# Exponents.

• Oct 2nd 2011, 11:02 AM
habibixox
Exponents.
Solve for http://webwork.rutgers.edu/webwork2_...dd0b8b8e91.png:
http://webwork.rutgers.edu/webwork2_...095126dd01.png

I know 9 is 3^2 and 4 is 2^2

but how can i simplify this while knowing the above?
• Oct 2nd 2011, 11:05 AM
Siron
Re: Exponents.
To make it easy you can take the $\log$ of both sides and use $\log(a^b)=b\log(a)$ and you can also use $a^{x+y}=a^{x}\cdot a^{y}$ in the first place
• Oct 2nd 2011, 11:11 AM
habibixox
Re: Exponents.
so would it be

9x/3 log 9 = 8x/2 log 4 ?
• Oct 2nd 2011, 11:15 AM
Siron
Re: Exponents.
If you use the rule I suggested then you get:
$9^{9x-3}=4^{8x-2}$
$\Leftrightarrow \log(9^{9x-3})=\log(4^{8x-2})$
$\Leftrightarrow (9x-3)\log(9)=(8x-2)\log(4)$
$\Leftrightarrow 9x\log(9)-3\log(9)=8x\log(4)-2\log(4)$
$\Leftrightarrow 9x\log(9)-8x\log(4)=3\log(9)-2\log(4)$
$\Leftrightarrow x[9\log(9)-8\log(4)]=3\log(9)-2\log(4)$
$\Leftrightarrow ...$
• Oct 2nd 2011, 11:19 AM
habibixox
Re: Exponents.
ohh!!! ah oki i gotcha now :)
all the rules of exponents just confuse me!

thanks so much! do you mind taking a look at my half life problem i posted earlier?
• Oct 2nd 2011, 11:41 AM
Siron
Re: Exponents.
You're welcome! :)

To solve exponential equations then you really have to know this basic rules.