3 years ago

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Mathematics

2 answers:

Contact [7]3 years ago

4 0

Answer:

A x=1.5y i think

Step-by-step explanation:

steposvetlana [31]3 years ago

4 0

The answer is: A. x=1.5y

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

skad [1K]

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

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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3 years ago

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padilas [110]

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Step-by-step explanation:

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7 0

3 years ago

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valentinak56 [21]