Tax inclusive pricing I have a system where is user can enter a price (without tax) and a tax rate. I then calculate the total cost of the item.
Example:
Price:100.00
Tax percent: 10%
Final price: 110.00 = (100 + (100* (10/100))
I have got a request to work backwards and start with the final price and tax and determine the price without tax.
In my system I store only price without tax and tax percent.
For example if a user wants a final price of 30.00 and a tax percent of 8.25
The starting price in this case has more than 2 decimals.
How many decimals do I need to store to allow for tax inclusive pricing for all possibilities?

I'm assuming you only need your output price accurate to two decimals. The meaning of "X is accurate to $n$ decimals" is that X is an approximation, but the difference between it and the true value is less than $\displaystyle\frac{5}{10^{n+1}}$ (we want to say $1/10^n$ but we have to account for rounding).

Technical details:
What you are saying in more technical language is that you want to calculate the error $\delta$ allowed on the input to have an error $\epsilon<\frac{1}{200}$ on the output.
Suppose the tax rate is $r$%, and the true price of a customer's purchase is $p$. A "perfect computer" would take $p$ as an input and give $p+\frac{r}{100}p$ as an output. Our real computer will use the same function but will not take in $p$ but instead some approimation $\hat p$. So the fully symbolic question is to find a $n$ such that
$$\left|p-\hat{p}\right|<\frac{5}{10^{n+1}} \qquad\Longrightarrow\qquad \left|p+\frac{r}{100}p-\hat{p}-\frac{r}{100}\hat p\right|<\frac{1}{200}$$
The left side of the right equation is:
$$\left|\left(p-\hat p\right)\left(1+\frac{r}{100}\right)\right| = \left|p-\hat p\right|\left(1+\frac{r}{100}\right)<\frac{5}{10^{n+1}}\left(1+\frac{r}{100}\right)$$
So if we could only get $\frac{5}{10^{n+1}}\left(1+\frac{r}{100}\right)<\frac{1}{200}$ then we'd be golden. Solving for $n$:
$$\left(1+\frac{r}{100}\right)<\frac{10^{n+1}}{1000}$$
$$\left(1+\frac{r}{100}\right)<10^{n-2}$$
$$\log_{10}\left(1+\frac{r}{100}\right)<n-2$$
$$2-\log_{10}\left(\frac{100+r}{100}\right)<n$$
$$\log_{10}\left(100+r\right)<n$$
Interest rates are almost surely less than $100$% so $n=3$ suffices here. Note that multiplying $p$ by a constant adds its logarithm to the left side. Summing over many choices for $p$ is bounded by multiplying the number of terms in the summation by the largest $p$, so this is also logarithmic.

The bottom line:
There's not quite enough information to solve the problem but there's enough for a recommendation. Start with three decimals, and add a few more according to these rules:


*

*How many of each kind item is a typical customer going to buy? If it's 1-8 then add nothing, if 9-98 then add one decimal, two decimals for 99-998 etc. 

*How many different items are they likely to buy? Use the same scale.

*Add one more if you have reasonably frequent bulk orders that exceed the above estimations.

