An investment initially worth $5300 earns 7.7% annual interest, and an investment initially worth $8000 earns 5.6% annual interest, both compounded annually.
How long will it take for the smaller investment to catch up with the larger one?
where is the work? I'm still lost
solve for t in
5300 * 1.077^t = 8000 * 1.056^t
t = 20.9 years
or, 21, since the interest is credited at year's end.
To determine how long it will take for the smaller investment to catch up with the larger one, we need to compare their values at different points in time.
Let's first calculate the value of the smaller investment after one year using the compound interest formula:
Value = Principal * (1 + Interest Rate)^Time
For the smaller investment:
Principal = $5300
Interest Rate = 7.7% = 0.077 (expressed as a decimal)
Time = 1 year
Value of smaller investment after one year = $5300 * (1 + 0.077)^1
Now, let's calculate the value of the larger investment after one year using the same formula:
Principal = $8000
Interest Rate = 5.6% = 0.056 (expressed as a decimal)
Time = 1 year
Value of larger investment after one year = $8000 * (1 + 0.056)^1
Now we need to compare these values. If the value of the smaller investment is greater, it has already caught up with the larger one. If not, we need to calculate the values for subsequent years until the smaller investment surpasses the larger one.
Repeat this process by increasing the time by 1 year each time until the value of the smaller investment is greater than the larger one.
For example, after 2 years:
Value of smaller investment = $5300 * (1 + 0.077)^2
Value of larger investment = $8000 * (1 + 0.056)^2
Continue this process until the value of the smaller investment exceeds the value of the larger investment. The number of years it takes for this to happen will be the answer to your question.