Nilai x yang memenuhi persamaan eksponensial : [tex]4^{1-x} + 2^{3-x} - 12 = 0 [/tex] adlah ..............
Matematika
agusrihartonohartono
Pertanyaan
Nilai x yang memenuhi persamaan eksponensial :
[tex]4^{1-x} + 2^{3-x} - 12 = 0
[/tex]
adlah ..............
[tex]4^{1-x} + 2^{3-x} - 12 = 0
[/tex]
adlah ..............
2 Jawaban
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1. Jawaban herrysahar
بِسْمِ اللَّهِ الرَّحْمَنِ الرَّحِيمِ
4^(1-x) + 2^(3-x) - 12 = 0
4^1 . (2^2)^(-x) + 2^3 . 2^(-x) - 12 = 0
------------------------------------------------- ( x (2^2x) )
4 + 8(2^x) - 12(2^2x) = 0
misal 2^2x = a^2
4 + 8a - 12a^2 = 0
12a^2 - 8a - 4 = 0
3a^2 - 2a - 1 = 0
(3a +1)(a-1) = 0
a = -1/3 atau a = 1
(*) 2^x = -1/3
x = ^2log-1/3 ( TM )
(*) 2^x = 1
2^x = 2^0
x = 0
SEMOGA MEMBANTU... -
2. Jawaban Anonyme
EksPonenSiaL
Misal :
2^x = a
4^(1 - x) + 2^(3 - x) - 12 = 0
4/a² + 8/a - 12 = 0
(4 + 8a - 12a²)/a² = 0
12a² - 8a - 4 = 0 ... ( : 4 )
3a² - 2a - 1 = 0
(3a + 1)(a - 1) = 0
a = -1/3 tdk memenuhi
a = 1
2^x = 1 ---> x = 0
Nilai x yg memenuhi :
x = 0