Rate of carbon-14 decay problem solved with logarithms

Carbon-14 Decay Solution
Use: *f(t) = f(0) e^kt* k is constant; t is time. First find value of the constant k. Enter value for 5730 years into: f(t) = f(0) e^kt 50% = 100% e^{k * 5730} ½ = e^5730k ln ½ = 5730k k = $\frac{ln ½}{5730}$ Now use k in original formula with t = 1000 years f(t) = f(0) e^kt f(1000) = f(0) e^{(( $\frac{ln ½}{5730}$ ) * 1000)} $\frac{f(1000)}{f(0)}$ = e^{(ln½ * $\frac{1000}{5730}$ )} $\frac{f(1000)}{f(0)}$ = ( $\frac{1}{2}$ )^{$\frac{1000}{5730}$} $\frac{f(1000)}{f(0)}$ = 0.886074125 ≈ 88.6074125%
333

Carbon-14 Decay Solution

Use: f(t) = f(0) e^kt k is constant; t is time.
First find value of the constant k.
Enter value for 5730 years into: f(t) = f(0) e^kt
50% = 100% e^{k * 5730}
½ = e^5730k
ln ½ = 5730k
k = $\frac{ln ½}{5730}$

Now use k in original formula with t = 1000 years
f(t) = f(0) e^kt
f(1000) = f(0) e^{(( $\frac{ln ½}{5730}$ ) * 1000)}

$\frac{f(1000)}{f(0)}$ = e^{(ln½ *
$\frac{1000}{5730}$ )}

$\frac{f(1000)}{f(0)}$ = ( $\frac{1}{2}$ )^{$\frac{1000}{5730}$}

$\frac{f(1000)}{f(0)}$ = 0.886074125 ≈ 88.6074125%

333