How do you solve $\sqrt[3]{3 x - 1} = 2$ $root 3(3x-1)=2$ ?

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Answer 1 · 2018-03-09T06:28:27

The solution is $x = 3$ $x=3$ .

Cube both sides of the equals sign, then solve for $x$ $x$ :

$\sqrt[3]{3 x - 1} = 2$ $root(3)(3x-1)=2$

$(root(color(red)cancelcolor(black)3)(3x-1))^color(red)cancelcolor(black)3=2^3$

$3x-1=2^3$

$3x-1=8$

$3x=9$

$x=3$

Answer 2 · 2018-03-09T06:31:00

$x=3$

$root(3)(3x-1)=2$
$(3x-1)=2^3$
$(3x-1)=8$
$3x=9$
$x=3$

How do you solve root 3(3x-1)=23√3x−1=2root 3(3x-1)=2?