Question

What are the potential solutions to the equation below?

2ln(x+3)=0
X=-3 and X=-4
X=-2 and x=-4
X= 2 and X=-3
X= 2 and X-4

Answer 1

2In(x+3) = 0

This is the same as:

`In(x+3)^2=0`

Take the exponent of both-side

`e^(In)(x+3)^2=e^0`
`(x+3)^2=1`

Take the square root of both-side

`x+3=\pm1`

subtract 3 from both-side

x = ± 1 - 3

Either x = -1 - 3 or x = +1 - 3

x = -4 or x = -2