Ask a first-order reaction how long it takes to lose HALF of itself and something remarkable falls out. Put $[\mathrm{A}] = \tfrac{1}{2}[\mathrm{A}]_0$ into $[\mathrm{A}] = [\mathrm{A}]_0 e^{-kt}$; the starting concentration cancels on both sides and leaves $t_{1/2} = \dfrac{\ln 2}{k}$ — read aloud t-half equals ell-en two over k. $t_{1/2}$, said t-half, is the half-life in seconds; $k$ is the first-order rate constant in $\mathrm{s^{-1}}$; and $\ln 2 = 0.693$ is just a number, the same one every time. The same relation runs backwards as $k = \dfrac{\ln 2}{t_{1/2}}$ when the half-life is what you were told.

Read what cancelled: $[\mathrm{A}]_0$ is GONE. A first-order half-life does not care whether you started with a tanker or a teaspoon — which is why an isotope can be quoted one half-life for all time, and why doctors can dose a drug by the clock. Do not confuse it with $1/k$, the tempting bare reciprocal: that is the time to fall to about 37 % (one over e), and 37 is not 50.

Once you hold $t_{1/2}$, counting down is arithmetic: $N = N_0\left(\tfrac{1}{2}\right)^{t/t_{1/2}}$, where $N$ is whatever is left, $N_0$ is what you started with (any unit you like — grams, moles, counts per minute — as long as both wear the same one), $t$ is the elapsed time and $t_{1/2}$ the half-life in the SAME time unit. The exponent $t/t_{1/2}$ is simply how many half-lives went by, and it does not have to be a whole number. Three half-lives leave an eighth, not a third — halving is repeated, never shared out.