[{"data":1,"prerenderedAt":-1},["ShallowReactive",2],{"doc-seo-508519-105":3,"doc-detail-508519-en":81,"detail-sidebar-cat-0-en-105":97},{"code":4,"msg":5,"data":6},0,"ok",{"site_id":7,"language":8,"slug":9,"title":10,"keywords":11,"description":12,"schema_data":13,"social_meta":74,"head_meta":76,"extra_data":78,"updated_unix":80},105,"en","compound-growth-and-decay-exponential-models-practice","Compound Growth and Decay - Exponential Models Practice","","Exponential growth and decay are taught through step-by-step worked examples and the use of core exponential models y = a(1 − r)t, y = ae−kt, and y = a(1 + r)t/ae^kt. The text applies the models to elimination of a chemical from a pool, radioactive half-life and determining k by logarithms, and standardized test style questions requiring computation of growth over time and comparison of continuously compounded savings accounts. Methods include algebraic substitution, logarithmic properties, and calculator-based evaluation.",{"@graph":14,"@context":73},[15,34,56],{"@type":16,"itemListElement":17},"BreadcrumbList",[18,23,27,31],{"item":19,"name":20,"@type":21,"position":22},"https://docshare.wps.com","Home","ListItem",1,{"item":24,"name":25,"@type":21,"position":26},"https://docshare.wps.com/document/","Document",2,{"item":28,"name":29,"@type":21,"position":30},"https://docshare.wps.com/document/exam/","Exam",3,{"item":32,"name":10,"@type":21,"position":33},"https://docshare.wps.com/document/compound-growth-and-decay-exponential-models-practice/508519/",4,{"url":32,"name":10,"@type":35,"image":36,"author":41,"headline":10,"publisher":44,"fileFormat":47,"inLanguage":8,"description":12,"dateModified":48,"datePublished":49,"encodingFormat":47,"isAccessibleForFree":50,"interactionStatistic":51},"DigitalDocument",{"url":37,"@type":38,"width":39,"height":40},"https://docshare.wps.com/thumbnails/compound-growth-and-decay-exponential-models-practice/508519.png","ImageObject",300,407,{"name":42,"@type":43},"eBook King","Person",{"url":19,"name":45,"@type":46},"DocShare","Organization","application/pdf","2026-11-11","2026-10-05",true,{"@type":52,"interactionType":53,"userInteractionCount":55},"InteractionCounter",{"@type":54},"ViewAction",8,{"@type":57,"mainEntity":58},"FAQPage",[59,65,69],{"name":60,"@type":61,"acceptedAnswer":62},"How is exponential decay modeled in the pool chemical problem?","Question",{"text":63,"@type":64},"The problem uses y = a(1 − r)t, where a is the initial amount and r is the hourly fraction eliminated. The time is found when y becomes three-fourths eliminated, leaving one-fourth remaining, then added to 8 am.","Answer",{"name":66,"@type":61,"acceptedAnswer":67},"How do you find the constant k using half-life?",{"text":68,"@type":64},"For half-life, when y = 0.5a, the equation y = ae−kt becomes 0.5 = e−k t. Taking natural logs yields ln 0.5 = −kt, so k is computed as −ln(0.5)/t.",{"name":70,"@type":61,"acceptedAnswer":71},"How is continuous compounding used to compare two savings accounts?",{"text":72,"@type":64},"Each account is written using A = Pert, with different principal P and rates r. The time t is solved from the inequality Sue’s amount > Norma’s amount by taking natural logs and rearranging.","https://schema.org",{"og:url":32,"og:type":75,"og:title":10,"og:site_name":45,"og:description":12},"article",{"robots":77,"canonical":32},"index,follow",{"doc_id":79,"site_id":7},508519,1791519142,{"code":4,"msg":82,"data":83},"success",{"doc_id":79,"user_id":84,"nickname":42,"user_avatar":85,"doc_module":4,"category_id":33,"category_name":29,"doc_title":10,"doc_description":12,"doc_content":86,"file_id":87,"file_url":88,"file_type":89,"file_size":90,"view_count":55,"is_deleted":4,"is_public":22,"is_downloadable":22,"audit_status":22,"page_count":33,"language":91,"language_code":8,"site_id":7,"html_lang":8,"table_of_contents":92,"faqs":93,"seo_title":94,"seo_description":12,"update_tm":95,"read_time":96,"preview_page_count":4},962088006270,"https://ap-avatar.wpscdn.com/davatar_085a072bc5b1113ac321206ff7593b45","Compound Growth and Decay  \nExample 1 Exponential Decay of the Formy = a(1 – r)t.  \nRECREATION A particular chemical must be added to a swimming pool at regular intervals because it is released from the water into the air at the rate of 5% per hour. Sixteen ounces of the chemical is added at 8 am. At what time will three–fourths ofthis chemical be gone from the pool?  \nExplore The problem gives the amount of chemical added to the pool and the rate at which the chemical is eliminated. It asks you to find the time it will take for three–fourths ofthe original amount of chemical to be eliminated from the pool. You then must add that time to 8 am to find the time of day.  \nPlan Use the formula y = a(1 – r)t. Let t be the number of hours since adding the chemical. The initial amount a is 16 ounces, and the percent of decrease r is 0.05. The amount remaining y is one–fourth of 16 or 4.  \nSolve y = a(1 – r)t Exponential decay formula  \n4 = 16(1 – 0 .05)t Replace y with 4, a with 16, and r with 0.05.  \n0.25 = (0 .95)t Divide each side by 16.  \nlog 0.25 = log (0 .95)t Property of Equality for Logarithms  \nlog 0.25 = t log (0.95) Product Property for Logarithms  \nlog 0.25  \n= t Divide each side by log 0.95.  \nlog 0.95  \n27.0268 ≈ t Use a calculator.  \nSince 27.0268 ≈ 27, it will take approximately 27 hours for three–fourths ofthe chemical tobe eliminated. The time will be 11 am of the next day.  \nCheck Use the formula to find how much of the original 16 ounces of chemical will remain after 27 hours.  \ny = a(1 – r)t Exponential decay formula  \ny = 16(1 – 0 .05)27 Replace a with 16, r with 0 .05, and t with 27.  \ny ≈ 4.0055 Use a calculator.  \nOne–fourth of 16 is 4, so the answer seems reasonable. Three-fourths ofthe chemical will be eliminated in about 27 hours. This will be at 11 am of the next day.  \nExample 2 Exponential Decay of the Formy = ae–kt.  \nCHEMISTRY The half–life of a radioactive substance is the time it takes for half of the atoms of the substance to decay. Each element has a unique half–life. Radon–222 has a half–life of about 3.8 days, while thorium–234 has a half–life of about 24 days. Find the value ofk for each element and compare their equations for decay.  \nThe equations will be of the formy = ae–kt, where t is in days. To determine the constant k for each element, let a be the initial amount of the substance. The amount y that remains after t days of the half– life is then represented by 0.5a. Use this idea to find the value ofk for each element and then to write their equations.  \nRadon–222  \ny = ae–kt 0.5a = a–e–k(3.8)  \n0.5 = e 3.8k– ln 0.5 = ln e 3.8k ln 0.5 = –3.8k  \nln 0.5 = k  \n–3.8 0.1824 ≈ k  \nThorium–234  \ny = ae–kt 0.5a = ae–k(24)  \n0.5 = e–24k ln 0.5 = ln e–24k ln 0.5 = –24k ln 0.5  \n= k –24  \n0.0289 ≈ k  \nExponential decay formula  \nReplace y with 0.5a and t with 3.8.  \nDivide each side by a.  \nProperty of Equality for Logarithmic Functions Inverse Property of Exponents and Logarithms Divide each side by –3.8.  \nUse a calculator.  \nExponential decay formula  \nReplace y with 0.5a and t with 24.  \nDivide each side by a.  \nProperty of Equality for Logarithmic Functions  \nInverse Property of Exponents and Logarithms Divide each side by –24.  \nUse a calculator.  \nThe equations for radon–222 and thorium–234 are y = ae–0.1824t andy = ae–0.0289t, respectively. For both equations, t represents time in days. In comparing the equations, it appears that the longer the half–life, the smaller the value ofk.  \nStandardized Test EXAMPLE  \nExample 3 Exponential Growth of the Formy = a(1 + r)t.  \nIn 1980, the value of farming land in a region of Wyoming was $150 per acre. Since then, the value has increased by exactly 0.75% per year. If the land continues to increase in value at this rate, what will the approximate value of the land per acre be in 2005?  \nA. $610 B. $330  \nC. $181 D. $175  \nRead the Test Item  \nYou need to find the value of land 2005 – 1980 or 25 years later. Since the land is increasing in value at a f","cbCaivIzenos3h1x","https://ap.wps.com/l/cbCaivIzenos3h1x","pdf",238304,"English","# Exponential Decay with Discrete Rate\n## Swimming Pool Chemical Elimination (5% per hour)\n# Exponential Decay with Half-Life\n## Comparing Radon-222 and Thorium-234\n# Exponential Growth Models\n## Farming Land Increase (1980 to 2005)\n# Continuous Compounding Growth\n## Comparing Sue and Norma Accounts","[{\"question\":\"How is exponential decay modeled in the pool chemical problem?\",\"answer\":\"The problem uses y = a(1 − r)t, where a is the initial amount and r is the hourly fraction eliminated. The time is found when y becomes three-fourths eliminated, leaving one-fourth remaining, then added to 8 am.\"},{\"question\":\"How do you find the constant k using half-life?\",\"answer\":\"For half-life, when y = 0.5a, the equation y = ae−kt becomes 0.5 = e−k t. Taking natural logs yields ln 0.5 = −kt, so k is computed as −ln(0.5)/t.\"},{\"question\":\"How is continuous compounding used to compare two savings accounts?\",\"answer\":\"Each account is written using A = Pert, with different principal P and rates r. The time t is solved from the inequality Sue’s amount \\u003e Norma’s amount by taking natural logs and rearranging.\"}]","Compound Growth and Decay - Exponential Models Practice | PDF",1791225487,10,{"code":4,"msg":82,"data":98},[99,103,107,110,115,120,125,129,134,137,140],{"id":22,"doc_module":4,"doc_module_name":25,"category_name":100,"show_sort_weight":101,"slug":102},"Story & Novel",90,"story-novel",{"id":26,"doc_module":4,"doc_module_name":25,"category_name":104,"show_sort_weight":105,"slug":106},"Literature",80,"literature",{"id":33,"doc_module":4,"doc_module_name":25,"category_name":29,"show_sort_weight":108,"slug":109},70,"exam",{"id":111,"doc_module":4,"doc_module_name":25,"category_name":112,"show_sort_weight":113,"slug":114},5,"Comic",60,"comic",{"id":116,"doc_module":4,"doc_module_name":25,"category_name":117,"show_sort_weight":118,"slug":119},6,"Technology",50,"technology",{"id":121,"doc_module":4,"doc_module_name":25,"category_name":122,"show_sort_weight":123,"slug":124},7,"Healthcare",40,"healthcare",{"id":55,"doc_module":4,"doc_module_name":25,"category_name":126,"show_sort_weight":127,"slug":128},"Research & Report",30,"research-report",{"id":130,"doc_module":4,"doc_module_name":25,"category_name":131,"show_sort_weight":132,"slug":133},9,"Religion & Spirituality",20,"religion-spirituality",{"id":132,"doc_module":4,"doc_module_name":25,"category_name":135,"show_sort_weight":132,"slug":136},"World Cup","world-cup",{"id":96,"doc_module":4,"doc_module_name":25,"category_name":138,"show_sort_weight":96,"slug":139},"Lifestyle","lifestyle",{"id":141,"doc_module":4,"doc_module_name":25,"category_name":142,"show_sort_weight":111,"slug":143},19,"General","general"]