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Step 1: Represent the modulus of the complex number
Let the modulus of the complex number be denoted by t, i.e., set
$t = |z| \ge 0$.
Step 2: Express the given inequality in terms of t
The inequality provided is
$ \exp \left( \frac{(t + 3)(t - 1)}{t + 1} \,\log_e 2 \right)\,\ge\,\log_{\sqrt{2}}\bigl|5\sqrt{7} + 9i\bigr|. $
First, compute
$\bigl|5\sqrt{7} + 9i\bigr| = \sqrt{(5\sqrt{7})^2 + 9^2} = \sqrt{25 \cdot 7 + 81} = \sqrt{175 + 81} = \sqrt{256} = 16.$
Hence,
$\log_{\sqrt{2}}(16).$
Since $\sqrt{2} = 2^{\tfrac{1}{2}}$ and $16 = 2^4,$ we have
$\log_{2^{1/2}}(2^4) = \frac{4}{\tfrac{1}{2}} = 8.$
This makes the inequality become
$\exp \Bigl(\frac{(t+3)(t-1)}{t+1} \ln(2)\Bigr) \,\ge\, 8.$
Step 3: Convert the exponential form to base 2
Recall
$\exp \bigl(\ln(2) \cdot x \bigr) = 2^x.$
Therefore,
$2^{\frac{(t+3)(t-1)}{t+1}} \,\ge\, 8 = 2^3.$
Hence, we get the simpler inequality:
$\frac{(t+3)(t-1)}{t+1} \,\ge\, 3.$
Step 4: Solve the inequality
First expand the numerator:
$(t+3)(t-1) = t^2 + 2t - 3.$
So we have
$\frac{t^2 + 2t - 3}{t+1} \,\ge\, 3.$
Multiply both sides by $(t+1)$ (noting $t+1>0$ for all $t\ge0$) to get:
$t^2 + 2t - 3 \,\ge\, 3(t+1) = 3t + 3.$
Bring all terms to one side:
$t^2 + 2t - 3 - 3t - 3 \,\ge\, 0,$
which simplifies to
$t^2 - t - 6 \,\ge\, 0.$
Factorize:
$t^2 - t - 6 = (t-3)(t+2).$
The solution to
$(t-3)(t+2) \,\ge\, 0$
is
$t \in (-\infty, -2] \cup [3, \infty).$
Since $t = |z| \ge 0,$ only
$t \in [3,\infty)$
is valid.
Step 5: Conclude the least value
Among all values $t \in [3,\infty)$, the smallest is $3.$ Therefore, the least possible value of $|z|$ that satisfies the given condition is
$\boxed{3}.$
