Rangpur Higher Secondary 2011/2

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BdMO
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Rangpur Higher Secondary 2011/2

Unread post by BdMO » Wed Feb 02, 2011 9:06 pm

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

Mehfuj Zahir
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Re: Rangpur Higher Secondary 2011/2

Unread post by Mehfuj Zahir » Wed Feb 02, 2011 10:57 pm

(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

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Avik Roy
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Re: Rangpur Higher Secondary 2011/2

Unread post by Avik Roy » Wed Feb 02, 2011 11:55 pm

@Mehfuj, nice solution. But it'd better if u use latex
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Tahmid Hasan
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Re: Rangpur Higher Secondary 2011/2

Unread post by Tahmid Hasan » Thu Feb 03, 2011 12:05 am

$9^x (9^9)^2=16^x$
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.)
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Tahmid Hasan
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Re: Rangpur Higher Secondary 2011/2

Unread post by Tahmid Hasan » Thu Feb 03, 2011 12:09 am

ahhh, ican't write it in latex,any moderator plz help :'(
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*Mahi*
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Re: Rangpur Higher Secondary 2011/2

Unread post by *Mahi* » Fri Feb 04, 2011 6:34 pm

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