3 years ago

I honestly don't know how to complete this problem. Please help may you include the steps also.

Mathematics

1 answer:

Arisa [49]3 years ago

5 0

Answer is B............

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

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Expanding the left side gives

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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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Y_Kistochka [10]

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