This is just a substitution problem.

A = A02^(-t/5760)

Rewriting that,

A = (Ao)*[2^(-t/5760)]

Let us say that is correct.

(It is. I am more used to A = (Ao)*[(1/2)^(t/5760)] in this case.)

Given:

Ao = 120 mg

A = 100 mg

So, just substitute them into the equation.

Remember, we need to isolate the t, so we do anything to isolate it.

100 = (120)*[2^(-t/5760)]

Divide both sides by 120,

100/120 = 2^(-t/5760)

0.83333333333 = 2^(-t/5760)

Get the logarithms (common log is okay) of both sides,

log(0.8333333333) = (-t/5760)*log(2)

-0.0791812 = (-t/5760)*(0.30103)

-0.0791812 = t(-0.30103/5760)

-0.0791812 = t(-0.00005226)

t = (-0.0791812)/(-0.00005226)

t = 1515 years

Therefore, to the nearest hundred, the specimen is 1,500 years old. ----answer.

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You can use the natural log, or Ln, if you like, and the answer will be the same.

No need to solve it that long if you can skip steps. If you're good with calculators, you need only one or two steps to get t at once.