A person invest $10,000 in a bank. The bank pays 7% interest compounded annually. To the nearest 10th of a year how long must a person leave the money in the bank until it reaches $18,900
To know the future value (FV) of certain amount of money that is going to stay in the bank for "n" periods of time, we apply the following formula (interest compounded anually, then interest rate "i" is annual): [tex]FV=PV\times(1+i)^n[/tex], where "PV" is the present value, in this case, PV=$10,000.
We know that the future value we want to get is FV=$18,900, that the interest rate i=0.07 and that PV=$10,000. We should find the amount of time that takes to reach the $18,900: we should clear "n" in the formula.
Let's do some algebra: [tex]FV=PV\times(1+i)^n[/tex]⇒ [tex]\frac{FV}{PV} =(1+i)^n[/tex] Applying logarithms we get⇒[tex]ln(\frac{FV}{PV})=n\times{ln(1+i)}[/tex] .Then, [tex]\frac{ ln(\frac{FV}{PV})}{{ln(1+i)}}=n[/tex]
Applying this last result we get n=9.4 years, which means 9 years and almost 5 months.
