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This question has come in DUET math Olympiad 2012
1.If 3\[{3^a}={4^{b}}= 36\]
Then find the value of \[2/a+1/b\]

\[3^{a}=36\]
\[\Rightarrow log3^{a} = log 36\]
\[\Rightarrow a log3 = log 36\]
\[\Rightarrow a = \frac{log 36}{log 3}\]

Similarly, \

Now \[\frac{2}{a} + \frac{1}{b}\]
\[\frac{2}{\frac{log36}{log3}} + \frac{1}{\frac{log36}{log4}}\]
\[\frac{2log3}{log36} + \frac{log4}{log36}\]
\[\frac{2log3+log4}{log36}\]
\[\frac{log3^{2}+log4}{log36}\]
\[\frac{log(3^{2}*4)}{log36}\]
\[\frac{log36}{log36}\]
\[=1 <====ANSWER\]

Essentially the same solution, without using logarithms:

\[\left\{\begin{align*}3^a=36&\Rightarrow 36^{\frac 2a}=3^2\\
4^b=36&\Rightarrow 36^{\frac 1b}=4\end{align*}\right\}\Rightarrow 36^{\frac 2a+\frac 1b}=3^2\times 4=36\Rightarrow \frac 2a+\frac 1b=1.\]

@Nayel vaia,
Nice solution .
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