Uninhibited Growth Model. This is an example problem from the book and need help figuring it out?
a) If $3500 is invested in an account that pays 5% interest compounded continuously, how long will it take to grow to $7000.
b) How much money will the account be worth in 20 years.
a. P = Po*e^rt.
P = 3500*e^rt = 7,000,
e^rt = 2.0,
rt*Lne = Ln 2,
r*t = 0.693,
0.05t = 0.693,
t = 13.9 Years.
b. P = Po*e^rt.
Po = $3500
r*t = 0.05 * 20,
P = ?
I understand part A perfectly well now. But I am confused with part B once I get $3500^0.05*20 would P just be $3500 or 1 as the answer?
To solve these problems, we can use the formula for continuous compound interest:
A = P*e^(rt)
Where:
A is the final amount
P is the principal amount (initial investment)
e is the base of the natural logarithm (approximately 2.71828)
r is the interest rate (in decimal form)
t is the time (in years)
a) If $3500 is invested in an account that pays 5% interest compounded continuously, and we want to find how long it takes to grow to $7000, we can set up the equation:
7000 = 3500 * e^(0.05t)
Dividing both sides of the equation by 3500, we have:
2 = e^(0.05t)
Now, take the natural logarithm of both sides to isolate the exponent:
ln(2) = ln(e^(0.05t))
Using the property of logarithms that ln(e^x) = x, we get:
ln(2) = 0.05t
Finally, divide both sides by 0.05:
t = ln(2) / 0.05
Using a calculator, we find that t is approximately 13.86 years. So, it will take around 13.86 years for the investment to grow from $3500 to $7000.
b) To find out how much money the account will be worth in 20 years, we can use the same formula, just plug in the values:
A = 3500 * e^(0.05 * 20)
Again, using a calculator, we get:
A = 3500 * e^(1)
A ≈ 3500 * 2.71828
A ≈ $9527.98
Therefore, after 20 years, the account will be worth approximately $9527.98.