Example:
Carbon-14 has a radioactive half-life of 5,730 years. Use the radioactive decay equation to find the age of a fossil that contains 20% of its original carbon-14.
[tex]\begin{array}{|l|c|} \cline{1-2}& \\ \text{Substitute the known values into the} & 0.2A_0 = A_0(0.5)^{\frac{t}{5730}}\\\text{equation} & \\ \cline{1-2}& \\\text{Isolate the exponential expression} & 0.2 = (0.5)^{\frac{t}{5730}}\\ & \\ \cline{1-2} & \\\text{Rewrite into log form} & \log_{0.5}(0.2) = \frac{t}{5730}\\ & \\ \cline{1-2}\end{array}[/tex]
[tex]\begin{array}{|l|c|} \cline{1-2}& \\ \text{Isolate the variable} & 5730\log_{0.5}(0.2) = t\\& \\ \cline{1-2}& \\ \text{Use the }\underline{\text{chan}}\text{g}\underline{\text{e}} \ \underline{\text{of}} \ \underline{\text{base}}\text{ formula to} & 5730\frac{\log(0.2)}{\log(0.5)} = t\\ \text{rewrite the logarithmic expression and solve} & \\ \text{for t} & 13,304.64798\approx t\\ & \\ \cline{1-2}\end{array}[/tex]
The fossil is approximately 13 thousand years old.
Side note: carbon-14 decays into nitrogen-14 and a beta particle.
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Question: How long will it take $2,000 to double if it is deposited in an account with an interest rate of 5% compounded continuously?
When interest is compounded continuously, the equation [tex]A = Pe^{rt}[/tex] represents the balance, A, after t years, where r is the rate of interest and P is the beginning balance or principal .
Twice $2,000 is $4,000. Substitute these variables into the compound interest formula, isolate the exponential expression and solve for t.
[tex]A = Pe^{rt}\\\\4000 = 2000e^{0.05t}\\\\4000/2000 = e^{0.05t}\\\\2 = e^{0.05t}\\\\\text{Ln}(2) = \text{Ln}(e^{0.05t})\\\\\text{Ln}(2) = 0.05t\\\\\frac{\ln(2)}{0.05} = t\\\\13.8629 \approx t[/tex]
In about 13.86 years, the amount doubles.
