Consider a function,

 

f (t) = t ^x

 

So

Here, f (5) = 5 ^x , f (4) = 4 ^x , f (3) = 3 ^x , f (2) = 2 ^x

 

Again, f ' (x) = x.t ^{x - 1}

 

The above function is continuous and differentiable throughout its domain.

 

Now, before I proceed, let me tell you that I will be using Lagrange's Mean value theorem to solve the problem.

 

Lagrange's mean value theorem for a function g (z) which is defined, continuous and differentiable througout the closed interval [ a, b ], there exists a point 'c' in the open interval of ( a,b ) where,

 

f ' (c) = [ f (b) - f (a) ] / (b - a)

 

Now,  5 ^x + 2 ^x = 4 ^x + 3 ^x

 

 5 ^x - 4 ^x = 3 ^x - 2 ^x 

 

 f (5) - f (4) = f (3) - f (2)

 

 [ f (5) - f (4) ] / (5 - 4) = [ f (3) - f (2) ] / (3 - 2)

 

 f ' (c) = f ' (d)       [ by Lagranges theorem, where 4 < c < 5  &  2 < d < 3 ]

 

  x. c ^{x - 1} = x . d ^{x - 1}

 

 x [ c ^{x - 1}  - d ^{x - 1} ] = 0

 

 x = 0 or c ^{x - 1} = d ^{x - 1}

 

But c lies in the interbal (4, 5) and d lies in the interval (2. 3)

 

For, c ^x-1 and d ^{x - 1} to be equal, x - 1 must be 0 i.e x = 1

 

So, the only 2 possible real solutions are 0,1

 

Hence, there are only 2 possible real solutions of the given equation and they are 0 and 1 only.

 

Cheers!!