{"id":236913,"date":"2025-06-16T16:06:45","date_gmt":"2025-06-16T16:06:45","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=236913"},"modified":"2025-06-16T16:06:47","modified_gmt":"2025-06-16T16:06:47","slug":"the-function-dh5e-0-4h-can-be-used-to-find-the-number-of-milligrams-d-of-a-certain-drug-that-is-in-a-patients-bloodstream-h-hours-after-the-drug-has-been-administered","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/06\/16\/the-function-dh5e-0-4h-can-be-used-to-find-the-number-of-milligrams-d-of-a-certain-drug-that-is-in-a-patients-bloodstream-h-hours-after-the-drug-has-been-administered\/","title":{"rendered":"The function d(h)=5e-0.4h can be used to find the number of milligrams d of a certain drug that is in a patient\u2019s bloodstream h hours after the drug has been administered."},"content":{"rendered":"\n

The function d(h)=5e-0.4h can be used to find the number of milligrams d of a certain drug that is in a patient\u2019s bloodstream h hours after the drug has been administered. When the number of milligrams reaches 2, the drug is to be administered again. a. How many milligrams will be present after 1 hour? b. What is the time between injections<\/p>\n\n\n\n

The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n

We are given the exponential decay function:d(h)=5e\u22120.4hd(h) = 5e^{-0.4h}d(h)=5e\u22120.4h<\/p>\n\n\n\n

This function gives the number of milligrams, ddd, of a drug in a patient\u2019s bloodstream hhh hours after administration.<\/p>\n\n\n\n

\n\n\n\n

Part a: How many milligrams will be present after 1 hour?<\/strong><\/h3>\n\n\n\n

To find the amount of drug after 1 hour, substitute h=1h = 1h=1 into the function:d(1)=5e\u22120.4(1)=5e\u22120.4d(1) = 5e^{-0.4(1)} = 5e^{-0.4}d(1)=5e\u22120.4(1)=5e\u22120.4<\/p>\n\n\n\n

Using a calculator:e\u22120.4\u22480.6703e^{-0.4} \\approx 0.6703e\u22120.4\u22480.6703d(1)=5\u00d70.6703\u22483.3515d(1) = 5 \\times 0.6703 \\approx 3.3515d(1)=5\u00d70.6703\u22483.3515<\/p>\n\n\n\n

Answer:<\/strong> About 3.35 milligrams<\/strong> of the drug will be present after 1 hour.<\/p>\n\n\n\n

\n\n\n\n

Part b: What is the time between injections?<\/strong><\/h3>\n\n\n\n

We are told the drug is re-administered when the amount in the bloodstream reaches 2 milligrams<\/strong>. We need to solve for hhh when:d(h)=5e\u22120.4h=2d(h) = 5e^{-0.4h} = 2d(h)=5e\u22120.4h=2<\/p>\n\n\n\n

Step 1: Divide both sides by 5<\/strong>e\u22120.4h=25=0.4e^{-0.4h} = \\frac{2}{5} = 0.4e\u22120.4h=52\u200b=0.4<\/p>\n\n\n\n

Step 2: Take the natural logarithm of both sides<\/strong>ln\u2061(e\u22120.4h)=ln\u2061(0.4)\\ln(e^{-0.4h}) = \\ln(0.4)ln(e\u22120.4h)=ln(0.4)\u22120.4h=ln\u2061(0.4)-0.4h = \\ln(0.4)\u22120.4h=ln(0.4)ln\u2061(0.4)\u2248\u22120.9163\\ln(0.4) \\approx -0.9163ln(0.4)\u2248\u22120.9163<\/p>\n\n\n\n

Step 3: Solve for hhh<\/strong>h=\u22120.9163\u22120.4\u22482.2908h = \\frac{-0.9163}{-0.4} \\approx 2.2908h=\u22120.4\u22120.9163\u200b\u22482.2908<\/p>\n\n\n\n

Answer:<\/strong> The time between injections is approximately 2.29 hours<\/strong>.<\/p>\n\n\n\n

\n\n\n\n

Explanation<\/strong><\/h3>\n\n\n\n

The function d(h)=5e\u22120.4hd(h) = 5e^{-0.4h}d(h)=5e\u22120.4h models exponential decay. The base of the natural exponential function is e\u22482.718e \\approx 2.718e\u22482.718, and the exponent \u22120.4h-0.4h\u22120.4h reflects a decay rate of 40 percent per hour. The initial dose is 5 milligrams, so at h=0h = 0h=0, the drug concentration is 5 milligrams. Over time, the concentration decreases.<\/p>\n\n\n\n

For part a, we substitute 1 into the function. We calculate the exponential term e\u22120.4e^{-0.4}e\u22120.4 which is roughly 0.6703. Multiplying this by 5 gives about 3.35 milligrams of the drug after 1 hour. This means a significant amount has metabolized or been cleared from the bloodstream in just one hour.<\/p>\n\n\n\n

For part b, we solve for when the drug level reaches 2 milligrams. This requires isolating hhh in the equation 5e\u22120.4h=25e^{-0.4h} = 25e\u22120.4h=2. By dividing and using logarithmic properties, we find that h\u22482.29h \\approx 2.29h\u22482.29 hours. This is the interval at which the drug concentration drops low enough to require another dose. This ensures that the drug remains effective in the body without dropping below therapeutic levels. It also prevents toxicity from overdosing by timing the next administration only after the previous dose has mostly worn off.<\/p>\n\n\n\n

\"\"<\/figure>\n","protected":false},"excerpt":{"rendered":"

The function d(h)=5e-0.4h can be used to find the number of milligrams d of a certain drug that is in a patient\u2019s bloodstream h hours after the drug has been administered. When the number of milligrams reaches 2, the drug is to be administered again. a. How many milligrams will be present after 1 hour? […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-236913","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236913","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=236913"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/236913\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=236913"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=236913"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=236913"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}