SOLUTION: Solve ln(3x+8)=ln(2x+2)+ln(x-2) for x Please help me

Question 1239: Solve ln(3x+8)=ln(2x+2)+ln(x-2) for x
Please help me
Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
Given ln(3x+8)=ln(2x+2)+ln(x-2)
Since the right hand side ln(2x+2)+ln(x-2) = ln(2x+2)(x-2) = ln(2x^2 -2x -4),
So,ln(3x+8) = ln(2x^2 -2x -4) and we have
3x+8 = 2x^2 -2x -4, or 2x^2 -5x -12 = 0.
Factor: (2x + 3)(x - 4) = 0.
So, x = -3/2 or 4
But by the definition of ln (log), 3x +8 , 2(x+1) and x-2 must be positive.
Hence, x > -8/3, x > -1 and x > 2. Therefore, x must be greater than 2 and
the solution -3/2 is invalid. The only solution is x =4.
Check,when x = 4, ln(3x+8)=ln(2x+2)+ln(x-2) becomes
ln 20 = ln 10 + ln 2 = ln 10*2 ,so 4 is a correct solution of the
given equation.

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