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$\frac{\sqrt[4]{7w-5}}{2}=\sqrt[4]{w-2}$

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Simplify the product of two radical expressions having same variable

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Q.

\frac{\sqrt[4]{7w-5}}{2}=\sqrt[4]{w-2}

Isolate fourth root: First, let's isolate the fourth root on one side by multiplying both sides of the equation by $2$ . $\newline$ $\frac{\sqrt[4]{7w-5}}{2} = \sqrt[4]{w-2}$ $\newline$ $2 \times \frac{\sqrt[4]{7w-5}}{2} = 2 \times \sqrt[4]{w-2}$ $\newline$ $\sqrt[4]{7w-5} = 2 \times \sqrt[4]{w-2}$
Remove fourth root: Now, to remove the fourth root, we raise both sides of the equation to the fourth power. $\newline$ $(\sqrt[4]{7w-5})^4 = (2 \cdot \sqrt[4]{w-2})^4$ $\newline$ $(7w-5) = (2^4) \cdot (\sqrt[4]{w-2})^4$ $\newline$ $(7w-5) = 16 \cdot (w-2)$
Distribute $16$ : Next, we distribute the $16$ to the terms inside the parentheses on the right side of the equation. $7w - 5 = 16(w - 2)$ $7w - 5 = 16w - 32$
Move terms: Now, we will move all terms involving $w$ to one side and constant terms to the other side. $7w - 16w = -32 + 5$ $-9w = -27$
Divide by $-9$ : Finally, we divide both sides by $-9$ to solve for $w$ . $\frac{-9w}{-9} = \frac{-27}{-9}$ $w = 3$

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