A stock analyst attempts to express the price p of a share of XYZ stock as an exponentially increasing function of the time since XYZ's initial public offering (IPO): $p = 10 e ^ { k m }$ , where m is the number of months since the IPO. The price of a share was $10.00 at the time of the IPO and $12.80 four months after the IPO. What is the approximate value of k? Round to the nearest thousandth.
A)0.637 per month
B)0.062 per month
C)3.200 per month
D)1.163 per month
E)16.204 per month

Question

Quizplus · Accepted Answer

Given $p = 10e^{km}$ and $p = 12.80$ when $m = 4$, we substitute these values to solve for $k$. $12.80 = 10e^{4k}$ leads to $e^{4k} = 1.28$. Taking the natural logarithm of both sides gives $4k = \ln(1.28)$, so $k = \frac{\ln(1.28)}{4}$. Calculating this value gives $k \approx 0.062$ per month.

A Stock Analyst Attempts to Express the Price P of a Share