# Solve for x. `ln(3x-1)= ln(2)+ln(x-2)+ln(x+1)` HELPP!

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### 1 Answer

`ln(3x-1)=ln2+ln(x-2)+ln(x+1)`

Condense right side. To do so, apply the product property of logarithm which is `ln M + ln N = ln (M*N)` .

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

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

Since both side have same base (natural logarithm), equate their argument equal to each other.

`3x - 1= 2 (x - 2)(x + 1)`

Then, expand right side. So FOIL (x-2)(x+1).

`3x - 1 = 2(x^2-x-2)`

Then, distribute 2 to x^2-x- 2.

`3x-1 = 2x^2-2x-4`

Then, set one side equal to zero by subtract 3x and adding 1 to both sides of the equation.

`3x-3x-1+1=2x^2-2x-3x-4+1`

`0=2x^2-5x-3`

Next, factor right side.

`0= (2x+1)(x-3)`

Then, set each factor equal to zero and solve for x.

For the first factor,

`2x+1 =0`

`2x=-1`

`x=-1/2`

And for the second factor,

`x-3=0`

`x=3`

Since in logarithm a negative argument is not allowed, x=-1/2 is not considered as a solution.

**Hence, the solution to the given logarithmic equation is `x=3` .**