What does $y = Ar^t$ mean?

Question

I keep coming up in this question and its variants in my math tests, but it has not been explained to me so I can't really answer it. As such, I am at risk of failing the year.

An example question is as below:

Bob bought a farm in 1980. In 2017 he sold it for $500,000 and calculated it rose exponentially at a rate of 4% over that time.

(i) Assuming that it increased in value according to the formula $y = Ar^t$, where $A$ is the purchase price and $^t$ is the time in years, what was his profit?

As you can see, the values of $A$ and $^t$ are pretty clearly defined, and $y$ is implicitly defined, but nowhere is the value $r$ defined.

So what is $r$ supposed to be, and how would I go about answering this question?

Answer 1

The $r$ means rate of increase of the price, which is $1 + 4\% = 1.04$.

$$t = 2017 - 1980 = 37 \text{ years}$$ $$y = $500000$$

Now we calculate the value of $A$, which is the purchase price of the farm in 1980.

$$\begin{array}{rcll} y &=& Ar^t\\ 500000&=&A\left(1.04\right)^{37}\\ A&=&\dfrac{500000}{(1.04)^{37}} \\ &\approx&$117148.42 \end{array}$$

Now, the profit is $\text{Selling price} - \text{Purchase price}$: $$\begin{array}{rcll} \text{Profit} &=& $500000 - $117148.42\\ &\approx&$382851.58 \end{array}$$

To clarify, $y = Ar^t$ is an exponential equation. Usually $r$ denotes the rate of increase, or the rate of decrease as $r$ stands for "rate".

Answer 2

I am not sure what is mean by exponential rate, but assuming that it means continuous compounding, we have the formula $y(t) = A e^{0.04 t}$, with $t$ in years.

Hence $y(2017) = y(1980) e^{0.04 (2017-1980)} = 500000$ from which can compute $y(1980)$. Then the profit is given by $y(2017)-y(1980) \approx 386181$.

Answer 3

The exponential function is: $$f(x)=ab^x,b>0.$$ Now $b$ can take various values, in particular: $$b=1+\frac{0.04}{m},m\ne 0$$ Note that when $m=1,2,4,12$, it is the annual, semi-annual, quarterly, monthly interest compounding, respectively. For example, for $x\in \mathbb N$: $$f(x)=a(1+0.04)^x=a\underbrace{(1+0.04)(1+0.04)\cdots (1+0.04)}_{x}\\ f(x)=a\left(1+\frac{0.04}{2}\right)^{2x}=a\underbrace{\left(1+\frac{0.04}{2}\right)\left(1+\frac{0.04}{2}\right)\cdots \left(1+\frac{0.04}{2}\right)}_{2x}.$$

When $m\to \infty$, the formula of continuous interest compounding becomes: $$f(x)=\lim_\limits{m\to \infty} a\left(1+\frac{0.04}{m}\right)^{mx}=ae^{0.04x}.$$

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The OP's problem suggests the exponential function $y=Ar^t$. However, it can be interpreted in two ways: $$y(t)=Ar^t=A(1+0.04)^t=Ag(t)$$ or: $$y(t)=Ar^t=A(e^{0.04})^t=Ag(t)$$ Note: The first is an ordinary exponential function, while the second is a natural exponential function.

Once it is clarified, the rest is a simple calculation with $$\begin{align}y(0)&=A,\\ t&=2017-1980=37,\\ y(37)&=500,000\\ \text{Profit}&=\frac{y(37)}{A}=\frac{500,000}{A}=g(37) \ \ \text{(in percentage of the purchase price)}\\ \text{Profit}&=y(37)-A=\\ &=500,000-\frac{y(37)}{g(37)}=\\ &=500,000-\frac{500,000}{g(37)}\ \ \text{(in dollars)}\end{align}$$