Question

How do you solve `Ln(7-2x) = Ln(3x+3)` and find any extraneous solutions?

Answer 1

`ln(7 - 2x) - ln(3x + 3) = 0`

`ln((7 - 2x)/(3x + 3)) = 0`

`(7 - 2x)/(3x + 3) = e^0`

`(7 - 2x)/(3x + 3) = 1`

`7 - 2x = 1(3x + 3)`

`7 - 2x = 3x + 3`

`7 - 3 = 5x`

`4 = 5x`

`4/5 = x`

Hopefully this helps!