The solution set of the equation( 1 - e^-x ) ( 25 - x^2 )ln( x I got ( 0, 2, 5, -5 )

We'll cancel the 1st
factor:

1 - e^-x = 0

We'll use
the negative power rule and we'll re-write the equation:

1
- 1/e^x = 0

e^x - 1 = 0 => e^x =
1

We'll create matching bases both sides. For this reason,
w'ell write 1 = e^0.

e^x =
e^0

Since the bases are matching, we'll equate the
exponents:

x =0

We'll cancel
the 2nd factor:

25 - x^2 =
0

-x^2 = -25

x^2 =
25

x1 = sqrt25 => x1 =
5

x2 = -5

We'll cancel the 3rd
factor:

ln (x-2) = 0

We'll
take antilogarithm:

x - 2 =
e^0

x - 2 = 1

x =
3

The complete set of solutions is: {-5 ; 0 ;
3 ; 5}.