How do you solve $\sqrt[4]{2 x} - 13 = - 9$ $root4(2x)-13=-9$ ?

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Answer 1 · 2017-04-25T13:17:54

$x = 128$ $x=128$

$\sqrt[4]{2 x} - 13 = - 9$ $root4(2x)-13=-9$

Add $13$ $13$ to both sides

$root4(2x)cancel(-13)cancel(+13)=-9+13$

$root4(2x)=4$

Now we are going to raise both sides of the equation to the power of $4$ so we can get rid of that $"root" 4$

$(root4(2x))^4=(4)^4$

[Exponent cancels with root of equal power]

$2x=256$

Divide both sides by $2$

$(2x)/2=256/2$

$(cancel2x)/cancel2=256/2$

$x=128$

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