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Difference between revisions of "1969 AHSME Problems/Problem 29"

(Created page with "== Problem == If <math>x=t^{1/(t-1)}</math> and <math>y=t^{t/(t-1)},t>0,t \ne 1</math>, a relation between <math>x</math> and <math>y</math> is: <math>\text{(A) } y^x=x^{1/y}...")
 
(Solution to Problem 29)
 
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
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<math>\fbox{C}</math>
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Plug in <math>t = 2</math> and test each expression (if the relation works, then both sides are equal).  After testing, options A, B, and D are out, but for option C, both sides are equal.  To check that C is a valid option, substitute the values of <math>x</math> and <math>y</math> and use exponent properties.
 +
<cmath>x^y = (t^{\frac{1}{t-1}})^{t^{\frac{t}{t-1}}} = t^{\frac{1}{t-1} \cdot t^{\frac{t}{t-1}}} </cmath>
 +
<cmath>y^x = (t^{\frac{t}{t-1}})^{t^{\frac{1}{t-1}}} = t^{\frac{t}{t-1} \cdot t^{\frac{1}{t-1}}}</cmath>
 +
In the second equation, the exponent can be rewritten as <math>\tfrac{1}{t-1} \cdot t \cdot t^{\frac{1}{t-1}}</math>.  Since <math>\tfrac{t-1}{t-1} = 1</math>, the exponent can be simplified as <math>\tfrac{1}{t-1} \cdot t^{\frac{t-1}{t-1}} \cdot t^{\frac{1}{t-1}} = \tfrac{1}{t-1} \cdot t^{\frac{t}{t-1}}</math>.  Since the base and exponent of both <math>x^y</math> and <math>y^x</math> are the same, we can confirm that <math>\boxed{\textbf{(C) } y^x = x^y}</math>.
  
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== See also ==
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== See Also ==
 
{{AHSME 35p box|year=1969|num-b=28|num-a=30}}   
 
{{AHSME 35p box|year=1969|num-b=28|num-a=30}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:49, 21 June 2018

Problem

If $x=t^{1/(t-1)}$ and $y=t^{t/(t-1)},t>0,t \ne 1$, a relation between $x$ and $y$ is:

$\text{(A) } y^x=x^{1/y}\quad \text{(B) } y^{1/x}=x^{y}\quad \text{(C) } y^x=x^y\quad \text{(D) } x^x=y^y\quad \text{(E) none of these}$

Solution

Plug in $t = 2$ and test each expression (if the relation works, then both sides are equal). After testing, options A, B, and D are out, but for option C, both sides are equal. To check that C is a valid option, substitute the values of $x$ and $y$ and use exponent properties. \[x^y = (t^{\frac{1}{t-1}})^{t^{\frac{t}{t-1}}} = t^{\frac{1}{t-1} \cdot t^{\frac{t}{t-1}}}\] \[y^x = (t^{\frac{t}{t-1}})^{t^{\frac{1}{t-1}}} = t^{\frac{t}{t-1} \cdot t^{\frac{1}{t-1}}}\] In the second equation, the exponent can be rewritten as $\tfrac{1}{t-1} \cdot t \cdot t^{\frac{1}{t-1}}$. Since $\tfrac{t-1}{t-1} = 1$, the exponent can be simplified as $\tfrac{1}{t-1} \cdot t^{\frac{t-1}{t-1}} \cdot t^{\frac{1}{t-1}} = \tfrac{1}{t-1} \cdot t^{\frac{t}{t-1}}$. Since the base and exponent of both $x^y$ and $y^x$ are the same, we can confirm that $\boxed{\textbf{(C) } y^x = x^y}$.

See Also

1969 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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