Jika [tex] x_{1} [/tex] dan [tex] x_{2} [/tex] memenuhi persamaan [tex] 2^{4x-1} - 5. 2^{2x+1} = -32 [/tex], maka [tex] x_{1}+ x_{2} =[/tex] ... A. [tex] \frac{

jawab

2⁴ˣ. 2⁻¹ - 5. 2²ˣ.2  = - 32

1/2 (2²ˣ)² - 10 (2²ˣ)  + 32 = 0 ....kalikan 2

(2²ˣ)² - 20 (2²ˣ)  + 64 = 0 .....misal (2²ˣ) = a

a² -20 a + 64 = 0
a1 . a2 = 64

(2²ˣ¹)(2²ˣ²) = 64 = 2⁶
2x₁+ 2x₂ = 6
x1 + x2 = 3

[tex] {2}^{4x - 1} - 5. {2}^{2x + 1} = - 32 \\ {2}^{4x} . {2}^{ - 1} - 5. {2}^{2x} . {2}^{1} = - 32 \\ \\ \frac{1}{2} . {2}^{4x} - 10. {2}^{2x} = - 32 \\ \\ \frac{1}{2}( {2}^{2x} ) {}^{2} - 10. {2}^{2x} + 32 = 0 \\ \\ {({2}^{2x} )}^{2} - 20. {2}^{2x} + 64 = 0 \\ \\ misal \: a = {2}^{2x} \\ \\ {a}^{2} - 20a + 64 = 0 \\ (a - 16)(a - 4) = 0 \\ a = 16 \: \: atau \: \: a = 4 \\ \\ untuk \: a = 16 \\ {2}^{2x} = 16 \\ {2}^{2x} = {2}^{4} \\ 2x = 4 \\ x_{1} = 2 \\ \\ untuk \: a = 4 \\ {2}^{2x} = 4 \\ {x}^{2x} = {2}^{2} \\ x_{2} = 1[/tex]

[tex] x_{ 1} = 2 \\ x_{2} = 1 \\ \\ x_{1} + x_{2} = 2 + 1 \\ = 3[/tex]
Jawaban 3 (E)