Show the 7-10 rule for fission product activity

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Fission product activity measured at the time t0 following the burst of a nuclear weapon is found to be a0. Show that the activity at the time t=7^nt0 is given approximately by a=a0/10^n. This is known as the 7-10 rule in civil defense.
 
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What equations do you have to work with?

How does it relate to a(t) = a0* exp(-0.693 t/t1/2)?
 
Yes, I was playing around with that one... Exponential decay for activity... I'm not sure what else I can use... At one point, I also used n = n0 *exp (-0.693t/t12), but wasn't sure if the n given in the problem was actually number of atoms or just some arbitrary variable.
 
I noticed that 0.693 is approximately 0.7 or 7/10.

The decaying exponential is a continuous function, and the rule of thumb uses discrete interval of time.

So one could pick a [itex]\Delta{T}[/itex] of one hour or one day.

According to an example, if activity is measured at 1 hr, then 8 hrs ([itex]\Delta{T}[/itex] = 7 hrs), activity should be reduced to 1/10 of the activity at 1 hr. Then at 15 hrs, 1 + 2*7, the activity should be 1/100 of the activity at 1 hr. And so on.
 
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