Contoh Soal 9 Fungsi Logaritma (Pemecahan Masalah Olimpiade)

$\begin{array}{l}\\ 41.& \text{Jika}x={}^{15}\log 75\text{dan}y={}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125},}\\ & \text{maka nilai}5x+3y-2xy\text{adalah}....\\ \\ & \text{KOMPETISI HARDIKNAS ONLINE}\\ & \text{POSI}(\text{Pelatihan Olimpiade Sain Indonesia})\\ & \text{Bidang Matematika 2020}\\ & \begin{array}{l}\\ \text{a}.& -1\\ \text{b}.& 1\\ \text{c}.& 3\\ \text{d}.& 5\\ \text{e}.& 7\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{e}}\\ & \begin{array}{rl}& 5x+3y-2xy\\ & =5({}^{15}\log 75)+3\left({}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125}}\right)\\ & -2({}^{15}\log 75)\left({}^{{\displaystyle {}^{\frac{3}{5}}}}\log {\displaystyle \frac{9}{125}}\right)\\ & =5\left({\displaystyle \frac{\log 75}{\log 15}}\right)+3\left({\displaystyle \frac{\log {\displaystyle \frac{9}{125}}}{\log {\displaystyle \frac{3}{5}}}}\right)\\ & -2\left({\displaystyle \frac{\log 75}{\log 15}}\right)\left({\displaystyle \frac{\log {\displaystyle \frac{9}{125}}}{\log {\displaystyle \frac{3}{5}}}}\right)\\ & =5\left({\displaystyle \frac{\log {3.5}^2}{\log 3.5}}\right)+3\left({\displaystyle \frac{\log -\log 125}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log {3.5}^2}{\log 3.5}}\right)\left({\displaystyle \frac{\log 9-\log 125}{\log 3-\log 5}}\right)\\ & =5\left({\displaystyle \frac{\log 3+\log 5^2}{\log 3-\log 5}}\right)+3\left({\displaystyle \frac{\log 3^2-\log 5^3}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log 3+\log 5^2}{\log 3+\log 5}}\right)\left({\displaystyle \frac{\log 3^2-\log 5^3}{\log 3-\log 5}}\right)\\ & =5\left({\displaystyle \frac{\log 3+2\log 5}{\log 3-\log 5}}\right)+3\left({\displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5}}\right)\\ & -2\left({\displaystyle \frac{\log 3+2\log 5}{\log 3+\log 5}}\right)\left({\displaystyle \frac{2\log 3-3\log 5}{\log 3-\log 5}}\right)\\ \\ & \text{Misalkan}\log 3=A,\log 5=B\end{array}\end{array}$

$.\begin{array}{rl}& \text{Selanjutnya}\\ & =5\left({\displaystyle \frac{A+2B}{A+B}}\right)+3\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & -2\left({\displaystyle \frac{A+2B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & =\left({\displaystyle \frac{5A+10B}{A+B}}\right)+\left({\displaystyle \frac{6A-9B}{A-B}}\right)\\ & -\left({\displaystyle \frac{2A+4B}{A+B}}\right)\left({\displaystyle \frac{2A-3B}{A-B}}\right)\\ & ={\displaystyle \frac{(5A+10B)(A-B)+(6A-9B)(A+B)}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{5A^2-5AB+10AB-10B^2}{A^2-B^2}}\\ & +{\displaystyle \frac{6A^2+6AB-9AB-9B^2}{A^2-B^2}}\\ & -\left({\displaystyle \frac{4A^2-6AB+8AB-12B^2}{A^2-B^2}}\right)\\ & ={\displaystyle \frac{7A^2-7B^2}{A^2-B^2}}\\ & ={\displaystyle \frac{7(A^2-B^2)}{A^2-B^2}}\\ & =7\end{array}$

$\begin{array}{l}\\ 42.& \text{Diberikan}A={}^6\log 16\text{dan}B={}^{12}\log 27\\ & \text{Terdapat bilangan-bilangan bulat positif}\\ & a,b,\text{dan}c\text{sehingga}(A+a)(B+b)=c\\ & \text{Nilai dari}a+b+c\text{adalah}....\\ \\ & \text{KOMPETISI HARDIKNAS ONLINE}\\ & \text{POSI}(\text{Pelatihan Olimpiade Sain Indonesia})\\ & \text{Bidang Matematika 2020}\\ & \begin{array}{l}\\ \text{a}.& 23\\ \text{b}.& 24\\ \text{c}.& 27\\ \text{d}.& 30\\ \text{e}.& 34\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{....}}\\ & \begin{array}{rl}& \text{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}}}\\ & \iff \log 2+\log 3={\displaystyle \frac{4\log 2}{A}...........(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}}\\ & \iff 2\log 2+\log 3={\displaystyle \frac{3\log 3}{B}.........(2)}\\ & \text{ELIMINASI}\\ & \text{Dari persamaan (1) dan (2) diperoleh}:\\ & \bullet \log 2={\displaystyle \frac{3\log 3}{B}-{\displaystyle \frac{4\log 2}{A}}}\\ & \iff \log 2={\displaystyle \frac{3A\log 3-4B\log 2}{AB}}\\ & \iff AB\log 2=3A\log 3-4B\log 2\\ & \iff AB\log 2+4B\log 2=3A\log 3\\ & \iff (AB+4B)\log 2=3A\log 3\\ & \iff {\displaystyle \frac{\log 2}{\log 3}={\displaystyle \frac{3A}{AB+4B}..........(3)}}\\ & \bullet \log 3={\displaystyle \frac{8\log 2}{A}-{\displaystyle \frac{3\log 3}{B}}}\\ & \iff \log 3={\displaystyle \frac{8B\log 2-3A\log 3}{AB}}\\ & \iff AB\log 3=8B\log 2-3A\log 3\\ & \iff AB\log 3+3A\log 3=8B\log 2\\ & \iff (AB+3A)\log 3=8B\log 2\\ & \iff \frac{\log 2}{\log 3}={\displaystyle \frac{AB+3A}{8B}...........(4)}\\ & \text{KESAMAAN}\\ & \frac{\log 2}{\log 3}=\frac{\log 2}{\log 3}\\ & {\displaystyle \frac{AB+3A}{8B}={\displaystyle \frac{3A}{AB+4B}}}\\ & \iff (AB+3A)(AB+4B)=(8B).(3A)\\ & \iff (B+3)(A+4)=24\\ & \iff (A+4)(B+3)=24\\ & \text{KESIMPULAN}\\ & a=4,b=3,\text{dan}c=24,\\ & \text{maka}a+b+c=4+3+24=31\end{array}\end{array}$