Persamaan Eksponen Bagian 5

Persamaan Eksponen Bagian 5

# Persamaan Eksponen Berbentuk $\left \{ h\left ( x \right ) \right \}^{f\left ( x \right )}=\left \{ h\left ( x \right ) \right \}^{g\left ( x \right )}$

$\begin {array}{lcl} \textup{Jika}\; \left \{ h\left ( x \right ) \right \}^{f\left ( x \right )}=\left \{ h\left ( x \right ) \right \}^{g\left ( x \right )},\; \textup{maka kemungkinannya adalah:}\\ 1.\; f\left ( x \right )=g\left ( x \right )\\ 2.\; h\left ( x \right )=1\\ 3.\; h\left ( x \right )=-1,\; \textup{asalkan}\; f\left ( x \right )\; \textup{dan}\; g\left ( x \right )\; \textup{keduanya ganjil atau keduanya genap}\\ 4.\; h\left ( x \right )=0,\; \textup{asalkan}\; f\left ( x \right )\; \textup{dan}\; g\left ( x \right )\; \textup{keduanya positif} \end{array}$

Soal dan Pembahasan

$\begin {array}{lcl} \textup{Tentukan penyelesaian dari}\; \left ( x-1 \right )^{2x-1}=\left ( x-1 \right )^{x+1} \end{array}$

Pembahasan:

$\begin {array}{lcl} \textup{Persamaan}\; \left ( x-1 \right )^{2x-1}\left ( x-1 \right )^{x+1},\; \textup{sepadan dengan persamaan }\\ \textup{eksponen berbentuk}\; \left \{ h\left ( x \right ) \right \}^{f\left ( x \right )}=\left \{ h\left ( x \right ) \right \}^{g\left ( x \right )},\; \textup{maka:}\\ h\left ( x \right )=x-1\\ f\left ( x \right )=2x-1\\ g\left ( x \right )=x+1\\ \textup{Kemungkinan penyelesaian:}\\ 1.\; \; \; \; \; \; \; \; \; \; f\left ( x \right )=g\left ( x \right )\\ \; \; \; \; \; \; \; \; \; \:2x-1=x+1\\ \; \; \; \; \; \; \; \; \; \; 2x-x=1+1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=2\\ \end{array}$

$\begin {array}{lcl} 2.\; \; \; \; \; \; \; \; \; \; h\left ( x \right )=1\\ \; \; \; \; \; \; \; \; \; \; \; \:x-1=1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=1+1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=2\\ 3.\; \; \; \; \; \; \; \; \; \; h\left ( x \right )=-1\\ \; \; \; \; \; \; \; \; \; \; \; \:x-1=-1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=-1+1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=0\\ \; \; \; \; \textup{Nilai}\; x=0,\; \textup{harus disubstitusikan ke}\; f\left ( x \right )\; \textup{dan}\; g\left ( x \right )\\ \; \; \; f\left ( x \right )=2x-1\\ \; \; \; f\left ( 0 \right )=2\left ( 0 \right )-1\\ \end{array}$

$\begin {array}{lcl} \; \; \; \; \; \; \; \; \; \; \; =-1\; \; \; \left ( \textup{ganjil} \right )\\ \; \; \; g\left ( x \right )=x+1\\ \; \; \; g\left ( 0 \right )=0+1\\ \; \; \; \; \; \; \; \; \; \; \; =1\; \; \; \left ( \textup{ganjil} \right )\\ \; \; \; \textup{Karena keduanya ganjil, maka}\; x=0\;\textup{ merupakan penyelesaiannya.}\\ \end{array}$

$\begin {array}{lcl} 4.\; \; \; \; \; \; \; \; \; \; h\left ( x \right )=0\\ \; \; \; \; \; \; \; \; \; \; \; \:x-1=0\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=0+1\\ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; x=1\\ \; \; \; \; \textup{Nilai}\; x=1,\; \textup{harus disubstitusikan ke}\; f\left ( x \right )\; \textup{dan}\; g\left ( x \right )\\ \end{array}$

$\begin {array}{lcl} \; \; \; f\left ( x \right )=2x-1\\ \; \; \; f\left ( 1 \right )=2\left ( 1 \right )+1\\ \; \; \; \; \; \; \; \; \; \; \; =3\; \; \; \left ( \textup{positif} \right )\\ \; \; \; g\left ( x \right )=x+1\\ \; \; \; g\left ( 1 \right )=1+1\\ \; \; \; \; \; \; \; \; \; \; \; =2\; \; \; \left ( \textup{positif} \right )\\ \textup{Karena keduanya positif, maka}\; x=1\; \textup{merupakan penyelesaiannya}.\\ \end{array}$