Or this one: A=A_{o}e^{-kt}

This one.

why is it necessary to adjust to base 10 log?

The natural log scale specifies the amount of time it takes for the concentration to drop from the initial concentration to a concentration of 1/e, not 1/2. This is the difference between half-life and the quantity τ, the reciprocal of the rate constant,

*k*. t

_{1/2} is equal to τ ln (2), or ln (2) / k. So, you solve the problem for the rate constant in the normal way, and then you just have to take this into account when you do your conversion to half-life.

Do you mean that after 5730 years, the rate should also be 460 decays/hr per gram?

This is what he means. With this type of estimation, you should be able to pick the right answer without really calculating anything at all.