Question

How do I find E(p)??????? Everytime I try I end up with q in the equation...
Shark Inc. has determined that demand for its newest netbook model is given by lnq−1lnp+0.004p=7,
where q is the number of netbooks Shark can sell at a price of p dollars per unit. Shark has determined that this model is valid for prices p≥100.

Answer 1

well start by rewriting the equation

\[\ln(q) = - \ln(p) - 0.004p + 7\]

or

\[\ln(q) = 7 - \ln(p) - 0.004p\]

the opposite og a base e log is the exponential e

so raise each term as a power of e

the log/ exponential law is
\[e^{\ln(a)} = a\]

so then you have

\[e^{\ln(q)} = e^7 - e^{0.004p} - e^{\ln(p)} \]

which simplifies to

\[q = e^7 - e^{0.004p} - p\]

Answer 2

does it matter if u simplify before you derive it?

Answer 3

well q can't be simplified....

Answer 4

no i know that, but I mean use the q= to derive rather then deriving from lnq=....

Answer 5

yes... use the equation of q to differentiate

Answer 6

so dq/dp would be -0.004p e^(0.004p) -1 ?

Answer 7

that looks ok...

Answer 8

but you shouldn't have the 0.004p, it should be just 0.004

Answer 9

oh right because the p would be a 1 right?

Answer 10

so then E(p)= (dq/dp) (p/q)
so i'm getting (-0.004(e^(0.004p))/(e^7- e^(0.004p))
but its saying i'm still wrong....

Answer 11

well aren't you missing a few things

\[E(p) = \frac{(-0.004e^{0.004p} -1)\times p}{e^7 - e^{0.004p} - p}\]