The equation ${log}_{e} x + {log}_{e} (1 + x) = 0$ can be written as

x^{2} + x - e = 0

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x^{2} + x - 1 = 0

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x^{2} + x + 1 = 0

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x^{2} + x e - e = 0

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Solution

The correct option is B $x^{2} + x - 1 = 0$
${log}_{e} x + {log}_{e} (1 + x) = 0$
${log}_{e} x (1 + x) = 0$
$x (x + 1) = 0$
$x^{2} + x = 1$
$x^{2} + x - 1 = 0$

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