Please show me how to solve this equation

**bilbobaggins** · December 14th 2007, 05:26 PM

500/1+25e^.3x=200
please show me the steps you would take to solve this equation.

**topsquark** · December 14th 2007, 05:32 PM

Originally Posted by bilbobaggins

500/1+25e^.3x=200
please show me the steps you would take to solve this equation.

I presume this is
$\frac{500}{1 + 25e^{0.3x}} = 200$

PLEASE use parenthesis!

$500 = 200(1 + 25e^{0.3x})$

$1 + 25e^{0.3x} = \frac{500}{200} = \frac{5}{2}$

$25e^{0.3x} = \frac{5}{2} - 1 = \frac{3}{2}$

Can you finish from here?

-Dan

**bilbobaggins** · December 14th 2007, 07:02 PM

Originally Posted by topsquark

I presume this is
$\frac{500}{1 + 25e^{0.3x}} = 200$

PLEASE use parenthesis!

$500 = 200(1 + 25e^{0.3x})$

$1 + 25e^{0.3x} = \frac{500}{200} = \frac{5}{2}$

$25e^{0.3x} = \frac{5}{2} - 1 = \frac{3}{2}$

Can you finish from here?

-Dan

25e^3= 3.55^x = 3/2?
And then from that point is there anyway to solve algebraically or is it calc only? Or do we use natural logs?

**topsquark** · December 15th 2007, 07:06 AM

Originally Posted by bilbobaggins

25e^3= 3.55^x = 3/2?
And then from that point is there anyway to solve algebraically or is it calc only? Or do we use natural logs?

You are trying to be too complicated with this. Divide both sides by 25 to get the exponential term alone and then take the natural log of both sides. Then solve for x.

-Dan

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