Penyelesaian:

Bentuk: \(g(x)^{f(x)} = h(x)^{f(x)}\) dengan \(f(x) = 5\), \(g(x) = 3x-2\), \(h(x) = x+6\)

Kasus 1: \(f(x) = 5 \neq 0\) (bukan variabel), maka langsung samakan basis:

\(g(x) = h(x)\)

\(3x – 2 = x + 6\)

\(2x = 8\)

\(\boxed{x = 4}\)

Verifikasi: \((3(4)-2)^5 = (4+6)^5 \Rightarrow 10^5 = 10^5\) ✓

Penyelesaian:

\(f(x) = 4\) (konstanta, bukan nol), maka:

\(x + 3 = 2x – 1\)

\(3 + 1 = 2x – x\)

\(\boxed{x = 4}\)

Catatan: Karena pangkat genap, kita juga perlu periksa \(g(x) = -h(x)\):

\(x + 3 = -(2x – 1)\)

\(x + 3 = -2x + 1\)

\(3x = -2\)

\(x = -\frac{2}{3}\)

Verifikasi \(x = -\frac{2}{3}\):

\((-\frac{2}{3}+3)^4 = (\frac{7}{3})^4\) dan \((2(-\frac{2}{3})-1)^4 = (-\frac{7}{3})^4 = (\frac{7}{3})^4\) ✓

\(\boxed{x = 4 \text{ atau } x = -\frac{2}{3}}\)

Penyelesaian:

Pangkat 7 (ganjil, konstanta ≠ 0), samakan basis:

\(x + 1 = 4x – 8\)

\(1 + 8 = 4x – x\)

\(9 = 3x\)

\(\boxed{x = 3}\)

Verifikasi: \((3+1)^7 = (4(3)-8)^7 \Rightarrow 4^7 = 4^7\) ✓

Penyelesaian:

Pangkat 3 (ganjil, konstanta ≠ 0):

\(2x = x + 5\)

\(\boxed{x = 5}\)

Verifikasi: \((10)^3 = (10)^3\) ✓

Penyelesaian:

Di sini \(f(x) = x+1\) adalah variabel, sehingga ada 2 kasus:

Kasus 1: \(g(x) = h(x)\)

\(x – 2 = 3x – 10\)

\(-2x = -8\)

\(x = 4\)

Kasus 2: \(f(x) = 0\)

\(x + 1 = 0 \Rightarrow x = -1\)

Cek: \(g(-1) = -1-2 = -3 \neq 0\) ✓ dan \(h(-1) = 3(-1)-10 = -13 \neq 0\) ✓

\(\boxed{x = 4 \text{ atau } x = -1}\)

Penyelesaian:

\(f(x) = x-3\), \(g(x) = x^2-4\), \(h(x) = x+2\)

Kasus 1: \(g(x) = h(x)\)

\(x^2 – 4 = x + 2\)

\(x^2 – x – 6 = 0\)

\((x-3)(x+2) = 0\)

\(x = 3\) atau \(x = -2\)

Verifikasi \(x=3\): \((9-4)^{0} = (5)^{0} \Rightarrow 1 = 1\) ✓

Verifikasi \(x=-2\): \((4-4)^{-5} = (0)^{-5}\) → Tidak terdefinisi ✗

Kasus 2: \(f(x) = 0\)

\(x – 3 = 0 \Rightarrow x = 3\)

Cek: \(g(3) = 5 \neq 0\) ✓, \(h(3) = 5 \neq 0\) ✓ (sudah termasuk)

\(\boxed{x = 3}\)

Penyelesaian:

\(f(x) = x^2-9\), \(g(x) = 2x+3\), \(h(x) = x+4\)

Kasus 1: \(g(x) = h(x)\)

\(2x+3 = x+4\)

\(x = 1\)

Kasus 2: \(f(x) = 0\)

\(x^2 – 9 = 0\)

\((x-3)(x+3) = 0\)

\(x = 3\) atau \(x = -3\)

Cek \(x=3\): \(g(3)=9 \neq 0\), \(h(3)=7 \neq 0\) ✓

Cek \(x=-3\): \(g(-3)=-3 \neq 0\), \(h(-3)=1 \neq 0\) ✓

\(\boxed{x = 1, \; x = 3, \; x = -3}\)

Penyelesaian:

Kasus 1: \(g(x) = h(x)\)

