Solve the following system:
\begin{align*} 4^{-x} + 27^{-y} &= \frac{5}{6} \\ 27^y - 4^x &\leq 1 \\ \log_{27}(y) - \log_{4}(x) &\geq \frac{1}{6} \end{align*}
Source: Romanian National Olympiad 1999
I tried to take the logarithm in base ten for the first 2 and denoted lg(x)=a and lg(y)=b, but I didn't get anywhere.
Using the first two statements, we have $$\frac56=4^{-x}+27^{-y}\geqslant \frac1{4^x}+\frac1{4^x+1} \implies 4^x\geqslant 2$$ Then using the last one, we have $$\log_{27}y\geqslant \log_4x+\frac16\geqslant \log_4(\log_42)+\frac16=-\frac13 \implies 27^y\geqslant 3$$ Hence $$\frac56=\frac1{4^x}+\frac1{27^y}\leqslant \frac12+\frac13=\frac56$$ compels all the inequalities above to become equalities, so $(x,y)=(\frac12, \frac13)$.