Question15. If an investment is growing continuously for $t$ years, its annual growth rate $r$ is given by the formula $r=\frac{1}{t} \ln \frac{P}{P_{0}}$ where $P$ is the current value and $P_{0}$ is the amount originally invested.
An investment of $\$ 13,400$ in a particular Internet company in 1992 was worth $\$ 8,040,000$ in 1998 . Find this investment's average annual growth rate during this period.

Studdy Solution

STEP 1

1. The investment grows continuously, so we use the continuous growth rate formula.
2. The initial investment amount $P_0$ is \$13,400.
3. The final investment amount \( P \) is \$8,040,000.
4. The time period \( t \) is from 1992 to 1998, which is 6 years.

STEP 2

1. Identify the known values and substitute them into the formula.
2. Calculate the natural logarithm of the ratio $\frac{P}{P_0}$ .
3. Compute the average annual growth rate $r$ .

STEP 3

Identify the known values and substitute them into the formula for the annual growth rate:
Given: - $P_0 = 13,400$ - $P = 8,040,000$ - $t = 1998 - 1992 = 6$
Substitute these values into the formula:
$r = \frac{1}{t} \ln \frac{P}{P_0}$ $r = \frac{1}{6} \ln \frac{8,040,000}{13,400}$

STEP 4

Calculate the natural logarithm of the ratio $\frac{P}{P_0}$ :
First, compute the ratio:
$\frac{P}{P_0} = \frac{8,040,000}{13,400}$
Calculate the value:
$\frac{P}{P_0} \approx 600$
Now, compute the natural logarithm:
$\ln 600$
Using a calculator, we find:
$\ln 600 \approx 6.3969$

STEP 5

Compute the average annual growth rate $r$ :
Substitute the value of $\ln 600$ into the formula:
$r = \frac{1}{6} \times 6.3969$
Calculate $r$ :
$r \approx 1.0662$
Convert $r$ to a percentage:
$r \approx 106.62\%$
The investment's average annual growth rate during this period is approximately:
$\boxed{106.62\%}$

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