Persamaan Logaritma 4

 D. Persamaan Logaritma Bentuk  ${}^{h(x)}\log f(x)={}^{h(x)}\log g(x)$.

Syarat penyelesaian dari bentuk:

$\begin{array}{rl}& \text{Jika}{}^{h(x)}\log f(x)={}^{h(x)}\log g(x)\\ & \text{dengan}f(x)\text{dan}g(x)\text{keduanya positif}\\ & \text{serta}h(x)>0,\text{dan}h(x)\ne 1,\\ & \text{maka}f(x)=g(x)\end{array}$.

$\begin{array}{rl}& \mathbf{\text{atau}}\\ & \text{Pernyataan}{}^{h(x)}\log f(x)={}^{h(x)}\log g(x)\\ & \text{akan bernilai benar jika}\\ & (1)h(x)>0,h(x)\ne 1\\ & (2)f(x)>0,g(x)>0\\ & (3)f(x)=g(x)\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^x\log (2x-3)={}^x\log (x-1)\\ & \text{b}.{}^x\log (2x^2+11x-6)={}^x\log (x^2+10x)\\ & \text{c}.{}^x\log (x-1)+{\displaystyle \frac{1}{{}^{x+6}\log x}=2+{\displaystyle \frac{1}{{}^2\log x}}}\\ & \text{d}.{}^{2x-1}\log (x^3+3x^2-4x-1)={}^{2x-1}\log (2x^2+3)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\mathbf{\text{a}}.\text{Dik}& \text{etahui}\\ & {}^x\log (2x-3)={}^x\log (x-1)\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& x>0\\ f(x)>0& \begin{array}{rl}2x& -3>0\\ \iff & 2x>3\\ \iff & x>{\displaystyle \frac{3}{2}}\end{array}\\ g(x)>0& \begin{array}{rl}x-& 1>0\\ \iff & x>1\end{array}\\ \hline\end{array}\\ & \text{Syarat numerusnya},x>{\displaystyle \frac{3}{2}}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff 2x-3=x-1\\ & \iff 2x-x=3-1\\ & \iff x=2\\ & \text{Karena}x>{\displaystyle \frac{3}{2},}\\ & \text{maka nilai}x=2\text{memenuhi}\\ (\ast )& \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{b}}.\text{Dik}& \text{etahui}\\ & {}^x\log (2x^2+11x-6)={}^x\log (x^2+10)\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& x>0\\ f(x)>0& \begin{array}{rl}2x^2& +11x-6>0\\ \iff & (x+6)(2x-1)>0\\ \iff & x<-6\text{atau}x>{\displaystyle \frac{1}{2}}\end{array}\\ g(x)>0& \begin{array}{rl}{x}^2& +10x>0\\ \iff & x(x+10)>0\\ \iff & x<-10\text{atau}x>0\end{array}\\ \hline\end{array}\\ & \text{Syarat numerusnya},x>{\displaystyle \frac{1}{2}}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff 2x^2+11x-6=x^2+10x\\ & \iff x^2+x-6=0\\ & \iff (x+3)(x-2)=0\\ & \iff x=-3\text{atau}x=2\\ & \text{Karena}x>{\displaystyle \frac{1}{2},}\\ & \text{maka nilai}x=2\text{memenuhi}\\ (\ast )& \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{c}}.\text{Dik}& \text{etahui}\\ & {}^x\log (x-1)+{\displaystyle \frac{1}{{}^{x+6}\log x}=2+{\displaystyle \frac{1}{{}^2\log x}}}\\ & \iff {}^x\log (x-1)+{}^x\log (x+6)={}^x\log x^2+{}^x\log 2\\ & \iff {}^x\log (x-1)(x+6)={}^x\log 2x^2\\ & \iff {}^x\log x^2+5x-6={}^x\log 2x^2\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& x>0\\ f(x)>0& \begin{array}{rl}{x}^2& +5x-6>0\\ \iff & (x+6)(x-1)>0\\ \iff & x<-6\text{atau}x>1\end{array}\\ g(x)>0& \begin{array}{rl}2x^2& >0\\ \iff & x>0\end{array}\\ \hline\end{array}\\ & \text{Syarat numerusnya},x>{\displaystyle 1}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff x^2+5x-6=2x^2\\ & \iff x^2-5x+6=0\\ & \iff (x-2)(x-3)=0\\ & \iff x=2\text{atau}x=3\\ & \text{Karena}x>{\displaystyle 1,}\\ & \text{maka nilai}x=2\text{dan}x=3\text{memenuhi}\\ (\ast )& \text{Jadi},\text{HP}=\{2,3\}\end{array}\\ & \begin{array}{rl}\mathbf{\text{d}}.\text{Dik}& \text{etahui}\\ & {}^{2x-1}\log (x^3+3x^2-4x-1)={}^{2x-1}\log (2x^2+3)\\ (\ast )& \text{Syarat numerus dan bilangan pokok}\\ & \begin{array}{|ll|}\hline \text{Syarat}& \text{hasil}\\ h(x)>0,h(x)\ne 1& 2x-1>0\iff x>{\displaystyle \frac{1}{2}}\\ f(x)>0& \begin{array}{rl}{x}^3& +3x^2-4x-1>0\\ \bullet & \text{Susah difaktorkan}\\ \bullet & \text{gunakan uji nilai}\end{array}\\ g(x)>0& \begin{array}{rl}2x^2& +3>0\\ \bullet & a>0,D<0\\ \bullet & \mathbf{\text{Definit positif}}\end{array}\\ \hline\end{array}\\ & \text{Syarat basis/bilangan pokoknya},x>{\displaystyle \frac{1}{2}}\\ (\ast )& \text{Syarat kedua},f(x)=g(x)\\ & \iff x^3+3x^2-4x-1=2x^2+3\\ & \iff x^3+x^2-4x-4=0\\ & \iff x^2(x+1)-4(x+1)=0\\ & \iff (x+1)(x^2-4)=0\\ & \iff (x+1)(x+2)(x-2)=0\\ & \iff x=-1\text{atau}x=-2\text{atau}x=2\\ & \text{Karena basisnya}x>{\displaystyle \frac{1}{2},}\\ & \text{maka nilai yang memenuhi hanya}x=2\text{saja}\\ & \text{dan nilai untuk numerusnya juga memenuhi}\\ & \text{yaitu}:(2{)}^3+3(2{)}^2-4(2)-1=11>0\\ & \text{demikian pula untuk}:2(2{)}^3+3=19>0\\ (\ast )& \text{Jadi},\text{HP}=\left\{2\right\}\end{array}\end{array}$.

Penjelasan untuk jawaban 1. d  tentang definit positif  di sini dan di sini

$\text{LATIHAN SOAL}$.

$\begin{array}{l}\\ 2.& \text{Tentukan himpunan penyelesaian dari}\\ & \text{a}.{}^x\log (2x+3)={}^x\log (x+7)\\ & \text{b}.{}^x\log (x+12)-{}^x\log (4x+1)=0\\ & \text{c}.{}^{x-2}\log (x^2-3)={}^{x-2}\log x\\ & \text{d}.{}^{3x-2}\log (x^2-2x+4)={}^{3x-2}\log (5-4x)\\ \end{array}$