# 2003 AMC 10A Problems/Problem 9

## Problem

Simplify $\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}$. $\mathrm{(A) \ } \sqrt{x}\qquad \mathrm{(B) \ } \sqrt{x^{2}}\qquad \mathrm{(C) \ } \sqrt{x^{2}}\qquad \mathrm{(D) \ } \sqrt{x}\qquad \mathrm{(E) \ } \sqrt{x^{80}}$

### Solution 1 $\sqrt{x\sqrt{x}}=\sqrt{(\sqrt{x})^{2}\cdot\sqrt{x}}=\sqrt{(\sqrt{x})^{3}}=\sqrt{x}$.

Therefore: $\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}=\sqrt{x\sqrt{x\sqrt{x}}}=\sqrt{x\sqrt{x}}= \sqrt{x} \Rightarrow \boxed{\mathrm{(A)}\ \sqrt{x}}$

### Solution 2

We know that $\sqrt{x\sqrt{x}} = \sqrt{x\cdot(x^\frac{1}{2})} = \sqrt{x^\frac{3}{2}} = x^\frac{1}{2}$ We plug this into $\sqrt{x\sqrt{x\sqrt{x\sqrt{x}}}}$ and get $\sqrt{x\sqrt{x\cdot(x^\frac{1}{2})}}$. Again, we substitute and get $\sqrt{x\cdot(x^\frac{1}{2})}$. We substitute one more time and get $x^\frac{1}{2} = \boxed{(A) \sqrt{x}}$.

### Solution 3 (Lame??)

WLOG, plug in some random square number like $9$ or $4$, and the output you get after simplifying the expression will always be the square root of the number, so the answer is just $\boxed{(A) \sqrt{x}}$.

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