I was doing a probability question where it was required to find out the least integral value of n with $$f(n)=\left(\frac 9 {10}\right)^n<\frac{1}{2}$$
A little calculation(by calculator of course) yields $\min(n)=7$.
as:
My question is there a good way to approximately tell which integer $n$ can be without computational software and log tables (even $\log 2,\log 3$ etc.)vCalculus seems to be an unlikely option
hint
For small $ X $, We can write
$$(1-X)^n\approx 1-nX$$
or, for more accuracy $$(1-X)^n\approx 1-nX+\frac{n(n-1)}{2}X^2$$ and
$$\frac{9}{10}=1-\frac{1}{10}$$
Rewrite what you want as
$$2\approx\left(10\over9\right)^n=\left(\left(1+{1\over9}\right)^9\right)^{n/9}\approx e^{n/9}$$
If you allow yourself to know that $\ln2\approx0.7$ (which is the basis for the "Rule of $70$" described in Rivers McForge's answer), then you get $n\approx9\ln2\approx6.3$, which accords with what you found.
As
$$
\left(\frac{9}{10}\right)^n \lt\frac{1}{2} \Rightarrow \left(\frac{10}{9}\right)^n \gt 2
$$
$$
\Rightarrow \left(1 +\frac{1}{9}\right)^n\gt 2
$$
Bernoulli's Inequality shows $n=9$ can be the solution.
Now we start descending from $n=9$ to $n=8$ and $n=7$ and eventually see $n=6$ does not suffice, hence the solution is $n=7$.
The informal heuristic you're looking for is called the "Rule of 70". In its most typical form, it says that if a quantity $A(t)$ increases by a percentage rate $r$ every unit of time, or $$A(t+1) = (1+r) A(t),$$ then the time it takes to double is approximately $70/r$. It can also be used for constant percentage decreases: if $$A(t+1) = (1 - r)A(t),$$ where $r$ is the percent decrease per unit of time, then the "half-life" of $A(t)$ (i.e. the time $\lambda$ it takes for $A(\lambda) = A(0)/2$) is again approximately $70/r$.
In this case, you have a constant percent decrease of $r = 0.10$, or 10%, so the Rule of 70 predicts the halving time should be $$70/10 = 7,$$ which is exactly what we found by checking each case.
