DUET math olympiad 2012

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Shapnil
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DUET math olympiad 2012

Unread post by Shapnil » 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 \[2/a+1/b\]

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Fahim Shahriar
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Joined: Sun Dec 18, 2011 12:53 pm

Re: DUET math olympiad 2012

Unread post by Fahim Shahriar » 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 \[\frac{2}{a} + \frac{1}{b}\]
\[\frac{2}{\frac{log36}{log3}} + \frac{1}{\frac{log36}{log4}}\]
\[\frac{2log3}{log36} + \frac{log4}{log36}\]
\[\frac{2log3+log4}{log36}\]
\[\frac{log3^{2}+log4}{log36}\]
\[\frac{log(3^{2}*4)}{log36}\]
\[\frac{log36}{log36}\]
\[=1 <====ANSWER\]
Name: Fahim Shahriar Shakkhor
Notre Dame College

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nayel
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Location: Dhaka, Bangladesh or Cambridge, UK

Re: DUET math olympiad 2012

Unread post by nayel » Sun Mar 25, 2012 5:56 pm

Essentially the same solution, without using logarithms:

\[\left\{\begin{align*}3^a=36&\Rightarrow 36^{\frac 2a}=3^2\\
4^b=36&\Rightarrow 36^{\frac 1b}=4\end{align*}\right\}\Rightarrow 36^{\frac 2a+\frac 1b}=3^2\times 4=36\Rightarrow \frac 2a+\frac 1b=1.\]
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

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Sazid Akhter Turzo
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Re: DUET math olympiad 2012

Unread post by Sazid Akhter Turzo » Sun Mar 25, 2012 11:00 pm

@Nayel vaia,
Nice solution :) :P .
Turzo

Shahriar tanvir
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Joined: Thu Mar 08, 2012 6:51 pm

Re: DUET math olympiad 2012

Unread post by Shahriar tanvir » Mon Mar 26, 2012 8:49 am

Mbl theke kisu bujha jai na. Help koren vaia...

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