25^(10x+8)=(1/125)^(4-2x) Solve the equation.

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) .