How would you set this problem up..I know to use A=Pe^rt
Larry has $2600 to invest and needs $3000 in 12 years. What annual rate of return wll he need to get in order to accomplish his goal, if interest is compounded continuously?
P = Po e^rt
3000 = 2600 e^12r
1.154 = e^12r
ln 1.154 = 12 r
.1431 = 12 r
r = .012
= 1.2 %
I assumed you knew the recipe for continuous compounding. You can get it as follows:
change in P = dP = P * r * change in T = r dt
dP/P = r dt
ln P = r t
integrate
P = c e^rt
when t = 0, P = Po
P = Po e^rt
To set up this problem using the formula A = Pe^(rt), we need to identify the given values and then substitute them into the formula.
Given:
P = $2600 (initial investment)
A = $3000 (amount needed after 12 years)
t = 12 years
We need to find the annual rate of return, represented by the variable r.
Now, let's substitute the values into the formula:
A = Pe^(rt)
$3000 = $2600 * e^(r * 12)
To solve for r, we need to isolate it on one side of the equation. To do this, divide both sides of the equation by $2600:
$3000 / $2600 = e^(r * 12)
To get rid of the exponential term, we can take the natural logarithm (ln) of both sides:
ln($3000 / $2600) = ln(e^(r * 12))
Using the property of logarithms, ln(e^(r * 12)) simplifies to r * 12:
ln($3000 / $2600) = r * 12
Next, divide both sides of the equation by 12:
ln($3000 / $2600) / 12 = r
Now, use a calculator to evaluate the left side of the equation, and you'll get the value of r. This will give you the annual rate of return that Larry needs to achieve in order to reach his goal.
It's important to note that the ln function represents the natural logarithm, which is logarithm base e.