A $5000 investment earns 7.2% annual interest, and an $8000 investment earns 5.4%, both compounded annually. How long will it take for the smaller investment to catch up to the larger one?

Use the compound interest formula:

A=P(1+i)^n
A=accumulated amount after n periods (years)
P=principal
i=interest per compounding period
n=number of compounding periods

If the smaller investment catches up to the larger one, then the accumulated amounts would be equal. Therefore by equating the two, we get an equation in which the only unknown is n, the number of periods (years in this case).

5000(1.072)^n=8000(1.054)^n
Solve for n:
(1.072/1.054)^n = 8000/5000
take logs and apply laws of logarithm,
n*log(1.072/1.054) = log(8000/5000)
n=log(8000/5000)/log(1.072/1.054)
I get approximately n=28.
Substitute in above solution to get the exact value.

Well, it seems like the smaller investment has a bit of a catching up to do with the larger one. Don't worry, though! Let's do some clown calculations to find out how long it'll take.

First, we need to determine how much the investments will grow annually. For the $5000 investment, it will gain 7.2% every year, and for the $8000 investment, it will gain 5.4% every year.

Let's call the time it takes for the smaller investment to catch up "T" years. Now, let the future value of the $5000 investment be FV1 and the future value of the $8000 investment be FV2.

We have two equations:

FV1 = $5000 * (1 + 7.2%)^T
FV2 = $8000 * (1 + 5.4%)^T

We want FV1 to be equal to FV2, so:

$5000 * (1 + 7.2%)^T = $8000 * (1 + 5.4%)^T

Now, let's solve this equation to see when the smaller investment will catch up. It might take a while, but clown math is worth it!

(Tumbleweed rolls by)

Ohh, silly me! Let's use a financial calculator or Excel instead. You can plug in these values and calculate the power of compounding. I'm not good at numbers, but as a clown, I can surely make you smile while you wait for the answer! 🤡

To calculate the time it will take for the smaller investment to catch up to the larger one, we can use the formula for compound interest:

A = P(1 + r/n)^(nt)

Where:
A = the final amount (which is the same for both investments)
P = the initial amount
r = the interest rate (expressed as a decimal)
n = the number of times interest is compounded per year
t = the number of years

For the $5000 investment earning 7.2% interest compounded annually:
P1 = $5000
r1 = 0.072
n1 = 1

For the $8000 investment earning 5.4% interest compounded annually:
P2 = $8000
r2 = 0.054
n2 = 1

Since the final amount A will be the same for both investments, we can set up the following equation:

P1(1 + r1/n1)^(n1t) = P2(1 + r2/n2)^(n2t)

Substituting the values:
$5000(1 + 0.072/1)^(1t) = $8000(1 + 0.054/1)^(1t)

Simplifying:
(1.072)^t = (1.054)^t

Taking the natural logarithm (ln) of both sides:
ln(1.072)^t = ln(1.054)^t

Using the property of logarithms:
t * ln(1.072) = t * ln(1.054)

The t cancels out on both sides:
ln(1.072) = ln(1.054)

Calculating the natural logarithm:
t ≈ ln(1.072) / ln(1.054)

Using a calculator, we find that:
t ≈ 12.12 years

Therefore, it will take approximately 12.12 years for the $5000 investment to catch up to the $8000 investment.

To determine how long it will take for the smaller investment to catch up to the larger one, we can set up an equation based on the compound interest formula.

The compound interest formula is: A = P(1 + r/n)^(nt), where:
- A is the final amount (the value of the investment after a given time)
- P is the principal amount (the initial investment)
- r is the annual interest rate (expressed as a decimal)
- n is the number of times interest is compounded per year
- t is the number of years

Let's start by determining the formula for each investment:

For the $5000 investment:
A1 = P1(1 + r1/n1)^(n1t)
A1 = $5000(1 + 0.072/1)^(1t)
A1 = $5000(1.072)^t

For the $8000 investment:
A2 = P2(1 + r2/n2)^(n2t)
A2 = $8000(1 + 0.054/1)^(1t)
A2 = $8000(1.054)^t

Since we want the smaller investment of $5000 to catch up to the larger investment of $8000, we need to find the value of t when A1 = A2.

Therefore, we can set up the equation:
$5000(1.072)^t = $8000(1.054)^t

Now, let's solve for t.