\(x^2 + 2x = 3x\)

\(x^2 – x = 0\)

\(x(x-1) = 0\)

\(x = 0\) atau \(x = 1\)

Cek \(x=0\): \(g(0) = 0\) → basis = 0, perlu hati-hati. \(0^{-1}\) tidak terdefinisi ✗

Cek \(x=1\): \(g(1) = 3\), \(h(1) = 3\), pangkat = 0 → \(3^0 = 3^0 = 1\) ✓

Kasus 2: \(f(x) = 0\)

\(x – 1 = 0 \Rightarrow x = 1\) (sudah termasuk)

Cek: \(g(1)=3 \neq 0\), \(h(1)=3 \neq 0\) ✓

\(\boxed{x = 1}\)

Penyelesaian:

Kasus 1: \(g(x) = h(x)\)

\(x + 5 = 3x + 1\)

\(4 = 2x\)

\(x = 2\)

Kasus 2: \(f(x) = 0\)

\(2x – 4 = 0 \Rightarrow x = 2\) (sudah termasuk)

Cek: \(g(2) = 7 \neq 0\), \(h(2) = 7 \neq 0\) ✓

\(\boxed{x = 2}\)

Penyelesaian:

Kasus 1: \(g(x) = h(x)\)

\(x^2 – x = 2x\)

\(x^2 – 3x = 0\)

\(x(x-3) = 0\)

\(x = 0\) atau \(x = 3\)

Cek \(x=0\): \(g(0) = 0\) dan \(h(0) = 0\) → \(0^3 = 0^3 = 0\) ✓ (karena pangkat \(0+3=3>0\))

Cek \(x=3\): \(g(3) = 6\), \(h(3) = 6\), pangkat = 6, \(6^6 = 6^6\) ✓

Kasus 2: \(f(x) = 0\)

\(x + 3 = 0 \Rightarrow x = -3\)

Cek: \(g(-3) = 9+3 = 12 \neq 0\) ✓, \(h(-3) = -6 \neq 0\) ✓

Verifikasi: \(12^0 = (-6)^0 \Rightarrow 1 = 1\) ✓

\(\boxed{x = 0, \; x = 3, \; x = -3}\)

Penyelesaian:

Faktorisasi: \(g(x) = (x-2)(x-3)\), \(h(x) = (x-1)(x-2)\), \(f(x) = (x-2)(x+2)\)

Kasus 1: \(g(x) = h(x)\)

\((x-2)(x-3) = (x-1)(x-2)\)

\(x^2 – 5x + 6 = x^2 – 3x + 2\)

\(-5x + 6 = -3x + 2\)

\(-2x = -4\)

\(x = 2\)

Cek \(x=2\): \(g(2) = 0\), \(h(2) = 0\), \(f(2) = 0\) → \(0^0\) tidak terdefinisi ✗

Kasus 2: \(f(x) = 0\)

\(x^2 – 4 = 0 \Rightarrow x = 2\) atau \(x = -2\)

\(x = 2\): sudah ditolak di atas ✗

\(x = -2\): \(g(-2) = 4+10+6 = 20 \neq 0\) ✓, \(h(-2) = 4+6+2 = 12 \neq 0\) ✓

Verifikasi: \(20^0 = 12^0 \Rightarrow 1 = 1\) ✓

\(\boxed{x = -2}\)

Penyelesaian:

\(g(x) = x^2+x-2 = (x+2)(x-1)\), \(h(x) = 2x+4 = 2(x+2)\), \(f(x) = x^2-x-6 = (x-3)(x+2)\)

Kasus 1: \(g(x) = h(x)\)

\((x+2)(x-1) = 2(x+2)\)

\((x+2)(x-1) – 2(x+2) = 0\)

\((x+2)(x-1-2) = 0\)

\((x+2)(x-3) = 0\)

\(x = -2\) atau \(x = 3\)

Cek \(x=-2\): \(g(-2)=0\), \(h(-2)=0\), \(f(-2)=0\) → \(0^0\) tidak terdefinisi ✗

Cek \(x=3\): \(g(3)=10\), \(h(3)=10\), \(f(3)=0\) → \(10^0 = 10^0 = 1\) ✓

Kasus 2: \(f(x) = 0\)

\((x-3)(x+2) = 0 \Rightarrow x = 3\) atau \(x = -2\)

\(x = 3\): sudah valid ✓

\(x = -2\): \(g(-2) = 0\) → tidak valid ✗

\(\boxed{x = 3}\)

Penyelesaian:

\(g(x) = (x-1)(x+1)\), \(h(x) = (x+1)(x+2)\), \(f(x) = (x+2)(x-1)\)

Kasus 1: \(g(x) = h(x)\)

\((x-1)(x+1) = (x+1)(x+2)\)

Jika \(x+1 \neq 0\), bagi kedua ruas dengan \((x+1)\):

\(x-1 = x+2\)

\(-1 = 2\) → Kontradiksi, tidak ada solusi dari sini.

