Difference between revisions of "1991 AHSME Problems/Problem 6"

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== Problem ==
 
== Problem ==
 
 
If <math>x\geq 0</math>, then <math>\sqrt{x\sqrt{x\sqrt{x}}}=</math>
 
If <math>x\geq 0</math>, then <math>\sqrt{x\sqrt{x\sqrt{x}}}=</math>
  
(A) <math>x\sqrt{x}</math> (B) <math>x \sqrt[4]{x}</math> (C) <math>\sqrt[8]{x}</math> (D) <math>\sqrt[8]{x^3}</math> (E) <math>\sqrt[8]{x^7}</math>
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<math>\textbf{(A) } x\sqrt{x}\qquad
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\textbf{(B) } x\sqrt[4]{x}\qquad
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\textbf{(C) } \sqrt[8]{x}\qquad
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\textbf{(D) } \sqrt[8]{x^3}\qquad
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\textbf{(E) } \sqrt[8]{x^7}</math>
  
 
== Solution ==
 
== Solution ==
<math>\fbox{}</math>
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Recall that square roots are one-half powers, namely <math>\sqrt y=y^{\frac12}</math> for all <math>y\geq0.</math>
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We have
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<cmath>\begin{align*}
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\sqrt{x\sqrt{x\sqrt{x}}} &= \sqrt{x\sqrt{x\cdot x^{\frac12}}} \\
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&= \sqrt{x\sqrt{x^{\frac32}}} \\
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&= \sqrt{x\cdot\left(x^{\frac32}\right)^{\frac12}} \\
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&= \sqrt{x^{\frac74}} \\
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&= \left(x^{\frac74}\right)^{\frac12} \\
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&= x^{\frac78} \\
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&= \boxed{\textbf{(E) } \sqrt[8]{x^7}}.
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\end{align*}</cmath>
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~Hapaxoromenon (Solution)
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~MRENTHUSIASM (Reformatting)
  
 
== See also ==
 
== See also ==

Latest revision as of 17:13, 5 September 2021

Problem

If $x\geq 0$, then $\sqrt{x\sqrt{x\sqrt{x}}}=$

$\textbf{(A) } x\sqrt{x}\qquad \textbf{(B) } x\sqrt[4]{x}\qquad \textbf{(C) } \sqrt[8]{x}\qquad \textbf{(D) } \sqrt[8]{x^3}\qquad \textbf{(E) } \sqrt[8]{x^7}$

Solution

Recall that square roots are one-half powers, namely $\sqrt y=y^{\frac12}$ for all $y\geq0.$

We have \begin{align*} \sqrt{x\sqrt{x\sqrt{x}}} &= \sqrt{x\sqrt{x\cdot x^{\frac12}}} \\ &= \sqrt{x\sqrt{x^{\frac32}}} \\ &= \sqrt{x\cdot\left(x^{\frac32}\right)^{\frac12}} \\ &= \sqrt{x^{\frac74}} \\ &= \left(x^{\frac74}\right)^{\frac12} \\ &= x^{\frac78} \\ &= \boxed{\textbf{(E) } \sqrt[8]{x^7}}. \end{align*} ~Hapaxoromenon (Solution)

~MRENTHUSIASM (Reformatting)

See also

1991 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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