Mathematics, 31.07.2020 03:01, justinhudson17
You have $1000 to invest in an account and need to have $2000 in one year. What interest rate would you need to have in order to have this if the amount is compounded weekly? Round your answer to the nearest percent.
Answers: 3
![1000\cdot (1+\dfrac{r}{5200})^{52}=2000\\\\\text{*** divide by 1000 both sides ***}\\\\(1+\dfrac{r}{5200})^{52}=\dfrac{2000}{1000}=2\\\\\text{*** take the ln **}\\\\52\cdot ln(1+\dfrac{r}{5200})=ln(2)\\\\\text{*** divide by 52 ***}\\\\ln(1+\dfrac{r}{5200})=\dfrac{ln(2)}{52}\\\\\text{*** take the exp ***}\\\\\displaystyle 1+\dfrac{r}{5200}=exp(\dfrac{ln(2)}{52})=2^{(\dfrac{1}{52})}=\sqrt[52]{2}\\\\r = 5200\cdot (\sqrt[52]{2}-1)=69.77875...](/tpl/images/0715/8592/d4599.png)
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You have $1000 to invest in an account and need to have $2000 in one year. What interest rate would...
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