<s>[INST] To find the logarithm of $1728$ to the base $2 \sqrt{3}$ 
Find the logarithm of: $1728$ to base $2\sqrt{3}$.

Let, $\log_{2\sqrt{3}} 1728 = y$, then 
$$\begin{align} (2\sqrt{3})^y &= 1728\\
2^y(\sqrt3)^y &= 1728\\2^y(3^\frac12)^y &= 1728\\2^y(3^\frac y2)
 &= 1728\\2^y × 3^\frac y2 &= 2^6 × 3^3 \end{align}$$
What should I do next to find the logarithm of $1728$?
 [/INST] $$12^3 = 1728 = (2\sqrt 3)^y = (\sqrt 12)^y = (12^\frac 1 2)^y = 12^\frac y 2 $$
$$ 3 = \frac y 2 $$
$$ y = 6 $$
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