Question

PLEASE HELP ME!!!

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=10e^km, 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 $14.60 four months after the IPO. What is the approximate value of k? Round to the nearest thousandth.

ANSWER CHOICES:
A) 1.295 per month
B) 0.670 per month
C) 1.570 per month
D) 0.095 per month
E) 3.650 per month

Answer 1

To find the value of k for the exponential growth function of a stock price, the equation 14.60 = 10e^(4k) is used. After solving for k, the value is approximately 0.095 per month, corresponding to choice D.

Step-by-step explanation:

The student is asking to find the approximate value of k in the exponential function p = 10ekm, which represents the price of a share of stock over time. Given that p is the price of the stock, e is the base of the natural logarithm, k is the constant growth rate, and m is the number of months since the IPO. With a starting price of $10 at the IPO and a price of $14.60 after four months, we can set up the equation as 14.60 = 10e4k to solve for k.

First, divide both sides by 10:

1.46 = e4k

Now, take the natural logarithm of both sides:

ln(1.46) = ln(e4k) = 4k

This simplifies to:

k = ln(1.46) / 4

Using a calculator, we find:

k ≈ 0.095

So the approximate value of k is 0.095 per month, which corresponds to answer choice D.

Answer 2

Answer: (d)

Step-by-step explanation:

Given

Price of the stock is given by the function

`P=10e^(km)`

after 4 months, price increases to $14.60

Insert the value in the function

`\Rightarrow 14.60=10e^(4k)\\\\\Rightarrow 1.46=e^(4k)\\\text{Taking natural log both sides}\\\Rightarrow \ln 1.46=\ln e^(4k)\\\Rightarrow 0.378=4k\\\Rightarrow k=0.0946\approx 0.095\ \text{per month}`

hence, option (d) is correct.