Lumberjack Admission Test

Let $\sqrt [x ]{ a } =u$ , then

$\sqrt { a } =u^{\frac { x}{ 2 } }$

$\log _{ \sqrt [ x ]{ a } }{ \sqrt { a } } =\log _{ u }{ u^{ \frac { x }{ 2 } } } =\frac { x}{ 2 };$

equivalently,

$\log _{ \sqrt [ y ]{ b } }{ \sqrt { b } } =\frac { y}{ 2 }.$

Then, the second equation may be written as:

$\frac { x }{2 }+ \frac { y}{2 }=\frac { a}{\sqrt{3} }$ .

Thus, we obtain the following system:

$\begin{cases} x^2+xy+y^2=a^2 (1)\\ x+y=\frac { 2a}{\sqrt{3}} (2)\end{cases}$

Squaring equation (2), we get

$x^2+2xy+y^2=\frac{ 4a^2 }{3 } (2\alpha)$ .

Subtracting (1) from (2 $\alpha)$ ,

$xy=\frac{a^2}{3}$ .

We arrive at the following system:

$\begin{cases} x+y=\frac{2a}{\sqrt{3}}\\ xy=\frac{a^2}{3} \end{cases}$ .

And thus,

$x=y=\frac{a}{\sqrt{3}}$

Now to the problem:

$\frac{(x+y)}{2a}=\sqrt{\frac{L}{3}}$

$\frac{(\frac{a}{\sqrt{3}}+\frac{a}{\sqrt{3}})}{2a}=\sqrt{\frac{L}{3}}$

$\frac{1}{\sqrt{3}}=\sqrt{\frac{L}{3}}$

$\boxed{L=1}$