AMC 10 Step by Step

2008 AMC 10B Problem 3

P1-10~1 minPrint

Assume that xx is a positive real number. Which is equivalent to xx3\sqrt[3]{x\sqrt{x}} ?

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Key insight

Turn every root into a fraction exponent. x times the square root of x is x to the 3/2, and a cube root divides that exponent by 3.

Solution

Roots are easiest to handle as exponents, so rewrite everything that way. Work from the inside out.

1. A square root is the power 12\tfrac12: x=x1/2\sqrt{x} = x^{1/2}.
2. Multiplying powers of the same base adds the exponents:
x⋅x=x1⋅x1/2=x1+1/2=x3/2.x \cdot \sqrt{x} = x^{1} \cdot x^{1/2} = x^{1 + 1/2} = x^{3/2}.
3. A cube root is the power 13\tfrac13, and a power of a power multiplies the exponents:
x3/23=x32⋅13=x1/2.\sqrt[3]{x^{3/2}} = x^{\frac32 \cdot \frac13} = x^{1/2}.

The answer is (D) x1/2\boxed{\textbf{(D)}\ x^{1/2}}.

Why this works

Once every root is written as a power of xx, only two rules are needed: multiplying adds exponents, and taking a root divides the exponent. Handling the inside first and the outside last keeps the steps in the right order. Whenever roots are stacked inside roots, switch to exponents.

Alternative approach

Try a number that makes every root come out whole. With x=64x = 64: 64=8\sqrt{64} = 8, so xx=64⋅8=512x\sqrt{x} = 64 \cdot 8 = 512, and 5123=8\sqrt[3]{512} = 8 because 8⋅8⋅8=5128 \cdot 8 \cdot 8 = 512. Now check the choices with x=64x = 64: 641/2=864^{1/2} = 8, while 641/6=264^{1/6} = 2, 641/464^{1/4} and 643/864^{3/8} are not whole numbers, and 641=6464^{1} = 64. Only (D) gives 88.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Write every radical as a fractional power. Inside the cube root,
xx=x1⋅x1/2=x3/2.x\sqrt{x} = x^{1} \cdot x^{1/2} = x^{3/2}.
Taking the cube root multiplies the exponent by 13\tfrac{1}{3}:
x3/23=x(3/2)⋅(1/3)=x1/2.\sqrt[3]{x^{3/2}} = x^{(3/2)\cdot(1/3)} = x^{1/2}.

The answer is (D) x1/2\boxed{\textbf{(D)}\ x^{1/2}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.