Solve for X, check domain of X

Question

ln(3x+8)=ln(2x+2)+ln(x-2)

Answer 1

ln(3x+8)=ln(2x+2)+ln(x-2)

ln a +ln b = ln (ab)

3x+8=(2x+2)(x-2)

3x+8=2x^2-4x+2x-4

3x+8=2x^2-2x-4

2x^2-5x-12=0

2x^2-8x+3x-12=0

2x(x-4)+3(x-4)=0

(x-4)(2x+3)=0

x=4 0r x=-3/2

Comments

The logarithm equation is ln(3x+8) = ln(2x+2)+ln(x-2).

ln(3x+8) = ln(2x+2)*(x-2)

ln(3x+8) - ln(2x+2)*(x-2) = 0

ln[(3x+8) / (2x+2)*(x-2)] = 0

From logarithm definition, y = ln(base a) x, where a>0, a ≠ 1 and x > 0, y ∈ R.

(3x+8) / (2x+2)*(x-2) > 0

Key numbers : x = -8/3, x = -1 and x = 2.

Test intervals : (-∞, -8/3), (-8/3, -1), (-1, 2) and (2, ∞).

Test Interval  x-Value    Polynomial value

(-∞, -8/3)        x = - 3        [3(-3)+8] / [2(-3)+2)*((-3)-2)] = - 1/20 < 0.

(-8/3, -1)         x = - 2        [3(-2)+8] / [2(-2)+2)*((-2)-2)] = 1/4 > 0.

(-1,  2)             x = 0          [3(0)+8] / [2(0)+2)*(0-2)] = -2 < 0.

(2,  ∞)             x = 3          [3(3)+8] / [2(3)+2)*((3)-2)] = 17/8 > 0.

From this we conclude that the inequality is satisfied on the open intervals (-8/3, -1) and (2,  ∞). So, the solution set is (-8/3, -1) U (2,  ∞).

The domain of function is (-8/3, -1) U (2,  ∞).