In a Martian civilization, all logarithms whose bases are not specified are assumed to be base $b$, for some fixed $b\ge2$. A Martian student writes down
\[3\log(\sqrt{x}\log x)=56\]
\[\log_{\log x}(x)=54\]
and finds that this system of equations has a single real number solution $x>1$. Find $b$.
Using change of base on the second equation to base b,
\[\frac{\log x}{\log \log_{b}{x}}=54\]
\[\log x = 54 \cdot \log\log_{b}{x}\]
\[x = (\log_{b}{x})^{54}\]
Note by dolphin7 - you could also just rewrite the second equation in exponent form.
Substituting this into the $\sqrt x$ of the first equation,
\[3\log_{b}{((\log_{b}{x})^{27}\log_{b}{x})} = 56\]
\[3\log_{b}{(\log_{b}{x})^{28}} = 56\]
\[\log_{b}{(\log_{b}{x})^{84}} = 56\]
We can manipulate this equation to be able to substitute $x = (\log_{b}{x})^{54}$ a couple more times:
\[\log_{b}{(\log_{b}{x})^{54}} = 56 \cdot \frac{54}{84}\]
\[\log_{b}{x} = 36\]
\[(\log_{b}{x})^{54} = 36^{54}\]
\[x = 6^{108}\]
However, since we found that $\log_{b}{x} = 36$, $x$ is also equal to $b^{36}$. Equating these,
\[b^{36} = 6^{108}\]
\[b = 6^3 = \boxed{216}\]