Question

nomor 2 pakai cara pakai cara pakai cara​

Answer 1

Verified answer

Jawaban:

x + 2y = -14

Penjelasan dengan langkah-langkah:

Diketahui:

[tex]\displaystyle\sf~ {4}^{y + 3x } = 64[/tex]

[tex]\displaystyle\sf~ {}^{x}log~ (x + 12) - 3 {}^{x}log ~ 4 = - 1 [/tex]

Ditanya:

x + 2y = ....?

Penyelesaian:

[tex]\begin{aligned}\displaystyle\sf~ {4}^{y + 3x} & = \displaystyle\sf~64 \\ \displaystyle\sf~ {\cancel{4}}^{y + 3x} & = \displaystyle\sf~ {\cancel{4}}^{3} \\ \displaystyle\sf~y + 3x& = \displaystyle\sf~3 \end{aligned}[/tex]

y + 3x = 3 .....(i)

[tex]\begin{aligned}\displaystyle\sf~ {}^{x}log~(x + 12) - 3 {}^{x}log ~4 & = \displaystyle\sf~ - 1 \\ \displaystyle\sf~ {}^{x}log~(x + 12) - {}^{x}log~ {4}^{3} & = \displaystyle\sf~ - 1. {}^{x}log ~x \\ \displaystyle\sf~ {}^{x}log~(x + 12) - {}^{x}log ~64 & = \displaystyle\sf~ {}^{x} log ~ {x}^{ - 1} \\ \\ \displaystyle\sf~ \cancel{{}^{x}log} ~ \frac{x + 12}{64} & = \displaystyle\sf~ \cancel{{}^{x}log} ~ \frac{1}{x} \\ \\ \displaystyle\sf~ \frac{x + 12}{64} & = \displaystyle\sf~ \frac{1}{x} \\ \\ \displaystyle\sf~x(x + 12)& = \displaystyle\sf~1.64 \\ \displaystyle\sf~ {x}^{2} + 12x& = \displaystyle\sf~64 \\ \displaystyle\sf~ {x}^{2} + 12x - 64 & = \displaystyle\sf~0 \\ \displaystyle\sf~(x + 16)(x - 4)& = \displaystyle\sf~0 \\ \displaystyle\sf~ x = - 16~atau~x = 4 \end{aligned}[/tex]

uji syarat : Basis > 0

untuk x = -16

-16 < 0 (tidak memenuhi)

untuk x = 4

4 > 0 (memenuhi)

Nilai x yang memenuhi adalah 4

Subtitusi nilai x ke persamaan (i)

y + 3x = 3

y + 3(4) = 3

y + 12 = 3

y = 3 - 12

y = -9

maka,

x + 2y

= 4 + 2(-9)

= 4 - 18

= -14

Opsi D