$$ x=\ln \left(\sqrt{y^{2}-1}-y\right) $$ where $$ -\frac{e^{2}+1}{2 e} \leq y \leq-1 $$

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To analyze the equation $$ x = \ln \left( \sqrt{y^2 - 1} - y \right) $$ for the given range of $$ y $$, we first need to ensure that the expression inside the logarithm is positive, as the logarithm is only defined for positive arguments. ### Step 1: Determine the domain of the logarithm We need to find when $$ \sqrt{y^2 - 1} - y > 0 $$. 1. Rearranging gives: $$ \sqrt{y^2 - 1} > y $$ 2. Squaring both sides (valid since both sides are non-negative in the given range): $$ y^2 - 1 > y^2 $$ This simplifies to: $$ -1 > 0 $$ which is not true. Therefore, we need to analyze the expression more carefully. ### Step 2: Analyze the expression $$ \sqrt{y^2 - 1} - y $$ We can rewrite the expression: $$ \sqrt{y^2 - 1} - y = \frac{(\sqrt{y^2 - 1} - y)(\sqrt{y^2 - 1} + y)}{\sqrt{y^2 - 1} + y} = \frac{(y^2 - 1) - y^2}{\sqrt{y^2 - 1} + y} = \frac{-1}{\sqrt{y^2 - 1} + y} $$ ### Step 3: Determine the sign of the expression Since $$ y $$ is in the range $$ -\frac{e^2 + 1}{2e} \leq y \leq -1 $$, we note that: - $$ \sqrt{y^2 - 1} $$ is defined and positive for $$ y < -1 $$. - The term $$ \sqrt{y^2 - 1} + y $$ is also positive because $$ \sqrt{y^2 - 1} > -y $$ for $$ y < -1 $$. Thus, $$ \sqrt{y^2 - 1} - y < 0 $$ in this range, which means that the logarithm $$ x = \ln \left( \sqrt{y^2 - 1} - y \right) $$ is not defined for the given range of $$ y $$. ### Conclusion The expression $$ \sqrt{y^2 - 1} - y $$ is negative for $$ -\frac{e^2 + 1}{2e} \leq y \leq -1 $$, which means that $$ x = \ln \left( \sqrt{y^2 - 1} - y \right) $$ is undefined in this range. Therefore, there are no valid values of $$ x $$ for the given range of $$ y $$. The expression $$ \sqrt{y^2 - 1} - y $$ is negative for $$ -\frac{e^2 + 1}{2e} \leq y \leq -1 $$, making $$ x = \ln \left( \sqrt{y^2 - 1} - y \right) $$ undefined in this range.

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