{"id":1191,"date":"2025-07-23T20:09:49","date_gmt":"2025-07-23T20:09:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=1191"},"modified":"2026-03-18T21:38:49","modified_gmt":"2026-03-18T21:38:49","slug":"exponential-and-logarithmic-models-learn-it-2-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/exponential-and-logarithmic-models-learn-it-2-2\/","title":{"raw":"Exponential and Logarithmic Models: Learn It 2","rendered":"Exponential and Logarithmic Models: Learn It 2"},"content":{"raw":"

Exponential Growth and Decay Cont.<\/h2>\r\n

Half-Life<\/h3>\r\nWe now turn to exponential decay<\/strong>. One of the common terms associated with exponential decay, as stated above, is half-life<\/strong>. \u00a0Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.\r\n\r\n
\r\n
\r\n

half-life<\/h3>\r\nHalf-life<\/strong> is the length of time it takes an exponentially decaying quantity to decrease to half its original amount.\r\n\r\n<\/div>\r\n

[latex]t = \\frac{\\ln\\left(\\frac{1}{2}\\right)}{r}[\/latex]<\/p>\r\n\r\n<\/section>[reveal-answer q=\"648283\"]Where does the half-life formula come from?[\/reveal-answer]\r\n[hidden-answer a=\"648283\"]To find the half-life of a function describing exponential decay, solve the following equation:<\/span>\r\n

\r\n

[latex]\\frac{1}{2}{A}_{0}={A}_{o}{e}^{rt}[\/latex]<\/p>\r\nWe find that the half-life depends only on the constant [latex]r[\/latex] and not on the starting quantity [latex]{A}_{0}[\/latex].\r\n\r\nThe formula is derived as follows\r\n

[latex]\\begin{align} \\frac{1}{2}A_{0} &= A_{0}e^{rt} \\\\ \\frac{1}{2} &= e^{rt} & \\text{Divide both sides by } A_{0}. \\\\ \\ln\\left(\\frac{1}{2}\\right) &= rt & \\text{Take the natural log of both sides.} \\\\ \\frac{\\ln\\left(\\frac{1}{2}\\right)}{r} &= t & \\text{Divide by } r. \\end{align}[\/latex]<\/p>\r\nThe formula of [latex]t = \\frac{\\ln\\left(\\frac{1}{2}\\right)}{r}[\/latex] can also be written as [latex]t = \\frac{\\ln\\left(\\frac{1}{2}\\right)}{r} = \\frac{\\ln(2^{-1})}{r}\u00a0 =\u00a0 -\\frac{\\mathrm{ln}\\left(2\\right)}{r}[\/latex].\r\n\r\nSince [latex]t[\/latex], the time, is positive, [latex]r[\/latex] must, as expected, be negative. So, you might see that the half-life formula is written as:\r\n

[latex]t=-\\frac{\\mathrm{ln}\\left(2\\right)}{r}[\/latex]<\/p>\r\n\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\nOne such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life<\/strong>. The table below\u00a0lists the half-life for several of the more common radioactive substances.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Substance<\/th>\r\nUse<\/th>\r\nHalf-life<\/th>\r\n<\/tr>\r\n<\/thead>\r\n
gallium-67<\/td>\r\nnuclear medicine<\/td>\r\n[latex]80[\/latex] hours<\/td>\r\n<\/tr>\r\n
cobalt-60<\/td>\r\nmanufacturing<\/td>\r\n[latex]5.3[\/latex] years<\/td>\r\n<\/tr>\r\n
technetium-99m<\/td>\r\nnuclear medicine<\/td>\r\n[latex]6[\/latex] hours<\/td>\r\n<\/tr>\r\n
americium-241<\/td>\r\nconstruction<\/td>\r\n[latex]432[\/latex] years<\/td>\r\n<\/tr>\r\n
carbon-14<\/td>\r\narcheological dating<\/td>\r\n[latex]5,715[\/latex] years<\/td>\r\n<\/tr>\r\n
uranium-235<\/td>\r\natomic power<\/td>\r\n[latex]703,800,000[\/latex] years<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:\r\n

[latex]\\begin{array}{l}A\\left(t\\right)={A}_{0}{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{T}t}\\hfill \\\\ A\\left(t\\right)={A}_{0}{e}^{\\mathrm{ln}\\left(0.5\\right)\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left({e}^{\\mathrm{ln}\\left(0.5\\right)}\\right)}^{\\frac{t}{T}}\\hfill \\\\ A\\left(t\\right)={A}_{0}{\\left(\\frac{1}{2}\\right)}^{\\frac{t}{T}}\\hfill \\end{array}[\/latex]<\/p>\r\nwhere\r\n

