D.
[tex] {5}^{ {x}^{2} - 3x - 4 } = {25}^{x + 1} [/tex]
[tex] {5}^{ {x}^{2} - 3x - 4 } = {5}^{2(x + 1)} [/tex]
[tex] {5}^{ {x}^{2} - 3x - 4 } = {5}^{2x + 2} [/tex]
sehingga :
[tex] {x}^{2} - 3x - 4 = 2x + 2 \\ {x}^{2} - 3x - 2x - 4 - 2 = 0 \\ {x}^{2} - 5x - 6= 0 \\ (x - 6)(x + 1) = 0 \\ x1 = 6 \: \: dan \: \: x2 = - 1 [/tex]
jadi, Hp = { –1, 6 }
E. silahkan di coba sendiri :')
Penjelasan dengan langkah-langkah:
Bagian d
[tex] \sf {5}^{ {x}^{2} - 3x - 4} = {25}^{x + 1} \\ \sf \: \: \: {5}^{ {x}^{2} - 3x - 4 } = {( {5}^{2}) }^{x + 1} \\ \sf \: {5}^{ {x}^{2} - 3x - 4 } = {5}^{2(x + 1)} \\ \sf {5}^{ {x}^{2} - 3x - 4 } = {5 }^{2x + 2} \: \\ \sf {x}^{2} - 3x - 4 = 2x + 2 \: \: \: \: \: \: \\ \sf {x}^{2} - 3x - 2x - 4 - 2 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf {x}^{2} - 5x - 6 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf (x + 1)(x - 6) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: [/tex]
-
x + 1 = 0
x = -1
-
x - 6 = 0
x = 6
-
Hp = -1 dan 6
-----------------
Bagian e
[tex] \sf {8}^{x + 3} = \sqrt{ {4}^{2x - 1} } \\ \sf {8}^{x + 3} = {4}^{ \frac{2x - 1}{2} } \: \: \: \: \: \\ \sf {( {2}^{3}) }^{x + 3} = {( {2}^{2}) }^{ \frac{2x - 1}{2} } \: \: \: \: \: \\ \sf {2}^{3(x + 3)} = {2}^{(2 \times \frac{2x - 1 }{2}) } \: \: \\ \sf {2}^{3x + 9} = {2}^{2x - 1} \: \: \: \: \: \: \\ \sf 3x + 9 = 2x - 1 \: \: \: \: \: \: \\ \sf 3x - 2x = - 1 - 9 \: \: \: \: \: \: \: \\ \sf x = - 10 \: [/tex]
[answer.2.content]
