Specifically, I am curious if there is a formula that models how much of a percentage increase would be required after a certain decrease to arrive back at the original number.
For example: A 70% decrease of 100 is 30. What formula shows us what the percentage increase will need to be in order to arrive back at our original value of 100?
Thanks
$$\Bigg(1-\frac{p}{100}\Bigg)\Bigg(1+\frac{\frac{100p}{100-p}}{100}\Bigg)=1$$
When $p=70$ you get $\frac{100p}{100-p}=\frac{7000}{30} \approx 233.33$. Note that many people get confused by large percentage increases and might not be able to tell whether this was more or less than tripling.
First, remember that a percentage is just a fraction, that is $70\%=70/100=0.7$ .
And a decrease of $70\%$ amounts to substracting $70/100$ times the original value, that is
$$y = x - \frac{70}{100}x= ( 1- 0.7)x =0.3x $$
In general, a decrease by some ratio $r$ amounts to multiplying the original number by $1-r$ (and an increase, by $1+r$). Hence, to get back the original number you must divide by $1-r$.
Now, if you have a decrease $d$ and want to calculate the "takeback" increase $i$ that takes the value back to the original, you want
$$ 1+ i = \frac{1}{1-d} \implies i = \frac{1}{1-d}-1 = \frac{d}{1-d}$$
In our example $d=0.7$, $i=0.7/0.3=2.33.3 = 233.33 \%$
Notice that the takeback increase is very different from the original decrease. However, if the decrease was small, then indeed $i \approx d$ (say, $d=1\%=0.01 \implies i=0.10101 = 1.01\% $)
