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?

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- Oct 2nd 2011, 11:02 AMhabibixoxExponents.
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 AMSironRe: Exponents.
To make it easy you can take the $\displaystyle \log$ of both sides and use $\displaystyle \log(a^b)=b\log(a)$ and you can also use $\displaystyle a^{x+y}=a^{x}\cdot a^{y}$ in the first place

- Oct 2nd 2011, 11:11 AMhabibixoxRe: Exponents.
so would it be

9x/3 log 9 = 8x/2 log 4 ? - Oct 2nd 2011, 11:15 AMSironRe: Exponents.
If you use the rule I suggested then you get:

$\displaystyle 9^{9x-3}=4^{8x-2}$

$\displaystyle \Leftrightarrow \log(9^{9x-3})=\log(4^{8x-2})$

$\displaystyle \Leftrightarrow (9x-3)\log(9)=(8x-2)\log(4)$

$\displaystyle \Leftrightarrow 9x\log(9)-3\log(9)=8x\log(4)-2\log(4)$

$\displaystyle \Leftrightarrow 9x\log(9)-8x\log(4)=3\log(9)-2\log(4)$

$\displaystyle \Leftrightarrow x[9\log(9)-8\log(4)]=3\log(9)-2\log(4)$

$\displaystyle \Leftrightarrow ...$ - Oct 2nd 2011, 11:19 AMhabibixoxRe: 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 AMSironRe: Exponents.
You're welcome! :)

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