Question

diketahui 6 pangkat 4x+16 =6 pangkat y
dan 9 pangakat (2x+12) =3 pangkat 6y
berapakah nilai x+y?​

Answer 1

Jawab: [tex]-\frac{10}{5}[/tex]

Penjelasan dengan langkah-langkah:

diketahui

[tex]6^{(4x+16)}=6^y = > 4x+16 = y\\\\9^{(2x+12)}=3^{6y}\\3^{2(2x+12)}=3^{6y} = > 2(2x+12) = 6y[/tex]

• mencari nilai x

[tex]2(2x + 12) = 6(4x + 16)\\4x + 24 = 24x + 96\\4x - 24x = 96 - 24\\-20x = 72\\x = -\frac{72}{20}\\x = -\frac{18}{5}[/tex]

• mencari nilai y

[tex]4x + 16 = y\\y = 4(-\frac{18}{5}) + 16\\y = \frac{80-72}{5}\\y = \frac{8}{5}[/tex]

maka

[tex]x + y = -\frac{18}{5} + \frac{8}{5} = -\frac{10}{5}[/tex]