Can decay percent be less than $-100\%$?

Question

I have this problem:

$$A=A_0\left(1+\frac{r}{n}\right)^{nt}$$

where $A=100$, $A_0=25$, $n=1$, and $t=2$.

This leads to

\begin{align*} (1+r)^2=4 &\quad\Rightarrow\quad |1+r|=2\\ &\quad\Rightarrow\quad 1+r = +2 \text{ or } -2 \\ &\quad\Rightarrow\quad r= 1 \text{ or } -3. \end{align*}

Can I have $r$ which is rate of decay $= -3$, which is $-300\%$?

Answer 1

Exponential functions $b^x$ imply $b\in(0,\infty)\setminus\{1\}$. Therefore, you do not have $-(1+r)$ is part of your solution.

Answer 2

Assuming that $-3$ is a legit value, the sequence over time would be

$$25,-50,100,-200,400,\cdots$$ which I would not call a decay.

You should question the a priori values of $r$.

Answer 3

There is a problem in your forumulas. Substituting your data in the formulas you get

$$25(1+r)^2=100$$

So is clearly meaning that $r$ is not a dacay rate but an increasing rate:

$r=\sqrt{4}-1=100$%

The result is correct and you can easily check it

Start: 25

1° period: 25+100%$\times$25=50

2° period: 50+100%$\times$50=100

If you want to have the decay rate you can modify your forumula in the following way:

$$100(1-r)^2=25$$

$r=1-\sqrt{\frac{1}{4}}=1-\frac{1}{2}=50$%

The result is correct and you can easily check it

Start: 100

1° period: 100-50%$\times$100=50

2° period: 50-50%$\times$50=25

Notes:

$n=1$ means that the result of $r$ is expressed as ratio per 1 period (one year, one month a.s.o.)

$r$ as a ratio, is always meant non negative