# Thread: repeated division ... what formula to use ?

1. ## repeated division ... what formula to use ?

im not sure what kind of formula to use for the below.

i want to divide a variable number by a constant number and need to know how many times it took to divide it to get below x number?
such as:
4650 is the variable number, 6 is the constant, and 20 is x.
4650 divided by 6, then the product of that divided by 6, etc etc.
(4650/6=775--> 775/6-129.167--> 129.167/6=21.528--> 21.528/6=3.588)
i need to know how many divisions it takes to get below x
so somewhere between the 3rd and 4th division is the number im looking for.

i need to be able to change the variable, but the constant and x would always be the same.

does that make sense? any idea as to what kind of equation i need to use for this type problem?

2. ## re: repeated division ... what formula to use ?

have you been taught indices and logs? if so, solve this equation for n:

$V \cdot (\frac{1}{6^n}) = 20$

Spoiler:

$\frac{1}{6^n} = \frac{20}{V}$

$\frac{1}{6^n} = \frac{20}{V}$

$-n = \log_6 \frac{20}{V}$

$-n = \frac{\ln \frac{20}{V}}{\ln 6}$

$n = -1 \times \frac{\ln \frac{20}{V}}{\ln 6}$

(optional: simplify using log rules)
$n = \frac{\ln V - \ln 20}{\ln 6}$

in the case V=4650 this gives n=3.04, which is "somewhere between 3 and 4" as you expect. The number wont fall below x until the 4th division is completed.

3. ## re: repeated division ... what formula to use ?

Originally Posted by SpringFan25
have you been taught indices and logs? if so, solve this equation for n:

$V \cdot (\frac{1}{6^n}) = 20$
i never was taught that or simply dont recall it. can you help educate me? do you know if this type equation can be solved in an excel spreadsheet formula?

4. ## re: repeated division ... what formula to use ?

The solution is given in the spoiler and is $n = \frac{\ln V - \ln 20}{\ln 6}$

You will need to round the answer up if it isn't an integer.