Question 242015
We use M(x) = c(.5^(x/h)), where c = initial mass, h = half-life and x = years.

NOTE: M(x) = the total amount of grams after certain years.

We first need to solve for h.

1.3 = 10(.5^(100/h))

1.3/10 = .5^(100/h)

0.13 = .5^(100/h)

Take log of both sides.

ln(0.13) = ln(.5^(100/h)

ln(0.13) = (100/h)(ln(.5))

Multiply both sides by h.

h(ln(0.13)) = 100(ln(.5))

Divide both sides by ln(0.13) to find h.

h ≈ 100(ln(.5))/ln(0.13)

h ≈ 33.97412529

We can round that off to the nearest year and the decimal number becomes 34 as the half-life.

I did enough.  All you have to do now is replace x with 100 and simplify and then do the same but using x = 1000 in the formula and simplify.

Can you take it from here?