if a^b =b^a and a not equal to b, find a and b
Answers
Answer:
a is not equal to b
a=4 and b=2
4²=2⁴
Answer:
a=2 and b=4 these are only integer solutions to the equation
Step-by-step explanation:
Try this: Substitute
t = b/a
into your equation. If a and b are integers, or if a and b are
rational numbers, then t will be rational. Substituting a*t for b, we get
t a a
(a ) = (at) .
Now, taking the (positive real) a'th root of both sides of the
equation, we get
t
a = at
and therefore
t-1
a = t
and
1/(t-1)
a = t .
In fact, we can get a parametrized form for all rational solutions
from this equation. Let
n = 1/(t-1),
so that
a = (1 + 1/n)^n
b = (1 + 1/n)^(n+1).
These are rational solutions for all (nonzero) integers n. It takes a
little more work to show that these (along with a=b) are *all*
rational solutions. And that work begins with proving the lemma:
If r/s and c/d are rational numbers in lowest terms, and (r/s)^(c/d)
is also rational, then r and s are both d'th powers of integers.
But you wanted integer solutions. And I'm done with my tangents. The
first step is to prove that t is an integer. Then all that's left is
a simple induction argument.
Of course, t is not an integer if b=2 and a=4. But if we permit
ourselves to switch a with b (note that this doesn't change the
equation) so that
a <= b
then we can prove that t will be an integer. You see, if a <= b, then
a b
a divides a .
But a^b = b^a, which means that
a a
a divides b ,
and this is enough to imply that a divides b (think about the previous
statement in terms of the prime factorizations of a and b), and
therefore t=b/a is an integer. And if a and b are positive, then t is
also positive.
Now t=1 gives the solutions with a=b. And if t=2, then
t-1
a = t
gives a = 2 (and therefore b = 4). And a=1 quickly implies b=1. But
if a >= 2, then
t-1 t-1
a >= 2 ,
and we can prove by induction that
t-1
2 > t
for all integers t >= 3. This is easily verified for t=3. Now
suppose that
t-2
2 > t-1
and then we have
t-1 t-2
2 = 2 * 2 > 2(t-1) = t + (t-2) > t.
Therefore, there are no solutions with t > 2, and so we have found all
of the integer solutions, namely (k,k), (2,4), and (4,2).