# The solution set of the equation( 1 - e^-x ) ( 25 - x^2 )ln( x - 2 )=0 is...please I need details I got ( 0, 2, 5, -5 )

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The equation given is : ( 1 - e^-x ) ( 25 - x^2 ) ln( x - 2 ) = 0

To find the solutions of this equation we need to equate each of the factors to 0.

( 1 - e^-x ) = 0

=> e^-x = 1

=> -x = 0

=> x = 0

( 25 - x^2 ) = 0

=> 25 = x^2

=> x = sqrt (25)

=> x = + 5 and x = -5

ln( x - 2 ) = 0

=> ln (x - 2) = ln 1

=> x - 2 = 1

=> x = 3

**The solution set is {-5 , 0, 3, 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}.**