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This question about finding the widest range of values the constant $a$ can take.

After completing the square for the expression inside the square root, and then solving the inequality that must satisfy (I did assume $2 < a\le 3$ in this step), I obtained $$\log_{e}(2-\sqrt{3-a})\le x\le \log_{e}(2+\sqrt{3-a}).$$

My question is, is there a bigger interval for $a$ for which the inequality holds true?

Wolfram gave $-1 < a\le 3$ if I remember correctly but I cannot justify why.

Thank you.

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$a\leq3$ is equivalent to the condition ${4e^{x}-e^{2x}-1-a}\geq0$ which means that the square root exists. This is done by the method mentioned in the question: $(e^x-2)^2\leq 3-a$. Equivalently this means that $2-\sqrt{3-a}\leq e^x\leq 2+\sqrt{3-a}$. The restriction $-1<a$ results if you left bound for $e^x$ is wanted to be positive. Otherwise any $a\leq 3$ is possible. The changes then apply to the range of the $x$ value only. For $a\leq -1$ this range is the interval $(-\infty,\ln(2+\sqrt{3-a}))$. For $a>-1$ the range is given by $(\ln(2-\sqrt{3-a}),\ln(2+\sqrt{3-a}))$.

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