Question

What are the potential solutions to the equation 2ln(x+3)=0

Answer 1

Answer:

The answer is B (edge)

Step-by-step explanation:

Answer 2

$2\ln(x+3)=0\implies \ln(x+3)^2=0\implies e^{\ln(x+3)^2}=e^0\implies (x+3)^2=1$

Expanding the left side gives

$x^2+6x+9=1\implies x^2+6x+8=0\implies (x+4)(x+2)=0$

which gives two solutions, $x=-4$ and $x=-2$. But if $x=-4$, then $\ln(x+3)=\ln(-1)$, but this number isn't real, so $x=-4$ is an extraneous solution. Meanwhile if $x=-2$, you get $\ln(-2+3)=\ln1=0$, so this solution is correct.

"Potential solutions" might refer to both possibilities, but there is only one actual (real) solution.