Quick questions on Exponential growth and decay: dN/dt = kN, Newton's law of cooling and applications

If a quantity $N(t)$ changes at a rate proportional to its current value,

If $T(t)$ is the temperature of a body and $T_a$ is the ambient temperature (assumed constant), then

Differentiate $N(t) = N_0 e^{k t}$ to get $\frac{dN}{dt} = k N_0 e^{k t} = k N$, which matches the equation. The same check works for the cooling solution, since $\frac{d}{dt}(T_a) = 0$ and $\frac{d}{dt}((T_0 - T_a) e^{k t}) = k (T - T_a)$.

For decay with rate constant $k < 0$, the half-life is $\tau = \frac{\ln 2}{|k|}$. Equivalently $N(t) = N_0 \cdot (1/2)^{t / \tau}$.

For growth, the doubling time is $T_d = \frac{\ln 2}{k}$.

Pure exponential change uses $\frac{dN}{dt} = k N$. Cooling involves a constant ambient term: $\frac{dT}{dt} = k(T - T_a)$. Do not forget the $-T_a$.

For decay and cooling, $k$ is negative. The exponent $k t$ should drive the function towards its limit, not away from it.

$N(t) = N_0 \cdot 2^{-t/\tau}$ and $N(t) = N_0 e^{k t}$ describe the same decay only if $k = -\frac{\ln 2}{\tau}$.

In a cooling problem the final temperature approaches $T_a$, not zero. The solution always includes a constant term equal to the ambient temperature.

In Maths Advanced, the natural logarithm $\ln$ (base $e$) is the inverse of $e^x$. Switching to $\log_{10}$ adds an unnecessary factor.