Supposing that the terms of equation are arrangements, hence, you may use factorial formula such that:

`A_(x+4)^x = (x!)/((x+4-x)!) =gt A_(x+4)^x = (x!)/(4!)`

`A_(x+3)^(x+2) = ((x+3)!)/((x+3-x-2)!)=gtA_(x+3)^(x+2) = ((x+3)!)`

Hence, you may write the factorial form of the equation such that:

`(x!)/(4!) = (0.5)((x+3)!)`

You may write `((x+3)!)` in terms of `x!` such that:

`((x+3)!)= x!*(x+1)(x+2)(x+3)`

`(x!)/(1*2*3*4) = (1/2)*x!*(x+1)(x+2)(x+3)`

Reducing like terms yields:

`1/12 = (x+1)(x+2)(x+3) =gt 12(x+1)(x+2)(x+3) - 1 = 0`

`(12x^2 + 36x + 24)(x+3) - 1 = 0`

`12x^3 + 36x^2 + 36x^2 + 108x + 24x + 72 - 1 = 0`

`12x^3 + 72x^2 + 132x + 71 = 0`

Evaluating the roots of polynomial yields `x_1 = -0.961 ; x_2 = -2.955 ; x_3 = -2.084.`

**Notice that none of the found roots are natural numbers, hence, the equation has no solution.**