    \r\n \t
  • [latex]{A}_{0}[\/latex] is the amount initially present<\/li>\r\n \t
  • [latex]T[\/latex]\u00a0is the half-life of the substance<\/li>\r\n \t
  • [latex]t[\/latex]\u00a0is the time period over which the substance is studied<\/li>\r\n \t
  • [latex]A[\/latex], or [latex]A(t)[\/latex],\u00a0is the amount of the substance present after time [latex]t[\/latex]<\/li>\r\n<\/ul>\r\n
    How To: Given the half-life, find the decay rate<\/strong>\r\n
      \r\n \t
    1. Write [latex]A={A}_{o}{e}^{kt}[\/latex].<\/li>\r\n \t
    2. Replace [latex]A[\/latex]\u00a0by [latex]\\frac{1}{2}{A}_{0}[\/latex] and replace [latex]t[\/latex]\u00a0by the given half-life.<\/li>\r\n \t
    3. Solve to find [latex]k[\/latex]. Express [latex]k[\/latex]\u00a0as an exact value (do not round).<\/li>\r\n<\/ol>\r\nNote: It is also possible to find the decay rate using<\/em> [latex]k=-\\frac{\\mathrm{ln}\\left(2\\right)}{t}[\/latex].\r\n\r\n<\/section>
      Uranium-235 has a half-life of [latex]703,800,000[\/latex] years. How long will it take for [latex]10\\%[\/latex] of a [latex]1000[\/latex]-gram sample of uranium-235 to decay?\r\n[reveal-answer q=\"536683\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"536683\"]\r\n

      [latex]\\begin{array}{l}\\text{ }y=\\text{1000}e\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\hfill \\\\ \\text{ }900=1000{e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{After 10% decays, 900 grams are left}.\\hfill \\\\ \\text{ }0.9={e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\hfill & \\text{Divide by 1000}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\mathrm{ln}\\left({e}^{\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t}\\right)\\hfill & \\text{Take ln of both sides}.\\hfill \\\\ \\mathrm{ln}\\left(0.9\\right)=\\frac{\\mathrm{ln}\\left(0.5\\right)}{\\text{703,800,000}}t\\hfill & \\text{ln}\\left({e}^{M}\\right)=M\\hfill \\\\ \\text{}\\text{}t=\\text{703,800,000}\\times \\frac{\\mathrm{ln}\\left(0.9\\right)}{\\mathrm{ln}\\left(0.5\\right)}\\text{years}\\hfill & \\text{Solve for }t.\\hfill \\\\ \\text{}\\text{}t\\approx \\text{106,979,777 years}\\hfill & \\hfill \\end{array}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nTen percent of [latex]1000[\/latex] grams is [latex]100[\/latex] grams. If [latex]100[\/latex] grams decay, the amount of uranium-235 remaining is [latex]900[\/latex] grams.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

      The half-life of carbon-14 is [latex]5,730[\/latex] years. Express the amount of carbon-14 remaining as a function of time, [latex]t[\/latex].[reveal-answer q=\"594419\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"594419\"]This formula is derived as follows.\r\n

      [latex]\\begin{array}{l}\\text{}A={A}_{0}{e}^{kt}\\hfill & \\text{The continuous growth formula}.\\hfill \\\\ 0.5{A}_{0}={A}_{0}{e}^{k\\cdot 5730}\\hfill & \\text{Substitute the half-life for }t\\text{ and }0.5{A}_{0}\\text{ for }f\\left(t\\right).\\hfill \\\\ \\text{}0.5={e}^{5730k}\\hfill & \\text{Divide both sides by }{A}_{0}.\\hfill \\\\ \\mathrm{ln}\\left(0.5\\right)=5730k\\hfill & \\text{Take the natural log of both sides}.\\hfill \\\\ \\text{}k=\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\hfill & \\text{Divide by the coefficient of }k.\\hfill \\\\ \\text{}A={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}\\hfill & \\text{Substitute for }r\\text{ in the continuous growth formula}.\\hfill \\end{array}[\/latex]<\/p>\r\nThe function that describes this continuous decay is [latex]f\\left(t\\right)={A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]. We observe that the coefficient of t<\/em>, [latex]\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\approx -1.2097x10^{-4}[\/latex] is negative, as expected in the case of exponential decay.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

      Radiocarbon Dating<\/h3>\r\nThe formula for radioactive decay is important in radiocarbon dating\u00a0<\/strong>which is used to calculate the approximate date a plant or animal died. Radiocarbon dating was discovered in 1949 by Willard Libby who won a Nobel Prize for his discovery. It compares the difference between the ratio of two isotopes of carbon in an organic artifact or fossil to the ratio of those two isotopes in the air. It is believed to be accurate to within about [latex]1\\% [\/latex] error for plants or animals that died within the last [latex]60,000[\/latex] years.\r\n\r\nCarbon-14 is a radioactive isotope of carbon that has a half-life of [latex]5,730[\/latex] years. It occurs in small quantities in the carbon dioxide in the air we breathe. Most of the carbon on Earth is carbon-12 which has an atomic weight of [latex]12[\/latex] and is not radioactive. Scientists have determined the ratio of carbon-14 to carbon-12 in the air for the last [latex]60,000[\/latex] years using tree rings and other organic samples of known dates\u2014although the ratio has changed slightly over the centuries.\r\n\r\nAs long as a plant or animal is alive, the ratio of the two isotopes of carbon in its body is close to the ratio in the atmosphere. When it dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-14 to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the plant or animal died can be approximated.\r\n\r\nSince the half-life of carbon-14 is [latex]5,730[\/latex] years, the formula for the amount of carbon-14 remaining after [latex]t[\/latex]\u00a0years is\r\n

