The supply function and the demand function for the sale of a type of DVD player are given by S(p)= 150e^(0.004p) and D(p)= 480e^(-0.003p) where S(p) is the number of DVD players that the company is willing to sell at the price p and D(p) is the quantity that the public is willing to buy at price p. Find p such that D(p)=S(p).
It's really long I know. Thanks so much for the help!
It's really long I know. Thanks so much for the help!
-
S(p)= 150e^(0.004p) =480e^(-0.003p) =D(p).
You need to solve for p in the equation 150e^(0.004p) =480e^(-0.003p)
So take ln of both sides:
ln(150e^(0.004p)) = ln(480e^(-0.003p))
use the ln rule that ln(x*y)=lnx+lny to get
ln150+ln(e^.004p)=ln480 +ln(e^-.003p)
we know that lne^x=x, so
ln 150 +.004p = ln480+( - .003p)
Now the hard part is done, solve for p!
You need to solve for p in the equation 150e^(0.004p) =480e^(-0.003p)
So take ln of both sides:
ln(150e^(0.004p)) = ln(480e^(-0.003p))
use the ln rule that ln(x*y)=lnx+lny to get
ln150+ln(e^.004p)=ln480 +ln(e^-.003p)
we know that lne^x=x, so
ln 150 +.004p = ln480+( - .003p)
Now the hard part is done, solve for p!
-
150e^(0.004p) = 480e^(-0.003p)
Take ln,
ln150 + 0.007p = ln480
p = 166.164
Take ln,
ln150 + 0.007p = ln480
p = 166.164