You currently have $3,000 (Present Value) in an account that has an interest rate of 7.5% per yearcompounded semi-annually (2 times per year). You want to withdraw all your mone…

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asked Apr 5, 2023 41.0k views

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Manuel Leuenberger · Answer 1 · 2023-04-08T12:59:34+0000

Step-by-step explanation

We can use the future value formula to solve the exercise.

$\begin{gathered} FV=PV(1+(r)/(n))^(nt) \\ \text{ Where} \\ FV\text{ is the future value} \\ PV\text{ is the present value} \\ \text{r is the yearly rate} \\ t\text{ is the number of years the money is invested} \end{gathered}$

In this case, we have:

$\begin{gathered} FV=6600 \\ PV=3000 \\ r=7.5\%=(7.5)/(100)=0.075 \\ n=2 \\ t=? \end{gathered}$
$\begin{gathered} FV=PV(1+(r)/(n))^(nt) \\ 6,600=3,000(1+(0.075)/(2))^(2t) \\ 6,600=3,000(1+0.0375)^(2t) \\ 6,600=3,000(1.0375)^(2t) \\ \text{ Divide by 3000 from both sides} \\ (6,600)/(3,000)=(3,000(1.037,5)^(2t))/(3,000) \\ 2.2=(1.0375)^(2t) \\ \text{ Apply natural logarithm from both sides} \\ \ln(2.2)=\ln((1.0375)^(2t)) \\ \text{ Apply the power property of natural logarithms }\ln(m^p)=p\ln(m) \\ \operatorname{\ln}(2.2)=2t\operatorname{\ln}(1.0375) \\ \text{ Divide by }2\operatorname{\ln}(1.0375)\text{ from both sides} \\ \frac{\operatorname{\ln}(2.2)}{2\operatorname{\ln}(1.0375)}=\frac{2t\operatorname{\ln}(1.0375)}{2\operatorname{\ln}(1.0375)} \\ 10.7\approx t\Rightarrow\text{ The symbol }\approx\text{ is read 'approximately'} \end{gathered}$ Answer

The number of years is 10.7.

You currently have $3,000 (Present Value) in an account that has an interest rate of 7.5% per yearcompounded semi-annually (2 times per year). You want to withdraw all your mone…

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