Question

The equation F = e^-kt describes the fraction F of an original isotope remaining after a period of t years; the exponent is negative because it refers to a decrease over time. The constant k provides a measure of how rapidly the original isotope decays. For the decay of carbon-14 to nitrogen-14, k = 0.00012097. To find t, rearrange the equation by following these steps: (a) Take the natural logarithm of both sides of the equation: ln(F ) = ln(e-^kt). Rewrite the right side of this equation by applying the following rule: ln(e^x) = x ln(e). (b) Since ln(e) = 1, simplify the equation. (c) Now solve for t and write the equation in the form “t = ________.”

Short answer

(a)\(\ln (F) = \, - kt\ln (e)\)

(b)\(\ln (F) = \, - kt\)

(c)\(t = - \frac{{\ln (F)}}{k}\)

Step-by-step solution

Natural log of the equation F = e-kt

The equation F = e^-ktdescribes the fraction of an isotope left after a specific period. By taking the natural logarithm of both the right and left sides of the equation, it is rearranged as:

\(\ln (F) = \ln ({e^{ - kt}})\)

Since\(\ln ({e^x}) = x\ln (e)\)

The equation can be written as:

\(\ln (F) = \, - kt\ln (e)\)

Simplified version of the equation

The equation\(\ln (F) = \, - kt\ln (e)\)can be further simplified as\(\ln (e) = 1\). By substituting the value of\(\ln (e) = 1\)in the equation of fraction:

\(\begin{aligned}{l}\ln (F) = - kt \times 1\\\ln (F) = \, - kt\end{aligned}\)

Thus, the simplified form of the equation is\(\ln (F) = \, - kt\).

Equation in the form of ‘t’

The equation of fraction\(\ln (F) = \, - kt\)is rearranged and written as:

\(\begin{aligned}{l}\ln (F) = \, - kt\\t = - \frac{{\ln (F)}}{k}\end{aligned}\)