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Posted: Sun Mar 25, 2012 10:08 am
This question has come in DUET math Olympiad 2012
1.If 3${3^a}={4^{b}}= 36$
Then find the value of $2/a+1/b$

### Re: DUET math olympiad 2012

Posted: Sun Mar 25, 2012 12:25 pm
$3^{a}=36$
$\Rightarrow log3^{a} = log 36$
$\Rightarrow a log3 = log 36$
$\Rightarrow a = \frac{log 36}{log 3}$

Similarly, \

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

### Re: DUET math olympiad 2012

Posted: Sun Mar 25, 2012 5:56 pm
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.

### Re: DUET math olympiad 2012

Posted: Sun Mar 25, 2012 11:00 pm
@Nayel vaia,
Nice solution .
Turzo

### Re: DUET math olympiad 2012

Posted: Mon Mar 26, 2012 8:49 am
Mbl theke kisu bujha jai na. Help koren vaia...