Question

Tentukan nilai x yang memenuhi
[tex]( \frac{1}{4}) ^{x - 1} = \sqrt[3]{2 {}^{3x + 1} } [/tex]
Thank you...​

Answer 1

[tex]{( \frac{1}{4} ) }^{x - 1} = \sqrt[3]{ {2}^{3x + 1} } [/tex]

[tex] {4}^{ - 1(x - 1)} = {2}^{ \frac{1}{3} (3x + 1)} [/tex]

[tex] {4}^{ - (x - 1)} = {2}^{ \frac{1}{3}(3x + 1) } [/tex]

[tex] {4}^{ - x + 1} = {2}^{ \frac{3x + 1}{3} } [/tex]

[tex] {2}^{2( - x + 1)} = {2}^{ \frac{3x + 1}{3} } [/tex]

[tex] {2}^{ - 2x + 2} = {2}^{ \frac{3x + 1}{3} } [/tex]

[tex] - 2x + 2 = \frac{3x + 1}{ 3 } [/tex]

[tex]3( - 2x + 2) = 3· \frac{3x + 1}{3} [/tex]

[tex] - 6x + 6 = 3x + 1[/tex]

[tex] - 6x +6 - 6= 3x + 1 - 6[/tex]

[tex] - 6x = 3x - 5[/tex]

[tex] - 6x - 3x = 3x - 5- 3x[/tex]

[tex] - 9x = - 5[/tex]

[tex] \frac{ - 9x}{ - 9} = \frac{ - 5}{ - 9} [/tex]

[tex]x = \frac{5}{9} [/tex]