How do you solve $\sqrt[4]{3 t} - 2 = 0$ $root4(3t)-2=0$ ?

Question

How do you solve $\sqrt[4]{3 t} - 2 = 0$ $root4(3t)-2=0$ ?

1 Answer

Impact of this question

1579 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-26T13:21:58

$5 \frac{1}{3}$ $5 1/3$

Explanation:

$\sqrt[4]{3 t} - 2 = 0$ $root(4)(3t) -2 = 0$

$\sqrt[4]{3 t} = 2$ $root(4)(3t) = 2$

Put to the power of four on both sides
(cause $\sqrt[4]{3 t} = 3 t^{\frac{1}{4}}$ $root(4)(3t) = 3t^(1/4)$ and to the power of four $3 t^{(\frac{1}{4}) \cdot 4} = 3 t$ $3t^((1/4) *4 )=3t$ )

$3 t = 2^{4}$ $3t = 2^4$

$3 t = 16$ $3t = 16$

$t = \frac{16}{3} = 5 \frac{1}{3}$ $t= 16/3 = 5 1/3$

Science

Math

Humanities

... and beyond

How do you solve $\sqrt[4]{3 t} - 2 = 0$ $root4(3t)-2=0$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 4√3t−2=0root4(3t)-2=0?

1 Answer

Explanation:

Related questions

Impact of this question

How do you solve $\sqrt[4]{3 t} - 2 = 0$ $root4(3t)-2=0$ ?