Determine the value of x so that the numbers a1 = In(x + 2). a2 = In(x + 3). a3 = In(4x + 8) form three consecutive members of an arithmetic sequence.
Select one:
a. Such x does not exist.
b. x = -+1
c. There are two possible solutions.
d. x = -1
e. None of the remaining possibilities is correct.

Step-by-step explanation:

hello :

the numbers a1 , a2 ,a3

form three consecutive members of an arithmetic sequence means : 2a2 = a1+,a3

2 In(x + 3) = In(x + 2)+In(4x + 8)(*)

now solve this equation use this note :

ln a^n = alna and : lna+lnb =lnab lna =lnb means : a=b ...a , b positif numbers

(*) equivaut : In(x + 3)² = In(x + 2)(4x + 8) and : x + 2 ,x + 3 , 4x + 8

are positifs numbers

(x + 3)² = (x + 2)(4x + 8)

x²+6x+9 = 4x²+16x+16

3x²+10x+7 =0

delta =10²-4(3)(7) 100 -84 = 4²

x1 = (-10-4)/6 = -7/3 reused

x 2 = (-10+4)/6 = -1 reused

conclusion : a. Such x does not exist.

$9.41

step-by-step explanation: