Problem 2:
If $9^{x+18} = 16^x$ and $b^x = 9^9$, what is the value of $b$?
Rangpur Higher Secondary 2011/2
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Please don't post problems (by starting a topic) in the "Higher Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
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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
(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
Re: Rangpur Higher Secondary 2011/2
@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
$9^x (9^9)^2=16^x$
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.)
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
- Location:Khulna,Bangladesh.
Re: Rangpur Higher Secondary 2011/2
ahhh, ican't write it in latex,any moderator plz help :'(
বড় ভালবাসি তোমায়,মা
Re: Rangpur Higher Secondary 2011/2
$(9^x)(9^{18})=16^x$
$\Rightarrow (9^x)((9^9)^2)=16^x$
$\Rightarrow (9^x)((b^2))^x)=16^x$
$\Rightarrow (9b^2)^x=16^x$
$\Rightarrow 9b^2=16$
$\Rightarrow b^2=\frac{16}{9}$
$\Rightarrow b=\pm \frac{4}{3}$
$\Rightarrow (9^x)((9^9)^2)=16^x$
$\Rightarrow (9^x)((b^2))^x)=16^x$
$\Rightarrow (9b^2)^x=16^x$
$\Rightarrow 9b^2=16$
$\Rightarrow b^2=\frac{16}{9}$
$\Rightarrow b=\pm \frac{4}{3}$
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Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi