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I have no idea how to set up this problem. I am aware of the formula $$A = Pe^{rt}$$ Assume the cost of a gallon of milk is $2.90. With continuous compounding, find the time it would take the cost to be 5 times as much (to the nearest tenth of a year), at an annual inflation rate of 6%.

I also know that the 6% goes in for the r (rate) as such

$$A = Pe^{.06t}$$

user161304
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  • Set $A=5P$ and solve. – T.J. Gaffney Jul 14 '14 at 04:35
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    I do not agree about $Pe^{0.06 t}$. – André Nicolas Jul 14 '14 at 04:36
  • Wait I'm confused so its $$ A = 5e^{rt}$$ I'm really confused on the responses here. $$A = Pe^{rt}$$ So if 6% is not my rate then where do I begin with this? – user161304 Jul 14 '14 at 04:57
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    There are many conventions for quoting interest rate. In the formula $A = Pe^{r_ct}$, the interest rate $r_c$ there is the one quoted in continuous compounding convention.

    In the market, when you are given an interest bearing security which matures $T$ year from now. the corresponding interest rate $r$ is typically quoted according the convention

    $$ e^{r_cT} = \begin{cases}1+ rT,&T < 1\ (1+r)^T,& T \ge 1\end{cases}$$

    In particular, when one say the annual inflation rate $r$ is $6%$, you have $e^{r_c} = 1.06$ and the formula you have becomes $A = P e^{r_ct} = P (1.06)^t$.

     – achille hui Jul 14 '14 at 05:18

2 Answers2

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You don't actually care what the current cost is. You are asked to find the $t$ that corresponds to $\frac AP=5$, which is when the price of anything has been multiplied by $5$. Your equation $A=Pe^{0.06t}$ gives an annual rate higher than $6\%$-plug in $t=1$ to get $\frac AP\approx 1.0618$ for a annual rate of $6.18\%$

Ross Millikan
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You can use $A=Pe^{rt}$ if you want. But then we have to find $r$. In one year, the price goes from $1$ to $1.06$, so $1.06=e^r$, giving $r=\ln(1.06)$. Now continue.

Alternately, use the equation $A=P(1.06)^t$. Let $q$ be the amount of time it takes for the price to quintuple. Put $A=5P$. Then we are solving the equation $(1.06)^q=5$. Take the logarithm of both sides, preferably the natural logarithm. We get $$q\ln(1.06)=\ln(5),$$ and now $q$ is easy to find.

André Nicolas
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  • The 1 in the equation $$A=P(1.06)^t$$ comes from the year right? So you did a separate function of $$A=5P$$. Based off of what I was taught in class I don't understand this $$A=5P$$. I understand where we can get $$A=P(1.06)^t$$ out of $$A=Pe^{rt}$$. I don't quite understand how to get to $$q\ln(1.06)=\ln(5)$$ or what to do with $$q\ln(1.06)=\ln(5)$$. – user161304 Jul 14 '14 at 14:53
  • We had $(1.06)^q=5$. In general, we have $\ln(a^b)=b\ln a$. That is how I got that $\ln((1-06)^q)=q\ln(1.06)$. Then, to finish, we divide both sides by $\ln(1.06)$, getting $w=\frac{\ln(5)}{\ln(1.06)}$. The numerical work is now for the calculator, or appropriate program. As to the $A=5P$, the question asked when the prices get multiplied by $5$. That asks for when an amount $P$ becomes $5P$. So we need to put $A=5P$. That gives $5P=P(1.06)^w$. Cancel the $P$ from both sides and we get the equation of the answer. – André Nicolas Jul 14 '14 at 15:05
  • The $1$ in the equation $A=P(1.06)^t$ comes from the fact that in $1$ year, the cost is increased by $6%$, so what used to cost $1$ dollar costs, after $1$ year, $1.06$ dollars. Or else use the PERT formula, but you need to find $r$, which is not $0.06$. – André Nicolas Jul 14 '14 at 15:08