Difference between revisions of "1985 AHSME Problems/Problem 15"
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==Problem== | ==Problem== | ||
| − | 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 | + | 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== | ||
| − | + | 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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==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}} | ||
Latest revision as of 20:35, 19 March 2024
Problem
If 
and 
are positive numbers such that 
and 
, then the value of is
![$\mathrm{(A)\ } 9 \qquad \mathrm{(B) \ }\frac{1}{9} \qquad \mathrm{(C) \ } \sqrt[9]{9} \qquad \mathrm{(D) \ } \sqrt[3]{9} \qquad \mathrm{(E) \ }\sqrt[4]{3}$](http://latex.artofproblemsolving.com/4/a/e/4aea90042a81db78f191c36dc2e706a2588a5f3e.png)
Solution
Substituting into 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*}](http://latex.artofproblemsolving.com/d/2/1/d21dc0bdf92356d7e1703825c93751b97a821abf.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
