Random equation

$2-x=log_{3}(10-3^{x})$

Rewrite the equation in exponential form

$3^{2-x} =10-3^{x}$

Now, $3^{2-x} =9 \cdot 3^{-x}$

$9 \cdot 3^{-x} = 10 - 3^{x}$

Multiply both sides by $3^{x}$

$9=10 \cdot 3^{x} - 3^{2x}$

Rearranging

$3^{2x} - 10 \cdot 3^{x} +9=0$

Factoring the left side

$(3^{x}-1)(3^{x}-9)=0$

Set each factor equal to $0$ and solve for x

$3^{x}-1=0$ or $3^{x} - 9=0$

$3^{x}=1$ or $3^{x}=9$

$x = 0$ or $x=2$

The sum of the solutions is:

$0+2= \boxed{2}$

Taking 3^x as t, the following quadratic equation is obtained : t²-10t+9 where t=1,9. This makes x=0,2 and hence the sum is 2.