# Y 5 2x Growth or Decay

Y 5 2x Growth or Decay

### Learning Outcomes

• Graph exponential growth and decay functions.
• Solve problems involving radioactive decay, carbon dating, and half life.

## Exponential Growth and Decay

In real-world applications, we need to model the behavior of a function. In mathematical modeling, we cull a familiar general part with properties that suggest that information technology volition model the real-earth phenomenon we wish to analyze. In the instance of rapid growth, nosotros may choose the exponential growth function:

$y={A}_{0}{eastward}^{kt}$

where ${A}_{0}$ is equal to the value at fourth dimension nada,
e is Euler’s constant, and
k is a positive constant that determines the rate (per centum) of growth. We may use the
exponential growth
function in applications involving
doubling time, the time it takes for a quantity to double. Such phenomena every bit wild animals populations, financial investments, biological samples, and natural resource may exhibit growth based on a doubling time. In some applications, however, every bit we will see when we discuss the logistic equation, the logistic model sometimes fits the information amend than the exponential model.

On the other hand, if a quantity is falling quickly toward zero, without always reaching goose egg, then we should probably cull the
exponential decay
model. Once more, we have the form $y={A}_{0}{e}^{-kt}$ where ${A}_{0}$ is the starting value, and
e is Euler’s abiding. At present
grand is a negative constant that determines the rate of disuse. We may use the exponential decay model when we are computing
half-life, or the time it takes for a substance to exponentially decay to one-half of its original quantity. We employ half-life in applications involving radioactive isotopes.

In our choice of a function to serve as a mathematical model, we often utilize data points gathered by careful ascertainment and measurement to construct points on a graph and hope we can recognize the shape of the graph. Exponential growth and disuse graphs accept a distinctive shape, as we can meet in the graphs below. It is important to call up that, although parts of each of the two graphs seem to lie on the
x-centrality, they are really a tiny distance above the
10-axis.

A graph showing exponential growth. The equation is $y=two{e}^{3x}$.

A graph showing exponential decay. The equation is $y=3{e}^{-2x}$.

Exponential growth and decay often involve very large or very small numbers. To describe these numbers, we often utilise orders of magnitude. The
order of magnitude
is the power of ten when the number is expressed in scientific notation with one digit to the left of the decimal. For example, the distance to the nearest star,
Proxima Centauri, measured in kilometers, is 40,113,497,200,000 kilometers. Expressed in scientific note, this is $4.01134972\times {10}^{13}$. We could describe this number as having guild of magnitude ${10}^{13}$.

### A General Notation: Characteristics of the Exponential Role $y=A_{0}e^{kt}$

An exponential office of the grade $y={A}_{0}{east}^{kt}$ has the post-obit characteristics:

• i-to-one office
• horizontal asymptote:
y= 0
• domain: $\left(-\infty , \infty \right)$
• range: $\left(0,\infty \right)$
• x intercept: none
• y-intercept: $\left(0,{A}_{0}\right)$
• increasing if
k> 0
• decreasing if
k< 0
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An exponential function models exponential growth when k > 0 and exponential decay when thousand < 0.

### Instance: Graphing Exponential Growth

A population of leaner doubles every hr. If the culture started with 10 bacteria, graph the population as a function of fourth dimension.

## Calculating Doubling Time

For growing quantities, we might want to observe out how long it takes for a quantity to double. As we mentioned above, the time it takes for a quantity to double is called the
doubling time.

Given the bones
exponential growth
equation $A={A}_{0}{east}^{kt}$, doubling time tin can be found by solving for when the original quantity has doubled, that is, by solving $2{A}_{0}={A}_{0}{eastward}^{kt}$.

The formula is derived every bit follows:

$\brainstorm{array}{l}2{A}_{0}={A}_{0}{e}^{kt}\hfill & \hfill \\ 2={e}^{kt}\hfill & \text{Divide both sides past }{A}_{0}.\hfill \\ \mathrm{ln}2=kt\hfill & \text{Take the natural logarithm of both sides}.\hfill \\ t=\frac{\mathrm{ln}two}{k}\hfill & \text{Divide by the coefficient of }t.\hfill \end{array}$

Thus the doubling time is

$t=\frac{\mathrm{ln}2}{k}$

### Example: Finding a Function That Describes Exponential Growth

According to Moore’s Law, the doubling time for the number of transistors that tin be put on a computer chip is approximately two years. Give a function that describes this behavior.

