diff --git a/C03-Growth-of-Functions/3.2.md b/C03-Growth-of-Functions/3.2.md
index 2e282097..7dbcabe3 100644
--- a/C03-Growth-of-Functions/3.2.md
+++ b/C03-Growth-of-Functions/3.2.md
@@ -82,6 +82,13 @@ Prove that for i ≥ 0, the (i + 2)nd Fibonacci number satisfies Fi+2 ≥ φi.
 ![](http://latex.codecogs.com/gif.latex?\\hspace{20mm}F_{i+2}\\ge\\phi^i\\\\~\\hspace{14mm}\\iff\\frac{\\phi^{i+2}-\\widehat\\phi^{i+2}}{\\sqrt5}\\ge\\phi^i\\\\~\\hspace{14mm}\\iff\(\\phi^2-\\sqrt5\)\\phi^i\\ge\\widehat\\phi^2\\widehat\\phi^i\\\\~\\hspace{14mm}\\iff\\phi^i\\ge\\widehat\\phi^i)
 
 
+### Exercise 3.2-8
+Show that ![](https://latex.codecogs.com/svg.latex?k%20\cdot%20ln(k)=\Theta{(n)}) implies ![](https://latex.codecogs.com/svg.latex?k=\Theta{(\frac{n}{%20ln(n)})})
+
+### `Answer`
+Symmetry: ![](https://latex.codecogs.com/svg.latex?f(n)%20=%20\Theta(g(n))%20\iff%20g(n)%20=%20\Theta(f(n)))
+![](https://latex.codecogs.com/svg.latex?\begin{align*}%20n%20&=%20\Theta%20(k%20\ln%20k)%20\%20,%20\\%20\ln%20n%20&=%20\Theta%20(\ln%20(k%20\ln%20k))%20\\%20&=%20\Theta%20(\ln%20k%20+%20\ln%20(\ln%20k))%20\\%20&=%20\Theta%20(\ln%20k)%20\%20,%20\\%20\frac{n}{\ln%20n}%20&=%20\Theta%20\left%20(%20\frac{k\ln%20k}{\ln%20k}%20\right%20)%20=%20\Theta%20(k)%20\%20.\\%20\\%20\therefore%20\%20k%20&=%20\Theta\left%20(%20\frac{n}{\ln%20n}%20\right%20)%20\end{align*})
+
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