can this equation be solved by algebra?

Question

a^x = x^b
find all possible x.
I can't find anyway to solve this equation,
is this equation not algebraically independent?
how can I find the result of this equation without a calculator?

a and b are constants.

Answer 1

To solve an equation of this type explicitly, you will need the Lambert W function. If you have it available, then the solution may be found as follows:

a^x = x^b
x ln a = b ln x
1/b ln a = 1/x ln x
-1/b ln a = 1/x ln (1/x)
-1/b ln a = e^(ln (1/x)) ln (1/x)
ln (1/x) = W(-1/b ln a)
1/x = e^(W(-1/b ln a))
x = e^(-W(-1/b ln a))

This can be simplified further:

x = e^(-W(-1/b ln a))
x = 1/e^(W(-1/b ln a))
x = W(-1/b ln a)/(W(-1/b ln a) e^(W(-1/b ln a)))
x = W(-1/b ln a)/(-1/b ln a)
x = -b W(-1/b ln a)/ln a

And we are done.

ETA: One note -- the Lambert W function is multivalued on the interval [-1/e, 0), so if you must find all solutions, you will have to take into account both values of the Lambert W function on that interval

Answer 2

a = x
x = b
a = b

answer is all reals?

Answer 3

im not sure you can find it since you dont know exactly wat the exponent is