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The Fourier transform of a continuous-time signal $x(t)$ is given by $X(\omega) = \frac{1}{(10+j \omega)^2}, – \infty < \omega < \infty$, where $j = \sqrt{-1}$ and $\omega$ denoes frequency. Then the value of $\mid \text{ln } x(t) \mid$ at $t=1$ is _________ (up to $1$ decimal place). ($\text{ln}$ denotes the logarithm base $e$)
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