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kow [346]
3 years ago
10

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

Mathematics
2 answers:
Korolek [52]3 years ago
7 0

Answer:

The answer is B (edge)

Step-by-step explanation:

skad [1K]3 years ago
4 0
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.
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