Philippines Mathematical Olympiad

Solution:

Let $y = \log_{3} x$. Then $x = 3^{y}$.

We compute the inner logarithms:

- $\log_{3} x = y$
- $\log_{27} x = \frac{\log_{3} x}{\log_{3} 27} = \frac{y}{3}$
- $\log_{\sqrt[3]{3}} x = \frac{\log_{3} x}{\log_{3} \sqrt[3]{3}} = \frac{y}{1/3} = 3y$

Now substitute into the equation:

$\log_{\sqrt[3]{3}}(\log_{3} x) + \log_{3}(\log_{27} x) + \log_{27}(\log_{\sqrt[3]{3}} x) = 1$

That is:

$\log_{\sqrt[3]{3}}(y) + \log_{3}\left(\frac{y}{3}\right) + \log_{27}(3y) = 1$

Compute each term:

1. $\log_{\sqrt[3]{3}}(y) = \frac{\log_{3} y}{\log_{3} \sqrt[3]{3}} = \frac{\log_{3} y}{1/3} = 3 \log_{3} y$

2. $\log_{3}\left(\frac{y}{3}\right) = \log_{3} y - \log_{3} 3 = \log_{3} y - 1$

3. $\log_{27}(3y) = \frac{\log_{3}(3y)}{\log_{3} 27} = \frac{\log_{3} 3 + \log_{3} y}{3} = \frac{1 + \log_{3} y}{3}$

Sum:

$3 \log_{3} y + (\log_{3} y - 1) + \frac{1 + \log_{3} y}{3} = 1$

Combine like terms:

$3 \log_{3} y + \log_{3} y + \frac{\log_{3} y}{3} = 1 + 1 - \frac{1}{3}$

But let's be careful:

$3 \log_{3} y + \log_{3} y = 4 \log_{3} y$

So:

$4 \log_{3} y + \frac{\log_{3} y}{3} - 1 + \frac{1}{3} = 1$

$\left(4 + \frac{1}{3}\right) \log_{3} y - 1 + \frac{1}{3} = 1$

$\frac{13}{3} \log_{3} y - \frac{2}{3} = 1$

$\frac{13}{3} \log_{3} y = 1 + \frac{2}{3} = \frac{5}{3}$

$\log_{3} y = \frac{5}{13}$

Recall $y = \log_{3} x$, so $\log_{3} y = \log_{3}(\log_{3} x)$.

Therefore,

$\boxed{\frac{5}{13}}$