Help me solve this please? ln(x+1) - ln(3-x) = ln(2-x)

Solve ln(x+1) - ln(3-x) = ln(2-x)

thank you!

ln(x+1) - ln(3-x) = ln(2-x)
ln[(x + 1)/((3 - x)(2 - x))] = 0
6 - 5x + x^2 = x + 1
x^2 - 6x + 5 = 0
(x - 1)(x - 5) = 0
x = 1 , 5 => reject x = 5 (not in domain)
x = 1

you are welcome.

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ln (x + 1) - ln (3 - x) = ln (2 - x) Use your rules or lows of logs in dealing with ln

ln ((x+1) /(3 -x )) = ln (2 - x) This states that

(x + 1) /(3 - x) = (2 - x) Multiply each side by (3 -x)

(x + 1) = (2 - x)(3 - x) Multiply it out
(x + 1) = x^2 - 5x + 6 rearrange into a standard quadratic equation

x^2 - 6x + 5 = 0
(x - 5)(x - 1) = 0
x = 5 cannot be used because (2 - 5) = -3 and you can't the ln of a negative number

x = 1 is the only answer

ln(x+1) - ln(3-x) = ln(2-x)
(x+1)/(3-x)=2-x
x+1=(2-x)(3-x)
x+1=x^2-5x+6
x^2-6x+5=0
(x-1)(x-5)=0
x-1=0; x-5=0
x=1; x=5 // latter is extraneous

x = 1