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③ HP dari
[tex]9 {}^{x {}^{2} + x} = 27 {}^{x {}^{2} - 1 } \\ 3 {}^{2x {}^{2} + 2x } = 3 {}^{3x {}^{2} - 3} \\ 2 x {}^{2} + 2x - 3x {}^{2} + 3 = 0 \\ 2x(x + 1) - 3( x{}^{2} - 1) = 0 \\ 2x(x + 1) - 3(x - 1)(x + 1) = 0 \\ (x + 1)(2x -3 (x - 1)) = 0 \\ \\ (x + 1)( - x + 3) = 0 \\ maka[/tex]
x = -1 dan x = 3

Jawaban : Ⓑ


④ HP dari

[tex]7 {}^{x {}^{2} - 5x + 6 } = 8 {}^{x {}^{2} - 5x + 6} \\ ( \frac{7}{8} ) {}^{x {}^{2} - 5x + 6} = 1 \\ \\ x {}^{2} - 5x + 6 = 0 \\ (x - 2)(x - 3) = 0[/tex]

maka x = 2 dan x = 3
HP : {2,3}
(tdk ada di opsi jwbn)

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Mata pelajaran : Matematika
Kelas : XII
Kategori : Eksponen dan Logaritma
Kata kunci : eksponen, bilangan berpangkat

Kode : 12.2.6 [Kelas 12 Matematika BAB 6 - Eksponen dan Logaritma]

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