Problem: If x≥0x \geq 0x≥0, then xxx=\sqrt{x \sqrt{x \sqrt{x}}}=xxx=
Answer Choices:
A. xxx \sqrt{x}xx
B. xx4x \sqrt[4]{x}x4x
C. x8\sqrt[8]{x}8x
D. x38\sqrt[8]{x^{3}}8x3
E. x78\sqrt[8]{x^{7}}8x7
Solution:
xxx=(x(x⋅x1/2)1/2)1/2=(x(x3/2)1/2)1/2=(x⋅x3/4)1/2=(x7/4)1/2=x7/8=x78.\begin{aligned} \sqrt{x \sqrt{x \sqrt{x}}} &= \left(x\left(x \cdot x^{1 / 2}\right)^{1 / 2}\right)^{1 / 2} = \left(x\left(x^{3 / 2}\right)^{1 / 2}\right)^{1 / 2} = \left(x \cdot x^{3 / 4}\right)^{1 / 2} \\ &= \left(x^{7 / 4}\right)^{1 / 2} = x^{7 / 8} = \sqrt[8]{x^{7}}. \end{aligned} xxx=(x(x⋅x1/2)1/2)1/2=(x(x3/2)1/2)1/2=(x⋅x3/4)1/2=(x7/4)1/2=x7/8=8x7.