1968 AHSME Problems/Problem 29: Difference between revisions

Revision as of 20:42, 17 July 2024

Problem

Given the three numbers $x,y=x^x,z=x^{x^x}$ with $.9<x<1.0$ . Arranged in order of increasing magnitude, they are:

$\text{(A) } x,z,y\quad \text{(B) } x,y,z\quad \text{(C) } y,x,z\quad \text{(D) } y,z,x\quad \text{(E) } z,x,y$

Solution

Seeing that we need to compare values with exponents, we think logarithms. Taking the logarithm base $x$ of each term, we obtain $1$ , $x$ , and $x^x$ . Because $0<x<1$ , $f(n)=\log_x(n)$ is monotonically decreasing, so the order of terms by magnitude in our new set of numbers will be reversed compared to the original set (i.e. if $a<b<c$ , then $\log_x(a)>\log_x(b)>\log_x(c))$ . However, the order of this set will be reversed again (back to the order of the original set) when we take the logarithm base $x$ a second time. After doing this operation, we find the values $0$ , $1$ , and $x$ , which correspond to $x$ , $y$ , and $z$ , respectively. Because $0.9<x<1$ , $0<x<1$ , and so, by the correspondence detailed above, $x<z<y$ , which yields us answer choice $\fbox{A}$ .

1968 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 28	Followed by Problem 30
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All AHSME Problems and Solutions

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 == Solution ==
-<math>\fbox{A}</math>
+Seeing that we need to compare values with exponents, we think [[logarithms]]. Taking the logarithm base <math>x</math> of each term, we obtain <math>1</math>, <math>x</math>, and <math>x^x</math>. Because <math>0<x<1</math>, <math>f(n)=\log_x(n)</math> is [[monotonic|monotonically decreasing]], so the order of terms by magnitude in our new set of numbers will be reversed compared to the original set (i.e. if <math>a<b<c</math>, then <math>\log_x(a)>\log_x(b)>\log_x(c))</math>. However, the order of this set will be reversed again (back to the order of the original set) when we take the logarithm base <math>x</math> a second time. After doing this operation, we find the values <math>0</math>, <math>1</math>, and <math>x</math>, which correspond to <math>x</math>, <math>y</math>, and <math>z</math>, respectively. Because <math>0.9<x<1</math>, <math>0<x<1</math>, and so, by the correspondence detailed above, <math>x<z<y</math>, which yields us answer choice <math>\fbox{A}</math>.
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1968 AHSME Problems/Problem 29: Difference between revisions

Revision as of 20:42, 17 July 2024

Problem

Solution

See also