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`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)` . 

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