Set a=2^x and b=3^x.
Then the equation becomes
a + b  a^2 + ab  b^2  1 = 0
a^2  a(b+1) + b^2  b + 1 = 0
This is quadratic in a and to stay in the reals the discriminant must be nonnegative.
disc = (b + 1)^2  4( b^2  b + 1) >= 0
Simplifying gives 3(b  1)^2 >= 0 which is satisfied only when b=1.
Then 1 = 3^x and x = 0 is the only solution.

Posted by xdog
on 20210605 12:09:08 