Tentukan himpunan penyelesaian persamaan \((x + 2)^{x + 4} = (x + 2)^{x^2 + 3x + 1}\)
Langkah 1: \(f(x) = g(x)\)
\(x + 4 = x^2 + 3x + 1\)
\(0 = x^2 + 3x \:-\: x + 1 – 4\)
\(x^2 + 2x \:-\: 3 = 0\)
\((x + 3)(x\:-\: 1) = 0\)
\(x_1 = -3\)
\(x_2 = 1\)
Langkah 2: \(h(x) = 1\)
\(x + 2 = 1\)
\(x = 1\:-\:2\)
\(x_3 = -1\)
Langkah 3: \(h(x) = 0\), \(f(x)\) dan \(g(x)\) harus positif
\(x + 2 = 0\)
\(x = -2\)
Substitusikan \(x = -2\) pada\(f(x)\) dan \(g(x)\)
\(f(-2) = -2 + 4 = 2\:\:\:\:\:\color{blue} \text{bernilai positif}\)
\(g(-2) = (-2)^2 + 3(-2) + 1\)
\(g(-2) = 4 \:-\: 6 + 1\)
\(g(-2) = -1\:\:\:\:\:\color{blue} \text{bernilai negatif}\)
Karena \(g(-2)\) bernilai negatif maka \(\xcancel{x = -2}\) bukan merupakan solusi
Langkah 4: \(h(x) = -1\), \(f(x)\) dan \(g(x)\) keduanya harus sama-sama genap atau sama-sama ganjil
\(x + 2 = -1\)
\(x = -1 \:-\: 2\)
\(x = -3\)
Substitusikan \(x = -3\) pada\(f(x)\) dan \(g(x)\)
\(f(-3) = -3 + 4 = 1\:\:\:\:\:\color{blue} \text{ganjil}\)
\(g(-3) = (-3)^2 + 3(-3) + 1\)
\(g(-3) = 9 \:-\: 9 + 1\)
\(g(-3) = 1\:\:\:\:\:\color{blue} \text{ganjil}\)
Karena \(f(x)\) dan \(g(x)\) sama-sama ganjil, maka \(x = -3\) merupakan solusi
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -3,\: -1, \:1\rbrace}\)
Tentukan himpunan penyelesaian persamaan \((x + 1)^{2x \:-\: 4} = (x + 1)^{x^2 + 3x \:-\: 10}\)
Langkah 1: \(f(x) = g(x)\)
\(2x \:-\: 4 = x^2 + 3x\:-\: 10\)
\(0 = x^2 + 3x \:-\: 2x -10 + 4\)
\(x^2 + x \:-\: 6 = 0\)
\((x + 3)(x \:-\: 2) = 0\)
\(x_1 = -3\)
\(x_2 = 2\)
Langkah 2: \(h(x) = 1\)
\(x + 1 = 1\)
\(x = 1-1\)
\(x_3 = 0\)
Langkah 3: \(h(x) = 0\), \(f(x)\) dan \(g(x)\) harus positif
\(x + 1 = 0\)
\(x = -1\)
Substitusikan \(x = -1\) pada\(f(x)\) dan \(g(x)\)
\(f(-1) = 2(-1)\:-\: 4 = -6\:\:\:\:\:\color{blue} \text{bernilai negatif}\)
\(g(-1) = (-1)^2 + 3(-1) \:-\:10\)
\(g(-1) = 1\:-\:3\:-\:10\)
\(g(-1) = -12\:\:\:\:\:\color{blue} \text{bernilai negatif}\)
Karena \(f(-1)\) dan \(g(-2)\) bernilai negatif maka \(\xcancel{x = -1}\) bukan merupakan solusi
Langkah 4: \(h(x) = -1\), \(f(x)\) dan \(g(x)\) keduanya harus sama-sama genap atau sama-sama ganjil
\(x + 1 = -1\)
\(x = -1 \:-\: 1\)
\(x = -2\)
Substitusikan \(x = -2\) pada\(f(x)\) dan \(g(x)\)
\(f(-2) = 2(-2) \:-\: 4 =-8\:\:\:\:\:\color{blue} \text{genap}\)
\(g(-2) = (-2)^2 + 3(-2) \:-\:10\)
\(g(-2) = 4 \:-\: 6\:-\:10\)
\(g(-2) = -12\:\:\:\:\:\color{blue} \text{genap}\)
Karena \(f(x)\) dan \(g(x)\) sama-sama genap, maka \(x = -2\) merupakan solusi
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -3,\: -2, \:0, \:2\rbrace}\)
Tentukan himpunan penyelesaian persamaan \((x\:-\: 6)^{x^2+1} = (x\:-\: 6)^{4x\:-\: 2}\)
Langkah 1: \(f(x) = g(x)\)
\(x^2 + 1= 4x\:-\: 2\)
\( x^2 \:-\: 4x +3=0\)
\((x\:-\: 3)(x \:-\: 1) = 0\)
\(x_1 = 3\)
\(x_2 = 1\)
Langkah 2: \(h(x) = 1\)
\(x \:-\: 6= 1\)
\(x = 1 + 6\)
\(x_3 = 7\)
Langkah 3: \(h(x) = 0\), \(f(x)\) dan \(g(x)\) harus positif
\(x \:-\: 6 = 0\)
\(x = 6\)
Substitusikan \(x = 6\) pada\(f(x)\) dan \(g(x)\)
\(f(6) = 6^2 + 1 = 37\:\:\:\:\:\color{blue} \text{bernilai positif}\)
\(g(6) = 4(6)\:-\:2\)
\(g(6) = 24\:-\: 2\)
\(g(6) = 22\:\:\:\:\:\color{blue} \text{bernilai positif}\)
Karena \(f(6)\) dan \(g(6)\) bernilai positif maka \(x = 6\) merupakan solusi
Langkah 4: \(h(x) = -1\), \(f(x)\) dan \(g(x)\) keduanya harus sama-sama genap atau sama-sama ganjil
\(x \:-\: 6 = -1\)
\(x = -1 +6\)
\(x = 5\)
Substitusikan \(x = 5\) pada\(f(x)\) dan \(g(x)\)
\(f(5) = 5^2 + 1 = 26\:\:\:\:\:\color{blue} \text{genap}\)
\(g(5) = 4(5)-2\)
\(g(5) = 20-2\)
\(g(5) = 18\:\:\:\:\:\color{blue} \text{genap}\)
Karena \(f(x)\) dan \(g(x)\) sama-sama genap, maka \(x = 5\) merupakan solusi
\(\bbox[yellow, 5px]{\text{HP} = \lbrace 1,\: 3, \:5, \:6, \:7\rbrace}\)
