**Problem Statement:**Using either logarithms or a graphing calculator, find the time required for the initial amount to be at least equal to the final amount.**Initial Amount:**$7,700, deposited at 5.7% compounded monthly, to reach at least $11,100.---**Solution:**The time required is \[ \_\_\_\_ \] year(s) and \[ \_\_\_\_ \] months.---**Explanation:**To solve this problem, you can use the compound interest formula:\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \]Where:- $ A $ is the final amount ($11,100),- $ P $ is the principal amount ($7,700),- $ r $ is the annual interest rate (5.7%, or 0.057),- $ n $ is the number of times interest is compounded per year (12 for monthly compounding),- $ t $ is the time in years.To find $ t $, rearrange the formula and solve using logarithms:\[ t = \frac{\log\left(\frac{A}{P}\right)}{n \cdot \log\left(1 + \frac{r}{n}\right)} \]This will give the time in years. To convert the decimal part of the years into months, multiply by 12. Fill in the blanks with the calculations from solving this equation.

The equation of compound interest is given by; A=P1+rnnt Where; A→Final amountP→Principal amountr→interest raten→number of times interest applied per time period t→number of time periods elapsedNow in the given question principal amount(P) is $7700 and the final amount (A) is $11,100. The percentage of interest is 5.7%. Therefore the rate of interest is; 5.7100=0.057 And it is also given that the interest compound monthly. That is it compound 12 times a year. Therefore the value of n is 12. Now applying all these in the above equation and solving for t as follows; A=P1+rnnt11100=77001+0.0571212t111007700=1+0.0571212tlog111007700=log1+0.0571212t=12t log1+0.05712log111007700log1+0.05712=12tt=112log111007700log1+0.05712=6.43 years. Now there are 12 months in a year .Therefore 0.43 years will be; 0.43·12 =5.16 months =6 months Therefore the initial amount will become equal to the given amount is 6 years and 6 months.