Difference between revisions of "1985 AHSME Problems/Problem 15"

Revision as of 13:01, 5 July 2013

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

Substitue $b=9a$ into $a^b=b^a$ to get $a^{9a}=(9a)^a$ . Since $x^{yz}=(x^y)^z$ , we have $a^{9a}=(a^9)^a$ , and $(a^9)^a=(9a)^a$ . Taking the

$a\text{th}$ root of both sides gives $a^9=9a$ . Dividing by $a$ yields $a^8=9\implies a=\sqrt[8]{9}=\sqrt[4]{3}, \boxed{\text{E}}$ .

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.

@@ Line 11: / Line 11: @@
 ==See Also==
 {{AHSME box|year=1985|num-b=14|num-a=16}}
+{{MAA Notice}}

Page

Toolbox

Search

Difference between revisions of "1985 AHSME Problems/Problem 15"

Revision as of 13:01, 5 July 2013

Problem

Solution

See Also