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avatar+1088 

Let a and b be positive real numbers such that a^b = b^a and b=9a. Then a can be expressed in the form where m and n are positive integers, and n is as small as possible. Find m+n.

 Mar 14, 2020
 #1
avatar
0

We can substitute b = 9a to get a^(9a) = (9a)^a.  Then a = 9^(1/3), so m + n = 3 + 9 = 12.

 Mar 16, 2020
 #2
avatar+1088 
+7

Sorry, but your answer is wrong, maybe you made a mistake when you solved for a.

 Mar 16, 2020
 #3
avatar+30397 
+1

a^b = b^a.     b = 9a

 

a^(9a) = (9a)^a

 

a*log(a^9) = a*log(9a)

 

log(a^9) = log(9a)

 

a^9 = 9a

 

a^8 = 9

 

a = 9^(1/8)

 

m = 8, n = 9

 Mar 16, 2020
 #4
avatar+1088 
+8

Sorry, but 17 is also wrong.

 Mar 17, 2020
 #5
avatar+117 
+1

From Alan #3,

\(\displaystyle a=9^{1/8}=(3^{2})^{1/8}=3^{1/4}=\sqrt[4]{3},\\ m+n=7.\)

Tiggsy  Mar 17, 2020

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