<s>[INST] Show that $e^{\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)}}= 2 + \sqrt{3}$ How from $e^{\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)}}$ do I get $2+\sqrt{3}$?
This is a simple problem that I am unable to solve. I tried using properties of logarithms, but I cannot find $2+\sqrt{3}$, as Wolfram says... how can I do it?
Thanks.
 [/INST] Note that, as Matti P.'s question comment indicates, you have
$$(2+\sqrt{3})(2-\sqrt{3}) = 4 - 3 = 1 \implies 2-\sqrt{3} = (2+\sqrt{3})^{-1} \tag{1}\label{eq1A}$$
To make the algebra simpler, use
$$x = 2+\sqrt{3} \implies x^{-1} = 2-\sqrt{3} \tag{2}\label{eq2A}$$
The exponent of $e$ in your expression now becomes
$$\begin{equation}\begin{aligned}
\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)} & = \frac{3\ln(x)\ln(x^{-1})}{2\ln(x^{-1}) - \ln(x)} \\
& = \frac{3\ln(x)(-1)\ln(x)}{-2\ln(x) - \ln(x)} \\
& = \frac{-3\ln^2(x)}{-3\ln(x)} \\
& = \ln(x)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This now results in
$$\begin{equation}\begin{aligned}
e^{\frac{3\ln(2+\sqrt3)\ln(2-\sqrt3)}{ 2\ln(2-\sqrt3)-\ln(2+\sqrt3)}} & = e^{\ln(x)} \\
& = x \\
& = 2 + \sqrt{3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
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