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Difference between revisions of "1985 AHSME Problems/Problem 15"
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==Problem==
 
==Problem==
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If <math> a </math> and <math> b </math> are positive numbers such that <math> a^b=b^a </math> and <math> b=9a </math>, then the value of <math> a </math> is:
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If <math>a</math> and <math>b</math> are positive numbers such that <math>a^b = b^a</math> and <math>b = 9a</math>, then the value of <math>a</math> is
  
 
<math> \mathrm{(A)\ } 9 \qquad \mathrm{(B) \ }\frac{1}{9} \qquad \mathrm{(C) \  } \sqrt[9]{9} \qquad \mathrm{(D) \  } \sqrt[3]{9} \qquad \mathrm{(E) \  }\sqrt[4]{3} </math>
 
<math> \mathrm{(A)\ } 9 \qquad \mathrm{(B) \ }\frac{1}{9} \qquad \mathrm{(C) \  } \sqrt[9]{9} \qquad \mathrm{(D) \  } \sqrt[3]{9} \qquad \mathrm{(E) \  }\sqrt[4]{3} </math>
  
 
==Solution==
 
==Solution==
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Substitue <math> b=9a </math> into <math> a^b=b^a </math> to get <math> a^{9a}=(9a)^a </math>. Since <math> x^{yz}=(x^y)^z </math>, we have <math> a^{9a}=(a^9)^a </math>, and <math> (a^9)^a=(9a)^a </math>. Taking the  
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Substituting <math>b = 9a</math> into <math>a^b = b^a</math> gives <cmath>\begin{align*}a^{9a} = \left(9a\right)^a &\iff \left(a^9\right)^a = \left(9a\right)^a \qquad \text{(using the identity } \left(x^y\right)^z = x^{yz}\text{ for } x > 0\text{)} \\ &\iff a^9 = 9a \qquad \text{(taking the } a\text{th root of both sides, as } a > 0\text{)} \\ &\iff a^8 = 9 \qquad \text{(as } a \neq 0\text{)} \\ &\iff a^4 = 3 \qquad \text{(taking square roots and noting that } a^4 \geq 0\text{}) \\ &\iff a = \boxed{\text{(E)} \ \sqrt[4]{3}} \qquad \text{(again as } a > 0\text{)}.\end{align*}</cmath>
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<math> a\text{th} </math> root of both sides gives <math> a^9=9a </math>. Dividing by <math> a </math> yields <math> a^8=9\implies a=\sqrt[8]{9}=\sqrt[4]{3}, \boxed{\text{E}} </math>.
 
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=14|num-a=16}}
 
{{AHSME box|year=1985|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:35, 19 March 2024

Problem

If $a$ and $b$ are positive numbers such that $a^b = b^a$ and $b = 9a$, then the value of $a$ is

$\mathrm{(A)\ } 9 \qquad \mathrm{(B) \ }\frac{1}{9} \qquad \mathrm{(C) \ } \sqrt[9]{9} \qquad \mathrm{(D) \ } \sqrt[3]{9} \qquad \mathrm{(E) \ }\sqrt[4]{3}$

Solution

Substituting $b = 9a$ into $a^b = b^a$ gives \begin{align*}a^{9a} = \left(9a\right)^a &\iff \left(a^9\right)^a = \left(9a\right)^a \qquad \text{(using the identity } \left(x^y\right)^z = x^{yz}\text{ for } x > 0\text{)} \\ &\iff a^9 = 9a \qquad \text{(taking the } a\text{th root of both sides, as } a > 0\text{)} \\ &\iff a^8 = 9 \qquad \text{(as } a \neq 0\text{)} \\ &\iff a^4 = 3 \qquad \text{(taking square roots and noting that } a^4 \geq 0\text{}) \\ &\iff a = \boxed{\text{(E)} \ \sqrt[4]{3}} \qquad \text{(again as } a > 0\text{)}.\end{align*}

See Also

1985 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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
All AHSME Problems and Solutions

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