} \right )^{x}\Leftrightarrow \log f(x)=x\log \left ( \displaystyle \frac{1}{3} \right )\\ &\Leftrightarrow x=\displaystyle \frac{\log y}{\log \left ( \displaystyle \frac{1}{3} \right )}=\, ^{.^{\frac{1}{3}}}\log y\\ &\textrm{Selanjutnya kita ganti}\: \: x\: \: \textrm{dengan}\: \: y^{-1},\: \: \textrm{dan}\\ & y\: \: \textrm{dengan}\: \: x,\: \: \textrm{sehingga menjadi bentuk}\\ &\Leftrightarrow y^{-1}=\, \color{red}^{.^{\frac{1}{3}}}\log x \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 91.&\textrm{Jika}\: \: x=\: ^{15}\log 75\: \: \textrm{dan}\: \: y=\: ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125},\\ &\textrm{maka nilai}\: \: 5x+3y-2xy\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&-1\\ \textrm{b}.&1\\ \textrm{c}.&3\\ \textrm{d}.&5\\ \color{red}\textrm{e}.&7 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{e}\\ &\begin{aligned}&\color{black}5x+3y-2xy\\ &=5\left ( ^{15}\log 75 \right )+3\left ( ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125} \right )\\ &\qquad -2\left ( ^{15}\log 75 \right )\left ( ^{\displaystyle ^{\frac{3}{5}}}\log \displaystyle \frac{9}{125} \right )\\ &=5\left ( \displaystyle \frac{\log 75}{\log 15} \right )+3\left ( \displaystyle \frac{\log \displaystyle \frac{9}{125}}{\log \displaystyle \frac{3}{5}} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 75}{\log 15} \right )\left ( \displaystyle \frac{\log \displaystyle \frac{9}{125}}{\log \displaystyle \frac{3}{5}} \right )\\ &=5\left ( \displaystyle \frac{\log 3.5^{2}}{\log 3.5} \right )+3\left ( \displaystyle \frac{\log -\log 125}{\log 3-\log 5} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 3.5^{2}}{\log 3.5} \right )\left ( \displaystyle \frac{\log 9-\log 125}{\log 3-\log 5} \right )\\ &=5\left ( \displaystyle \frac{\log 3+\log 5^{2}}{\log 3-\log 5} \right )+3\left ( \displaystyle \frac{\log 3^{2}-\log 5^{3}}{\log 3-\log 5} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 3+\log 5^{2}}{\log 3+\log 5} \right )\left ( \displaystyle \frac{\log 3^{2}-\log 5^{3}}{\log 3-\log 5} \right )\\ &=5\left ( \displaystyle \frac{\log 3+2\log 5}{\log 3-\log 5} \right )+3\left ( \displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5} \right )\\ &\qquad -2\left ( \displaystyle \frac{\log 3+2\log 5}{\log 3+\log 5} \right )\left ( \displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5} \right )\\\\ &\color{red}\textrm{Misalkan}\: \: \color{black}\log 3=A,\: \: \log 5=B \end{aligned} \end{array}$
$.\qquad\begin{aligned} &\color{red}\textrm{Selanjutnya}\\ &=5\left ( \displaystyle \frac{A+2B}{A+B} \right )+3\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &\qquad -2\left ( \displaystyle \frac{A+2B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\left ( \displaystyle \frac{5A+10B}{A+B} \right )+\left ( \displaystyle \frac{6A-9B}{A-B} \right )\\ &\qquad -\left ( \displaystyle \frac{2A+4B}{A+B} \right )\left ( \displaystyle \frac{2A-3B}{A-B} \right )\\ &=\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{5A^{2}-5AB+10AB-10B^{2}}{A^{2}-B^{2}}\\ &\quad +\displaystyle \frac{6A^{2}+6AB-9AB-9B^{2}}{A^{2}-B^{2}}\\ &\qquad -\left (\displaystyle \frac{4A^{2}-6AB+8AB-12B^{2}}{A^{2}-B^{2}} \right )\\ &=\displaystyle \frac{7A^{2}-7B^{2}}{A^{2}-B^{2}}\\ &=\displaystyle \frac{7\left ( A^{2}-B^{2} \right )}{A^{2}-B^{2}}\\ &=\color{red}7 \end{aligned}$
