Exponents.

**habibixox** · October 2nd 2011, 11:02 AM

Solve for

:

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

but how can i simplify this while knowing the above?

**Siron** · October 2nd 2011, 11:05 AM

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

**habibixox** · October 2nd 2011, 11:11 AM

so would it be

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

**Siron** · October 2nd 2011, 11:15 AM

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

**habibixox** · October 2nd 2011, 11:19 AM

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?

**Siron** · October 2nd 2011, 11:41 AM

You're welcome!

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

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