tentukan nilai x yang memenuhi persamaan
9^2×1=(1/27)^×-2![]()
9^2×1=(1/27)^×-2
Penjelasan dengan langkah-langkah:
[tex] {9}^{2} \times 1 = ( \frac{1}{27} {)}^{x - 2} [/tex]
[tex] = {3}^{2} (2x - 1) = {3}^{ - 3} (x - 2) \\ [/tex]
[tex] = 4x - 2 = - 3x + 6[/tex]
[tex] = 4x + 3x = 6 + 2[/tex]
[tex] = 7x = 8[/tex]
[tex]x = \frac{8}{7} [/tex]
x = 4/7
Penjelasan dengan langkah-langkah:
[tex] {9}^{2x + 1} = { (\frac{1}{27} )}^{x - 2} [/tex]
[tex] {( {3}^{2}) }^{2x + 1} = {( {3}^{ - 3}) }^{ x - 2} [/tex]
[tex] {3}^{2(2x + 1)} = {3}^{ - 3( x - 2)} [/tex]
[tex] {3}^{4x + 2} = {3}^{ - 3x + 6} [/tex]
[tex]4x + 2 = - 3x + 6[/tex]
4x + 3x = 6 - 2
7x = 4
x = 4/7
[answer.2.content]
