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 $1728$?

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I think you're done. What value of $y$ satisfies $y=6$ and $y/2=3$? – Ian Coley Apr 14 '13 at 17:26

It might have been easier if you tried manipulating the right hand side instead.

$$(2\sqrt3)^y=1728$$ $$(2\sqrt3)^y=2^6\times3^3$$ $$(2\sqrt3)^y=2^6\times\sqrt3^6=(2\sqrt3)^6$$

Alternatively, you could have changed the base of the logarithm.

$$\log_{2\sqrt3}1728=\frac{\log 1728}{\log 2\sqrt3}=\frac{6\log2+3\log3}{\log2+\frac12\log3}=\frac{6(\log2+\frac12\log3)}{\log2+\frac12\log3}$$

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Hint: by the very definition of logarithm

$$\log_{2\sqrt 3}1728=y\iff (2\sqrt 3)^y=1728=12^3=2^6\cdot 36\ldots$$

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I don't understand, @user43418: you already accepted an answer there... – DonAntonio Apr 14 '13 at 18:01
Well then: post a new question. I won't try to answer a question's that's already been accepted. – DonAntonio Apr 14 '13 at 18:03
After your final equality, I think instead of $36$ you should have $3^3$ (this even agrees with the $12^3$). – Clayton Apr 14 '13 at 18:39

$$12^3 = 1728 = (2\sqrt 3)^y = (\sqrt 12)^y = (12^\frac 1 2)^y = 12^\frac y 2$$

$$3 = \frac y 2$$

$$y = 6$$

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Or: $1728 = 12^{3}$ and $12 = (2\sqrt{3})^{2},$ so $1728 = (2\sqrt{3})^{6}.$

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