1. ## Please show me how to solve this equation

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

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

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

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

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

Can you finish from here?

-Dan

3. Originally Posted by topsquark
I presume this is
$\frac{500}{1 + 25e^{0.3x}} = 200$

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

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

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

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?

4. 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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### if the marginal revenue (MR) function is defined as the rate of change of total revenue (R) with respect to the quantity demanded at an instant .if the marginal revenue of a function is given by : MR= 25e​-x/400(1-x/100)

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