+0

+13
104
5
+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
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
+1088
+7

Mar 16, 2020
#3
+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
+1088
+8

Sorry, but 17 is also wrong.

Mar 17, 2020
#5
+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