The half-life [tex]t_{1/2}[/tex] is the amount of time it takes for some quantity [tex]N_0[/tex] of carbon-14 to decay to half the original amount, or [tex]N=\frac{N_0}2[/tex].
In terms of the formula, it's the time such that
[tex]\dfrac{N_0}2=N_0e^{-kt_{1/2}}[/tex]
and we can divide both sides by the original amount to get
[tex]\dfrac12=e^{-kt_{1/2}}[/tex]
We want to find the time [tex]t[/tex] it takes for 57%, or 0.57, of the original amount to remain. This means we solve for [tex]t[/tex] in
[tex]0.57N_0=N_0e^{-kt}[/tex]
or
[tex]0.57=e^{-kt}[/tex]
We're given [tex]k=0.0001[/tex]; plug this in and solve for [tex]t[/tex]:
[tex]0.57=e^{-kt}\implies\ln0.57=-kt\implies t=-\dfrac{\ln0.57}k\approx\boxed{5621\,\mathrm{years}}[/tex]
