Carbon-14 dating problem solved with logarithms

Carbon-14 Dating Solution
Use: *f(t) = f(0) e^kt* k is constant; t is time. First find value of k. Enter value for 5730 years into: f(t) = f(0) e^kt 50% = 100% e^{k * 5730} ½ = e^5730k ln½ = 5730k k = $\frac{ln ½}{5730}$ Now use k in original formula with C¹⁴ at $\frac{1}{10}$ f(t) = f(0) e^kt $\frac{f(1000)}{f(0)}$ = $\frac{1}{10}$ = e^{( $\frac{ln ½}{5730}$ t)} $\frac{1}{10}$ = e^{(ln½ * $\frac{t}{5730}$ )} $\frac{1}{10}$ = ( $\frac{1}{2})$ ^{$\frac{t}{5730}$} ln $\frac{1}{10}$ = ln(( $\frac{1}{2})$ ^{$\frac{t}{5730}$}) ln $\frac{1}{10}$ = $\frac{t}{5730}$ * ln½ $\frac{ln(1/10)}{ln(1/2)}$ = $\frac{t}{5730}$ t = 5730 * $\frac{ln(1/10)}{ln(1/2)}$ t = 5730 * 3.321928 ≈ 19,034.65 years.
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Carbon-14 Dating
Solution

Use: f(t) = f(0) e^kt k is constant; t is time.
First find value of k.
Enter value for 5730 years into: f(t) = f(0) e^kt
50% = 100% e^{k * 5730}
½ = e^5730k
ln½ = 5730k
k = $\frac{ln ½}{5730}$

Now use k in original formula with C¹⁴ at

\frac{1}{10}

f(t) = f(0) e^kt

$\frac{f(1000)}{f(0)}$ = $\frac{1}{10}$ = e^{( $\frac{ln ½}{5730}$ t)}

$\frac{1}{10}$ = e^{(ln½ * $\frac{t}{5730}$ )}

$\frac{1}{10}$ = ( $\frac{1}{2})$ ^{$\frac{t}{5730}$}

ln $\frac{1}{10}$ = ln(( $\frac{1}{2})$ ^{$\frac{t}{5730}$})

ln $\frac{1}{10}$ = $\frac{t}{5730}$ * ln½

$\frac{ln(1/10)}{ln(1/2)}$ = $\frac{t}{5730}$

t = 5730 * $\frac{ln(1/10)}{ln(1/2)}$

t = 5730 * 3.321928 ≈ 19,034.65 years.

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Carbon-14 DatingSolution

Carbon-14 Dating
Solution