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Difference between revisions of "1968 AHSME Problems/Problem 29"

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\text{(E) } z,x,y</math>
 
\text{(E) } z,x,y</math>
  
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== Solution ==
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== Solution 1 ==
  
 
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>.
 
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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== Solution 2 ==
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Because <math>0<x<1</math>, taking <math>x</math> to the <math>x</math>th power will bring it closer to <math>1</math>, thereby raising its value. Because we have established that <math>x<x^x</math>, and <math>f(n)=x^n</math> is a [[monotonic|monotonically decreasing]] function, we know that <math>x^x>x^{x^x}</math>. However, because <math>x^x<1</math>, <math>x^{x^x}</math>, compared to <math>x</math>, will be closer to (but still less than) <math>1</math>. Thus, <math>x^{x^x}>x</math>. Putting this all together, we see that <math>x<x^{x^x}<x^x</math>, or <math>x<z<y</math>, which is answer choice <math>\fbox{A}</math>.
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== See also ==
 
== See also ==

Latest revision as of 07:56, 18 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 1

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 $\fbox{A}$.


Solution 2

Because $0<x<1$, taking $x$ to the $x$th power will bring it closer to $1$, thereby raising its value. Because we have established that $x<x^x$, and $f(n)=x^n$ is a monotonically decreasing function, we know that $x^x>x^{x^x}$. However, because $x^x<1$, $x^{x^x}$, compared to $x$, will be closer to (but still less than) $1$. Thus, $x^{x^x}>x$. Putting this all together, we see that $x<x^{x^x}<x^x$, or $x<z<y$, which is answer choice $\fbox{A}$.


See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 28 		Followed by
Problem 30
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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