How do you solve $\sqrt[4]{x - 7} = 2$ $root4(x-7)=2$ ?

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Answer 1 · 2016-05-13T22:38:04

$x = 23$ $x=23$

Given:

$\sqrt[4]{x - 7} = 2$ $root(4)(x-7) = 2$

Raise both sides to the $4$ $4$ th power to get:

$x - 7 = 16$ $x-7 = 16$

Add $7$ $7$ to both sides to find:

$x = 23$ $x = 23$

Check:

$\sqrt[4]{23 - 7} = \sqrt[4]{16} = \sqrt[4]{2^{4}} = 2$ $root(4)(23 - 7) = root(4)(16) = root(4)(2^4) = 2$

How do you solve 4√x−7=2root4(x-7)=2?