Question 373981
{{{p=5000(1-(4/(4+e^(-0.002x))))}}}
To find x for p = 600:
{{{600 = 5000(1-(4/(4+e^(-0.002x))))}}}
We need "peel away" from the outside on the right side. First we get rid of the 5000 by dividing both sides be 5000:
{{{600/5000 = 1-(4/(4+e^(-0.002x)))}}}
or
{{{0.12 = 1-(4/(4+e^(-0.002x)))}}}
Next the 1 must go. Subtract 1 from each side:
{{{-0.88 = -(4/(4+e^(-0.002x)))}}}
Multiply (or divide) both sides by -1 to eliminate the minus sign:
{{{0.88 = 4/(4+e^(-0.002x))}}}
Next we'll eliminate the fraction by multiplying both sides by {{{(4+e^(-0.002x))}}}:
{{{0.88)*(4+e^(-0.002x)) = 4}}}
Using the Distributive Property on the left side:
{{{0.88)*(4)+(0.88)*e^(-0.002x) = 4}}}
{{{3.52 + (0.88)*e^(-0.002x) = 4}}}
Subtracting 3.52 from each side:
{{{(0.88)*e^(-0.002x) = 0.48}}}
Divide by 0.88:
{{{e^(-0.002x) = (0.88)*0.48}}}
{{{e^(-0.002x) = 0.4224}}}
Now that the exponential part of the equation is isolated, we use logarithms to proceed. Since the base of the exponent is e, it is best to use base e logarithms (aka lm):
{{{ln(e^(-0.002x)) = ln(0.4224)}}}
On the left side we can use a property of logarithms, {{{log(a, (p^q)) = q*log(a, (p))}}}, to "move" the exponent out in front of the logarithm. (It is this property which is the very reason we use logarithms. It gives us a way to get the variable out of the exponent.) Using the property on the left side we get:
{{{(-0.002x)ln(e) = ln(0.4224)}}}
By definition, {{{ln(e) = 1}}} so this becomes:
{{{-0.002x = ln(0.4224)}}}
Last of all we divide by -0.002:
{{{x = ln(0.4224)/(-0.002)}}}
This is an exact expression for your answer. Use your calculator if you want a decimal approximation:
{{{x = (-0.8618025465900853)/(-0.002)}}}
x = 430.9012732950426674
So for a price of $699 the demand will be approximately 431.<br>
For a price of $400, replace the p with 400 and repeat these steps to find the demand at that price.