      [latex]A\\approx {A}_{0}{e}^{\\left(\\frac{\\mathrm{ln}\\left(0.5\\right)}{5730}\\right)t}[\/latex]<\/p>\r\nwhere\r\n

        \r\n \t
      • [latex]A[\/latex]\u00a0is the amount of carbon-14 remaining<\/li>\r\n \t
      • [latex]{A}_{0}[\/latex] is the amount of carbon-14 when the plant or animal began decaying.<\/li>\r\n<\/ul>\r\nTo find the age of an object we solve this equation for [latex]t[\/latex]:\r\n

        [latex]t=\\frac{\\mathrm{ln}\\left(\\frac{A}{{A}_{0}}\\right)}{-0.000121}[\/latex]<\/p>\r\nFrom the equation [latex]A\\approx {A}_{0}{e}^{-0.000121t}[\/latex] we know the ratio of the percentage of carbon-14 in the object we are dating to the percentage of carbon-14 in the atmosphere is [latex]r=\\frac{A}{{A}_{0}}\\approx {e}^{-0.000121t}[\/latex]. We solve this equation for [latex]t[\/latex], to get\r\n

        [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex]<\/p>\r\n\r\n

        How To: Given the percentage of carbon-14 in an object, determine its age<\/strong>\r\n
          \r\n \t
        1. Express the given percentage of carbon-14 as an equivalent decimal [latex]r[\/latex].<\/li>\r\n \t
        2. Substitute for [latex]r[\/latex] in the equation [latex]t=\\frac{\\mathrm{ln}\\left(r\\right)}{-0.000121}[\/latex] and solve for the age, [latex]t[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
          A bone fragment is found that contains [latex]20\\%[\/latex] of its original carbon-14. To the nearest year, how old is the bone?[reveal-answer q=\"850584\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"850584\"]To solve for the age of the bone fragment, we can use the exponential decay formula:[latex]A = A_0e^{rt}[\/latex]Given that the bone fragment contains [latex]20\\% = 0.20[\/latex] of its original carbon-14, this means that [latex]A_0 = 100\\% = 1[\/latex] and [latex]A = 0.20[\/latex].\r\n

          [latex]\\begin{align} T_{1\/2} &= 5730 \\text{ years} \\\\ r &= \\frac{\\ln(2)}{5730} \\\\ 0.20 &= e^{rt} \\\\ \\ln(0.20) &= rt & \\text{Take the natural log of both sides.} \\\\ t &= \\frac{\\ln(0.20)}{r} \\\\ t &= \\frac{\\ln(0.20)}{\\frac{\\ln(2)}{5730}} \\\\ t &= \\frac{\\ln(0.20) \\cdot 5730}{\\ln(2)} \\\\ t &\\approx \\frac{-1.60944 \\cdot 5730}{0.69315} \\\\ t &\\approx \\frac{-9219.5192}{0.69315} \\\\ t &\\approx 13299.69 \\approx 13300 \\text{ years old.} \\end{align}[\/latex]<\/p>\r\nAnalysis of the Solution<\/strong>\r\n\r\nThe instruments that measure the percentage of carbon-14 are extremely sensitive and, as we mention above, a scientist will need to do much more work than we did in order to be satisfied. Even so, carbon dating is only accurate to about [latex]1\\%[\/latex], so this age should be given as [latex]\\text{13,300 years}\\pm \\text{1% or 13,300 years}\\pm \\text{133 years}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

          [ohm_question hide_question_numbers=1]321547[\/ohm_question]<\/section>
          [ohm_question hide_question_numbers=1]321548[\/ohm_question]<\/section>","rendered":"

          Exponential Growth and Decay Cont.<\/h2>\n

          Half-Life<\/h3>\n

          We now turn to exponential decay<\/strong>. One of the common terms associated with exponential decay, as stated above, is half-life<\/strong>. \u00a0Every radioactive isotope has a half-life, and the process describing the exponential decay of an isotope is called radioactive decay.<\/p>\n

          \n
          \n

          half-life<\/h3>\n

          Half-life<\/strong> is the length of time it takes an exponentially decaying quantity to decrease to half its original amount.<\/p>\n<\/div>\n

          [latex]t = \\frac{\\ln\\left(\\frac{1}{2}\\right)}{r}[\/latex]<\/p>\n<\/section>\n