$\begin{array}{ll}\\ 92.&\textrm{Diberikan}\: \: A=\: ^{6}\log 16\: \: \textrm{dan}\: \: B=\: ^{12}\log 27\\ &\textrm{Terdapat bilangan-bilangan bulat positif}\\ &a,\: b,\: \: \textrm{dan}\: \: c\: \: \textrm{sehingga}\: \: (A+a)(B+b)=c\\ &\textrm{Nilai dari}\: \: a+b+c\: \: \textrm{adalah}\: ....\\\\ &\color{blue}\textrm{KOMPETISI HARDIKNAS ONLINE}\\ &\color{blue}\textrm{POSI}(\textrm{Pelatihan Olimpiade Sain Indonesia})\\ &\color{blue}\textrm{Bidang Matematika 2020}\\ &\begin{array}{llll}\\ \textrm{a}.&23\\ \textrm{b}.&24\\ \textrm{c}.&27\\ \textrm{d}.&30\\ \textrm{e}.&34 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{....}\\ &\begin{aligned}&\color{black}\textrm{Diketahui}\\ &A=\: ^{6}\log 16=\displaystyle \frac{\log 16}{\log 6}=\displaystyle \frac{\log 2^{4}}{\log 2.3}=\frac{4\log 2}{\log 2+\log 3}\\ &\Leftrightarrow \color{black}\log 2+\log 3=\displaystyle \frac{4\log 2}{A}\: ...........\color{red}(1)\\ &B=\: ^{12}\log 27=\displaystyle \frac{\log 27}{\log 12}=\frac{\log 3^{3}}{\log 2^{2}.3}=\frac{3\log 3}{2\log 2+\log 3}\\ &\Leftrightarrow \color{black}2\log 2+\log 3=\displaystyle \frac{3\log 3}{B}\: .........\color{red}(2)\\ &\color{black}\textrm{ELIMINASI}\\ &\textrm{Dari persamaan (1) dan (2) diperoleh}:\\ &\bullet \quad \log 2=\displaystyle \frac{3\log 3}{B}-\displaystyle \frac{4\log 2}{A}\\ &\qquad \Leftrightarrow \log 2=\displaystyle \frac{3A\log 3-4B\log 2}{AB}\\ &\qquad \Leftrightarrow AB\log 2=3A\log 3-4B\log 2\\ &\qquad \Leftrightarrow AB\log 2+4B\log 2=3A\log 3\\ &\qquad \Leftrightarrow (AB+4B)\log 2=3A\log 3\\ &\qquad \Leftrightarrow \displaystyle \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{3A}{AB+4B}\: ..........\color{red}(3)\\ &\bullet \quad \log 3=\displaystyle \frac{8\log 2}{A}-\displaystyle \frac{3\log 3}{B}\\ &\qquad \Leftrightarrow \log 3=\displaystyle \frac{8B\log 2-3A\log 3}{AB}\\ &\qquad \Leftrightarrow AB\log 3=8B\log 2-3A\log 3\\ &\qquad \Leftrightarrow AB\log 3+3A\log 3=8B\log 2\\ &\qquad \Leftrightarrow (AB+3A)\log 3=8B\log 2\\ &\qquad \Leftrightarrow \frac{\log 2}{\log 3}=\color{purple}\displaystyle \frac{AB+3A}{8B}...........\color{red}(4)\\ &\color{black}\textrm{KESAMAAN}\\ &\qquad\quad \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ &\color{purple}\displaystyle \frac{AB+3A}{8B}=\color{purple}\displaystyle \frac{3A}{AB+4B}\\ &\qquad \Leftrightarrow (AB+3A)(AB+4B)=(8B).(3A)\\ &\qquad \Leftrightarrow (B+3)(A+4)=24\\ &\qquad \Leftrightarrow (A+4)(B+3)=24\\ &\color{black}\textrm{KESIMPULAN}\\ &a=4,\: b=3,\: \: \textrm{dan}\: \: c=24,\\ &\color{purple}\textrm{maka}\: \: \color{black}a+b+c=4+3+24=\color{red}31 \end{aligned} \end{array}$.
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