1. ## Exponents.

Solve for :

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

but how can i simplify this while knowing the above?

2. ## Re: 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

3. ## Re: Exponents.

so would it be

9x/3 log 9 = 8x/2 log 4 ?

4. ## Re: Exponents.

If you use the rule I suggested then you get:
$\displaystyle 9^{9x-3}=4^{8x-2}$
$\displaystyle \Leftrightarrow \log(9^{9x-3})=\log(4^{8x-2})$
$\displaystyle \Leftrightarrow (9x-3)\log(9)=(8x-2)\log(4)$
$\displaystyle \Leftrightarrow 9x\log(9)-3\log(9)=8x\log(4)-2\log(4)$
$\displaystyle \Leftrightarrow 9x\log(9)-8x\log(4)=3\log(9)-2\log(4)$
$\displaystyle \Leftrightarrow x[9\log(9)-8\log(4)]=3\log(9)-2\log(4)$
$\displaystyle \Leftrightarrow ...$

5. ## 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?

6. ## Re: Exponents.

You're welcome!

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