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My attempt.

$\sqrt[4]{16y} = 2y$

so

$\sqrt[4]{y}+2y = 6$ after moving the constants.

I'm not too sure about this, but i think i can get rid of the radical by raising both sides of the equation to the 4th power?

$y+16y = 1,296$

After simplification I'm left with this.

$y = \frac{1,296}{17}$

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  • $\begingroup$ $$\sqrt[4]{16y} = 2\sqrt[4]y$$ $\endgroup$ – J. W. Tanner Nov 17 '19 at 21:10
  • $\begingroup$ It is not true that $(16y)^{\frac{1}{4}}=2y$. It should be $2y^{\frac{1}{4}}$. $\endgroup$ – IamWill Nov 17 '19 at 21:10
  • $\begingroup$ If you raise both sides of the eqation $\sqrt[4]y+2y=6$ to the fourth power you obtain $y+8y\sqrt[4]{y^3}+24y^2\sqrt[4]{y^2}+32y^3\sqrt[4]y+16y^4=1296$. Then again, identity $\sqrt[4]{16y}=2y$ is not a thing, so whatever. $\endgroup$ – Gae. S. Nov 17 '19 at 21:14
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Recognize $\sqrt[4]{16}=2$ and simplify the equation

$$3\sqrt[4]{y}= 6$$

Then solve

$$y = 16$$

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Let $x = \sqrt[4]{y}$. Then your equation is equivalent to $$x+2x-2=4$$ Thus: $$3x = 6 \Rightarrow x = 2.$$ Now, we know $x = 2 = \sqrt[4]{y}$,so it follows $y = 16$.

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