Jika \(x+1 = 0\), yaitu \(x = -1\):

\(g(-1) = 0\), \(h(-1) = 0\), \(f(-1) = (1)(-2) = -2\)

\(0^{-2}\) tidak terdefinisi ✗

Kasus 2: \(f(x) = 0\)

\((x+2)(x-1) = 0 \Rightarrow x = -2\) atau \(x = 1\)

Cek \(x = -2\): \(g(-2) = 3 \neq 0\) ✓, \(h(-2) = (-1)(0) = 0\) ✗

Cek \(x = 1\): \(g(1) = 0\) ✗

\(\boxed{\text{Tidak ada solusi}}\)

Penyelesaian:

\(g(x) = 2x^2-x-1 = (2x+1)(x-1)\)

\(h(x) = x^2+2x+1 = (x+1)^2\)

\(f(x) = x^2-2x-3 = (x-3)(x+1)\)

Kasus 1: \(g(x) = h(x)\)

\((2x+1)(x-1) = (x+1)^2\)

\(2x^2 – x – 1 = x^2 + 2x + 1\)

\(x^2 – 3x – 2 = 0\)

Diskriminan: \(9 + 8 = 17\)

\(x = \frac{3 \pm \sqrt{17}}{2}\)

Kedua nilai ini valid (basis tidak nol saat disubstitusi).

Kasus 2: \(f(x) = 0\)

\((x-3)(x+1) = 0 \Rightarrow x = 3\) atau \(x = -1\)

Cek \(x=3\): \(g(3) = 14 \neq 0\) ✓, \(h(3) = 16 \neq 0\) ✓

Cek \(x=-1\): \(g(-1) = 2-(-1)-1 = 0\) ✗

\(\boxed{x = \frac{3+\sqrt{17}}{2}, \; x = \frac{3-\sqrt{17}}{2}, \; x = 3}\)

Penyelesaian:

\(g(x) = (x-1)(x-3)\), \(h(x) = (x-2)(x-4)\), \(f(x) = (x-1)(x-4)\)

Kasus 1: \(g(x) = h(x)\)

\(x^2 – 4x + 3 = x^2 – 6x + 8\)

\(2x = 5\)

\(x = \frac{5}{2}\)

Cek: \(g(\frac{5}{2}) = \frac{25}{4} – 10 + 3 = -\frac{3}{4}\), \(f(\frac{5}{2}) = (\frac{3}{2})(-\frac{3}{2}) = -\frac{9}{4}\)

Perlu \(g(x)>0\) untuk basis negatif dengan pangkat non-integer? Sebenarnya pangkat \(-\frac{9}{4}\) dari basis negatif tidak terdefinisi di bilangan real ✗

Kasus 2: \(f(x) = 0\)

\((x-1)(x-4) = 0 \Rightarrow x = 1\) atau \(x = 4\)

Cek \(x=1\): \(g(1) = 0\) → \(0^0\) tidak terdefinisi ✗

Cek \(x=4\): \(g(4) = 3 \neq 0\) ✓, \(h(4) = 0\) → \(0^0\) tidak terdefinisi ✗

Hmm, tidak ada solusi valid? Mari periksa ulang kasus 1 dengan \(x = \frac{5}{2}\) lebih hati-hati.

Sebenarnya dalam konteks persamaan eksponen sekolah, kita biasanya mensyaratkan basis positif. Maka kita perlu \(g(x) > 0\) dan \(h(x) > 0\).

\(x = \frac{5}{2}\): \(g(\frac{5}{2}) = \frac{25-20+12}{4} = -\frac{3}{4} < 0\) ✗

\(\boxed{\text{Tidak ada solusi real}}\)

Jumlah semua solusi = tidak ada (himpunan penyelesaian kosong)