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Exponential and Logarithm

Exponential Decay

Nabil Akbarazzima Fatih

Mathematics

Definition of Exponential Decay

Exponential functions don't only represent significant growth, but also consistent decline over specific periods of time. This consistent decline is called exponential decay.

Exponential decay is the process of value reduction at a consistent rate. Mathematically, an exponential decay function can be written as:

f(x)=n×axf(x) = n \times a^x

With the conditions:

  • 0<a<10 < a < 1 (base)
  • nn is a non-zero real number (coefficient)
  • xx is any real number (variable)

When 0<a<10 < a < 1, the value of axa^x becomes smaller as the value of xx increases, causing the function graph to decrease.

Characteristics of Exponential Decay Graph

The graph of an exponential decay function shows a decreasing rate of decline. Initially, the function value decreases rapidly, then the rate of decrease slows down as it approaches zero but never actually reaches zero.

Unlike exponential growth graphs that rise at an increasing rate, exponential decay graphs fall at a decreasing rate.

The graph below shows an exponential decay function with base a=0.5a = 0.5 and coefficient n=50n = 50, which can be expressed as:

f(x)=50×(12)xf(x) = 50 \times \left(\frac{1}{2}\right)^x
Exponential Decay Graph
Graph decreases at a progressively slower rate.

Applications of Exponential Decay

Exponential decay has many applications in real life. Here are some examples:

Drug Elimination in the Body

A patient receives a 50 microgram dose of pain medication. Every hour, half of the remaining drug dose will decay and be eliminated from the body.

If we model the amount of drug remaining after xx hours as f(x)f(x), then:

f(x)=50×(12)xf(x) = 50 \times \left(\frac{1}{2}\right)^x

Calculating the remaining dose:

  • Initial dose: f(0)=50f(0) = 50 micrograms
  • After 1 hour: f(1)=50×12=25f(1) = 50 \times \frac{1}{2} = 25 micrograms
  • After 2 hours: f(2)=25×12=12.5f(2) = 25 \times \frac{1}{2} = 12.5 micrograms
  • After 3 hours: f(3)=12.5×12=6.25f(3) = 12.5 \times \frac{1}{2} = 6.25 micrograms

With this pattern, we can calculate that after 9 hours, the amount of the drug in the patient's body will be less than 0.1 micrograms.

Radioactive Decay

Radioactive substances decay at a certain rate, typically expressed as a percentage per unit of time. If we have an initial mass M0M_0 and a decay rate rr per unit time, then the mass remaining after time tt can be expressed as:

M(t)=M0×(1r)tM(t) = M_0 \times (1-r)^t

Ball Bounce

When a ball is dropped, each time it bounces, the bounce height reaches a portion of its previous height. If the initial height is h0h_0 and each bounce reaches a proportion pp of the previous height, then the height of the nn-th bounce is:

hn=h0×pnh_n = h_0 \times p^n

For example, if a basketball is dropped from a height of 3 meters and each bounce reaches 35\frac{3}{5} of the previous height, then:

  • Initial height: h0=300h_0 = 300 cm
  • First bounce: h1=300×35=180h_1 = 300 \times \frac{3}{5} = 180 cm
  • Second bounce: h2=180×35=108h_2 = 180 \times \frac{3}{5} = 108 cm
  • Third bounce: h3=108×35=64.8h_3 = 108 \times \frac{3}{5} = 64.8 cm

The bounce height function can be expressed as:

h(n)=300×(35)nh(n) = 300 \times \left(\frac{3}{5}\right)^n

The ball will eventually stop bouncing when the bounce height becomes very small, around 0.65 cm at the 12th bounce.

Exercises

Here are some practice problems to deepen your understanding of exponential decay:

  1. Two hundred mg of a substance is injected into the body of a patient suffering from lung cancer. The substance is excreted from the body through the kidneys every hour. If 50% of the substance is eliminated from the patient's body every hour, how many mg of the substance remains in the patient's body after 5 hours?

  2. The mass of a radioactive substance is 0.3 kg at 10 am. The decay rate of the radioactive substance is 15% per hour. How much radioactive substance will remain 8 hours later?

  3. A basketball is dropped from a height of 3 meters. The ball hits the ground and then bounces back to a height of 35\frac{3}{5} of its previous height. The ball continues to bounce with the same height ratio until it finally stops bouncing and falls to the ground. a. Draw a graph of the function showing the change in bounce height until the ball finally touches the ground. b. At which bounce does the ball finally stop bouncing?

Answer Key

  1. Let f(x)f(x) be the amount of substance remaining after xx hours. We have:

    f(x)=200×(12)xf(x) = 200 \times \left(\frac{1}{2}\right)^x

    After 5 hours:

    f(5)=200×(12)5=200×0.03125=6.25f(5) = 200 \times \left(\frac{1}{2}\right)^5 = 200 \times 0.03125 = 6.25

    So, the substance remaining after 5 hours is 6.25 mg.

  2. Let f(x)f(x) be the mass of the radioactive substance after xx hours. We have:

    f(x)=0.3×(0.85)xf(x) = 0.3 \times (0.85)^x

    After 8 hours:

    f(8)=0.3×(0.85)80.083 kg or about 83 grams.f(8) = 0.3 \times (0.85)^8 \approx 0.083 \text{ kg} \text{ or about } 83 \text{ grams}.
  3. a. The bounce height function is:

    h(n)=300×(35)nh(n) = 300 \times \left(\frac{3}{5}\right)^n
    Exponential Decay Graph
    Graph decreases at a progressively slower rate.

    b. The ball will stop bouncing when its height approaches zero. Based on calculations, at the 12th bounce, the ball's height reaches 0.653035 cm, which is small enough to consider the ball has stopped bouncing. However, considering the ball's mass, it might have stopped bouncing by the 10th bounce, when its height is about 1.8 cm.

    BounceHeight
    n=0n = 0300300
    n=1n = 1180180
    n=2n = 2108108
    n=3n = 364.864.8
    n=4n = 438.8838.88
    n=5n = 523.32823.328
    n=6n = 613.996813.9968
    n=7n = 78.398088.39808
    n=8n = 85.0388485.038848
    n=9n = 93.0233093.023309
    n=10n = 101.8139851.813985
    n=11n = 111.0883911.088391
    n=12n = 120.6530350.653035
    n=13n = 130.3918210.391821
    n=14n = 140.2350920.235092
    n=15n = 150.1410550.141055

Application of Concepts in Life

In mathematical modeling, exponential decay is used to model various phenomena such as:

  • Radioactive decay
  • Cooling of hot objects
  • Depreciation of goods
  • Drug elimination in the body

Beyond scientific applications, exponential concepts can also be found in social contexts. For example, the relationship between charity (giving) and blessing (receiving) can be illustrated with an exponential growth function. The more we share with others in need, the more goodness will return, at an increasingly faster rate.