Alfred is saving up money for a down payment on a house. He currently has $3908, but knows he can get a loan at a lower interest rate if he can put down $4861. If he invests the $3908 in an account that earns 4.8% annually, compounded continuously, how long will it take Alfred to accumulate the $4861? Round your answer to two decimal places, if necessary.
3908*e^.048t = 4861
t = 4.55 years
Well, Alfred is trying to be a serious adult and save up for a house. Good for him! Let's see how long it will take him to accumulate that extra amount he needs.
To solve this problem, we can use the continuous compound interest formula:
A = P * e^(rt)
Where:
A is the desired amount ($4861),
P is the initial investment ($3908),
r is the interest rate (4.8% or 0.048),
t is the time we're trying to find.
Plugging in the values, we get:
4861 = 3908 * e^(0.048t)
Now, to solve for t, let's get rid of that pesky e. We'll take the natural logarithm of both sides:
ln(4861) = ln(3908 * e^(0.048t))
Using the properties of logarithms, we can simplify this to:
ln(4861) = ln(3908) + 0.048t
Now, isolate t:
0.048t = ln(4861) - ln(3908)
Finally, divide both sides by 0.048:
t = (ln(4861) - ln(3908)) / 0.048
After plugging this into a calculator, we find that t ≈ 2.29
So, it will take Alfred approximately 2.29 years to accumulate the additional amount he needs for the down payment on the house.
Well, that wasn't very funny, was it? I guess saving for a house isn't exactly "clowning" around. But hey, at least Alfred will have a cozy place to call "clown" sweet home!
To find out how long it will take Alfred to accumulate $4861, we can use the continuous compound interest formula:
A = P * e^(rt)
Where:
A is the future value (amount Alfred wants to accumulate)
P is the present value (initial amount Alfred has)
e is the base of the natural logarithm (approximately 2.71828)
r is the annual interest rate (4.8% or 0.048)
t is the time in years (what we need to solve for)
Given:
P = $3908
A = $4861
r = 0.048
We can rearrange the formula to solve for t:
t = ln(A/P) / r
Substituting the given values:
t = ln(4861/3908) / 0.048
t = ln(1.244654) / 0.048
Using a calculator, we find:
t ≈ 0.2147 years
So, it will take approximately 0.21 years for Alfred to accumulate $4861.
To find out how long it will take Alfred to accumulate the required amount of $4861, we need to use the compound interest formula:
A = P * e^(rt)
Where:
A = the final amount
P = the initial amount
r = interest rate per year (in decimal form)
t = time in years
e = Euler's number, approximately 2.71828
In this case, Alfred has $3908 initially and needs to accumulate $4861. The interest rate is 4.8% or 0.048 in decimal form.
Let's plug in the values and solve for t:
4861 = 3908 * e^(0.048t)
Now, we can divide both sides by 3908:
4861/3908 = e^(0.048t)
Simplifying further:
1.2440116 = e^(0.048t)
To find t, we need to take the natural logarithm of both sides:
ln(1.2440116) = ln(e^(0.048t))
Using the property of logarithms:
ln(1.2440116) = 0.048t * ln(e)
Since ln(e) equals 1, we can simplify further:
ln(1.2440116) = 0.048t
Now, divide both sides by 0.048:
t = ln(1.2440116) / 0.048
Using a calculator, we find that ln(1.2440116) is approximately 0.21892, so:
t = 0.21892 / 0.048
t ≈ 4.56 years
Therefore, it will take approximately 4.56 years for Alfred to accumulate the required amount of $4861.