To evaluate the given equation `10^(3x-10)=(1/100)^(6x-1)` , we may apply `100=10^2` . The equation becomes:

`10^(3x-10)=(1/10^2)^(6x-1)`

Apply Law of Exponents: `1/x^n = x^(-n)` .

`10^(3x-10)=(10^(-2))^(6x-1)`

Note:` 1/100= 10^(-2)`

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

`10^(3x-10)=10^((-2)*(6x-1))`

`10^(3x-10)=10^(-12x+2)`

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

`3x-10=-12x+2`

Add `12x` on both sides of the equation.

`3x-10+12x=-12x+2+12x`

`15x-10=2`

Add `10` on both sides of the equation.

`15x-10+10=2+10`

`15x=12`

Divide both sides by `15` .

`(15x)/15=12/15`

`x=12/15`

Simplify.

`x=4/5`

Checking: Plug-in `x=4/5` on `10^(3x-10)=(1/100)^(6x-1)` .

`10^(3*(4/5)-10)=?(1/100)^(6*(4/5)-1)`

`10^(12/5-10)=?(1/100)^(24/5-1)`

`10^(12/5-50/5)=?(1/100)^(24/5-5/5)`

`10^((-38)/5)=?(1/100)^(19/5)`

`10^((-38)/5)=?(10^(-2))^(19/5)`

`10^((-38)/5)=?10^((-2)*19/5)`

`10^((-38)/5)=10^((-38)/5) ` **TRUE**

Thus, the `x=4/5` is the **real exact solution** of the equation `10^(3x-10)=(1/100)^(6x-1)` .

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