[vaz360]

I understand how to use the rate law for first-order reactions but what is the exact reason for it being ln([A]_o/[A]_t) = kt? Why is the natural log there? How was this formula derived?

[UG]

So you start with your 'normal' rate law: d[A]/dt = -k[A]
Then using a technique called separation of variables you get d[A]/[A] = -k dt
Now you integrate both sides. From calculus, the integral of d[A]/[A] is ln [A].
So your equation becomes ln[A] = -kt + C
C is some constant of integration which needs to be found. You can find it because you know one initial condition, that is, the concentration of [A] at the start (time = 0) is [A]₀. You substitute this into the equation: ln[A]₀ = -k(0) + C and you end up with ln[A]₀ = C.
Now the equation becomes ln[A] = -kt + ln[A]₀. With a little bit more rearranging you end up with the equation you have shown, where [A]_t is equivalent to [A] in the equation I just derived.