Solve the equation `2 ln(x + 3) -lnx = ln(2x -2)` .

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Correction: `((x+3)^2 - 2x^2 + 2x)/(2x^2 - 2x) = 0 => x^2 + 6x + 9 - 2x^2 + 2x = 0 => -x^2 + 8x + 9 = 0 => x^2 - 8x - 9 = 0`

`x_(1,2) = (8+-sqrt(64 + 36))/2 => x_(1,2) = (8+-10)/2`

`x_1 = 18/2 => x_1 = 9`

`x_2 = -2/2 => x_2 = -1`

Hence, the solution to equation is `x = 9. `

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`2 ln(x + 3) -lnx = ln(2x -2)`

Using logarithm rules we know that;

`ln A + ln B=ln(AB)`

`ln A-ln B=ln(A/B)`

`alnA=ln(A^a)`

Apply these rules to the above equation.

`2 ln (x+3) = ln (x+3)^2`

`2 ln (x+3) - ln x = ln (x+3)^2 - ln x = ln[(x+3)^2/x]`

`2 ln (x+3) – ln x = ln (2x-2)`

`ln[(x+3)^2/x] = ln(2x-2)`

Remove ln in both sides.

`(x+3)^2/x=2x-2`

`x^2+6x+9 = 2x^2-2x`

`0=x^2-8x-9`

`0=x^2-9x+x-9`

`0=(x-9)(x+1)`

`x=9` or `x=-1`

Since ln are not derived for negative values of x the answer is x=9.

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