## Rangpur Higher Secondary 2011/2

Problem for Higher Secondary Group from Divisional Mathematical Olympiad will be solved here.
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BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

### Rangpur Higher Secondary 2011/2

Problem 2:
If $9^{x+18} = 16^x$ and $b^x = 9^9$, what is the value of $b$?

Mehfuj Zahir
Posts: 78
Joined: Thu Jan 20, 2011 10:46 am

### Re: Rangpur Higher Secondary 2011/2

(9^x)(9^18)=16^X
(9^X).(9^9)^2=16^x
(9^x).(b^2x)=4^2x
9^x=(4/b)^2x
9=(4/b)^2
3=4/b
b=4/3

Avik Roy
Posts: 156
Joined: Tue Dec 07, 2010 2:07 am

### Re: Rangpur Higher Secondary 2011/2

@Mehfuj, nice solution. But it'd better if u use latex
"Je le vois, mais je ne le crois pas!" - Georg Ferdinand Ludwig Philipp Cantor

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Rangpur Higher Secondary 2011/2

$9^x (9^9)^2=16^x$
or,$\frac{9^x (b^x)^2}{ 16^x}=1^x$
or,$b^2=\frac{16}{9}$

Mod edit: most of the cases you need to add some spaces, and brackets to clarify. (LaTeX will just ignore extra spaces.)
বড় ভালবাসি তোমায়,মা

Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Rangpur Higher Secondary 2011/2

ahhh, ican't write it in latex,any moderator plz help :'(
বড় ভালবাসি তোমায়,মা

*Mahi*
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### Re: Rangpur Higher Secondary 2011/2

$(9^x)(9^{18})=16^x$
$\Rightarrow (9^x)((9^9)^2)=16^x$
$\Rightarrow (9^x)((b^2))^x)=16^x$
$\Rightarrow (9b^2)^x=16^x$
$\Rightarrow 9b^2=16$
$\Rightarrow b^2=\frac{16}{9}$
$\Rightarrow b=\pm \frac{4}{3}$
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