1/2 does not = 1/4 , yet (1/2)^1/2 = (1/4)^1/4 . how many pairs (a,b) there with a^a = b^b?

Question

justify your answer!

Answer 1

Let's consider the function f:(0, oo) --> R defined by f(x) = x^x. Then, f'(x) = x^x (1 + ln(x)). It follows that f'(x) <0 for x in (0, 1/e), f'(1/e) = 0 and f'(x) >0 for x in (1/e, oo). This implies f is stricty decreasing on (0, 1/e], has a global minimum at x* =1/e with f(x*) = 1/(e^(1/e)) and is strictly increasing on [1/e, oo). In addition, f is continuous, f(x) --> 1 as x --> 0+ and f(x) --> oo as x --> oo. We also have f(1) =1

So, for every real number y in (1/(e^(1/e)) , 1), the equatin x^x = y has exactly 2 solutions, one in (0, 1/e), the other in (1/e, 1), which is the same as to say that there are 2 distinct numbers x1 and x2. such that x1^x1 = x2^x^2 = y. Since f is strictly decreasing in (0, 1/e] and strictly increasing in [1/e, oo), it follows that to each y in (1/(e^(1/e), 1) there corresponds one, and only one pair of distinct numbers (x1, x2) such that x1^x1 = x2^x2. Since there are infinitely - actually, uncountably - many y's in 1/(e^(1/e), 1), it follows there are infinitely many (actually uncountably many) pairs (x1, x2) with x1 <> x2 such that x1^x1 = x2^x2.

Answer 2

There are infinitely many such pairs, and here's the general solution for all positive a,b
(scythian below then generalizes further for negative a,b):

Assume without loss of generality that b>a
Then take logs to base a:
a^a = b^b
a ln a = b ln b
a/b =ln b / ln a

This wasn't so helpful, so alternatively, write b =ka:
a^a = (ka)^(ka)
a ln a = ka ln (ka) = ka (ln k + ln a)
a ln a = ka ln k + ka ln a
0 = ka ln k + (k-1)a ln a
- k ln k = (k-1) ln a
- (k)/(k-1) ln k = ln a
Now exponentiate again:
k^ (-(k)/(k-1)) = a

So for each real-valued choice of k (excluding 0 or 1), that has precisely one solution in a,
so this generates infinitely many pairs if we take the domain of k as {Real} \ {0,1}

k^ (-(k)/(k-1)) = a allows us to generate a directly from k, or get k iteratively from a

Your example (a=1/4, b=1/2) has b/a = k = 2:
k^ (-(k)/(k-1)) = a
2^ (-(2)/(2-1)) = a
2^(-2) = a = 1/4
b = ka = 2* 1/4 = 1/2

So here's another one, the k = 3 solution:
k^ (-(k)/(k-1)) = a
3^ (-(3)/(3-1)) = a
3^ (-3/2) = a = 1/3√3 = 0.19245
b = ka = 3 * 1/3√3 = 1/√3 = 0.5773
Verify: a^a ?= b^b
0.19245^ 0.19245 ?= 0.5773^0.5773 = 0.72823

And here's another one, the k = 1.5 solution:
k^ (-(k)/(k-1)) = a
1.5^ (-1.5/0.5) = a
1.5^(-3) = a
(3/2)^(-3) = a
(2/3)^(3) = a
8/27 = a = 0.2963

b = ka = 1.5* 8/27 = 4/9 = 0.4444

Verify: a^a ?= b^b
(8/27)^(8/27) ?= (4/9)^(4/9)
0.6974 = 0.6974

etc.

Answer 3

The function a^a-b^b for positive a, b is zero where a = b, and where

a = bLog(b) / ProductLog(bLog(b))

which a curve running from (0,1) through (1/e, 1/e) and ending in (1,0). All the points that lie on this curve meets this equation. See link to a similiar Y!A question and answer to this:

For negative a, b, the expression a^a - b^b usually involves non-zero complex numbers.

Answer 4

(1/2)^1/2 = (2^ -1)^(1/2)
= (2 ^ -1/2)
= 0.7071 left equation:
(1/4)^(1/4)=0.7071 right equation


a^a=b^b
assume a=2 then b =4
2^2= 4^4
4 # 256

lets try again :assume a=2.2 b= 4.4
2.2^2.2 =5.6667
4.4^4.4 = 677.94

so a^a =b^b is true for fission with 1/x where x is any real number ..