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> | ||
| − | == Solution == | + | == Solution 1 == |
| − | <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 06:56, 18 July 2024
Contents
Problem
Given the three numbers 
with 
. Arranged in order of increasing magnitude, they are:
Solution 1
Seeing that we need to compare values with exponents, we think logarithms. Taking the logarithm base 
of each term, we obtain 
, , and 
. Because 
, 
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 
, then 
. However, the order of this set will be reversed again (back to the order of the original set) when we take the logarithm base a second time. After doing this operation, we find the values 
, , and , which correspond to , 
, and 
, respectively. Because 
, , and so, by the correspondence detailed above, 
, which yields us answer choice 
.
Solution 2
Because , taking to the th power will bring it closer to , thereby raising its value. Because we have established that 
, and 
is a monotonically decreasing function, we know that 
. However, because 
, 
, compared to , will be closer to (but still less than) . Thus, 
. Putting this all together, we see that 
, or , which is answer choice .
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
