To evaluate the given equation `25^(10x+8)=(1/125)^(4-2x)` , we may apply `25=5^2` and `1/125=5^(-3)` . The equation becomes:

`(5^2)^(10x+8)=(5^(-3))^(4-2x)`

Apply Law of Exponents: `(x^n)^m = x^(n*m)` .

`5^(2*(10x+8))=5^((-3)*(4-2x))`

`5^(20x+16)=5^(-12+6x)`

Apply the theorem: If `b^x=b^y` then `x=y` , we get:

`20x+16=-12+6x`

Subtract `6x` from both sides of the equation.

`20x+16-6x=-12+6x-6x`

`14x+16=-12`

Subtract `16` from both sides of the equation.

`14x+16-16=-12-16`

`14x=-28`

Divide both sides by `14` .

`(14x)/14=(-28)/14`

`x=-2`

Checking: Plug-in `x=-2` on `25^(10x+8)=(1/125)^(4-2x)` .

`25^(10*(-2)+8)=?(1/125)^(4-2*(-2))`

`25^(-20+8)=?(1/125)^(4+4)`

`25^(-12)=?(1/125)^(8)`

`(5^2)^(-12)=?(5^(-3))^(8)`

`5^(2*(-12))=?5^((-3)*8)`

`5^(-24)=5^(-24) ` **TRUE**

Thus, there is no extraneous solution. The `x=-2` is the **real exact solution** of the equation `25^(10x+8)=(1/125)^(4-2x)` .