### Endeavor It

Recent data suggests that, every bit of 2013, the rate of growth predicted by Moore’south Law no longer holds. Growth has slowed to a doubling time of approximately 3 years. Discover the new function that takes that longer doubling fourth dimension into account.

$f\left(t\right)={A}_{0}{e}^{\frac{\mathrm{ln}2}{3}t}$

## Half-Life

We now turn to
exponential decay. One of the common terms associated with exponential decay, equally stated to a higher place, is
half-life, the length of time information technology takes an exponentially decaying quantity to decrease to half its original corporeality. Every radioactive isotope has a half-life, and the procedure describing the exponential decay of an isotope is called radioactivity.

To find the one-half-life of a function describing exponential decay, solve the following equation:

$\frac{one}{two}{A}_{0}={A}_{o}{e}^{kt}$

Nosotros discover that the one-half-life depends only on the constant
k and not on the starting quantity ${A}_{0}$.

The formula is derived as follows

$\begin{array}{l}\frac{1}{2}{A}_{0}={A}_{o}{e}^{kt}\hfill & \hfill \\ \frac{1}{2}={e}^{kt}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(\frac{1}{2}\right)=kt\hfill & \text{Take the natural log of both sides}.\hfill \\ -\mathrm{ln}\left(2\correct)=kt\hfill & \text{Utilise backdrop of logarithms}.\hfill \\ -\frac{\mathrm{ln}\left(2\right)}{yard}=t\hfill & \text{Separate by }1000.\hfill \finish{assortment}$

Since
t, the time, is positive,
one thousand must, every bit expected, be negative. This gives united states of america the half-life formula

$t=-\frac{\mathrm{ln}\left(two\right)}{chiliad}$

In previous sections, nosotros learned the properties and rules for both exponential and logarithmic functions. Nosotros have seen that any exponential function can be written as a logarithmic function and vice versa. We take used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model existent-world situations, whether the unknown is in an exponent or in the statement of a logarithm.

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One such awarding is in scientific discipline, in computing the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its
half-life. The table below lists the half-life for several of the more common radioactive substances.

Substance Utilise One-half-life
gallium-67 nuclear medicine 80 hours
cobalt-lx manufacturing v.3 years
technetium-99m nuclear medicine vi hours
americium-241 structure 432 years
carbon-fourteen archeological dating 5,715 years
uranium-235 atomic ability 703,800,000 years

We 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 afterwards a specified fourth dimension. We can use the formula for radioactive decay:

$\brainstorm{array}{l}A\left(t\right)={A}_{0}{e}^{\frac{\mathrm{ln}\left(0.5\correct)}{T}t}\hfill \\ A\left(t\right)={A}_{0}{due east}^{\mathrm{ln}\left(0.5\right)\frac{t}{T}}\hfill \\ A\left(t\right)={A}_{0}{\left({e}^{\mathrm{ln}\left(0.5\right)}\right)}^{\frac{t}{T}}\hfill \\ A\left(t\right)={A}_{0}{\left(\frac{1}{ii}\correct)}^{\frac{t}{T}}\hfill \terminate{assortment}$

where

• ${A}_{0}$ is the amount initially present
• $T$ is the half-life of the substance
• $t$ is the time flow over which the substance is studied
• $A$, or $A(t)$, is the amount of the substance present after fourth dimension $t$

### How To: Given the one-half-life, find the decay charge per unit

1. Write $A={A}_{o}{e}^{kt}$.
2. Supplant
A by $\frac{1}{ii}{A}_{0}$ and replace
t by the given half-life.
3. Solve to discover
k. Limited
thousand as an exact value (do not round).

Note:
It is also possible to find the decay rate using
$one thousand=-\frac{\mathrm{ln}\left(2\right)}{t}$.

### Example: Using the Formula for Radioactivity to Find the Quantity of a Substance

How long will it have for 10% of a g-gram sample of uranium-235 to decay?

### Try It

How long will information technology take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

$t=703,800,000\times \frac{\mathrm{ln}\left(0.eight\right)}{\mathrm{ln}\left(0.five\right)}\text{ years }\approx \text{ }226,572,993\text{ years}$.

### Case: Finding the Function that Describes Radioactivity

The half-life of carbon-fourteen is 5,730 years. Express the amount of carbon-14 remaining as a function of time,
t.

### Endeavor Information technology

The half-life of plutonium-244 is 80,000,000 years. Discover a function that gives the amount of plutonium-244 remaining every bit a function of time measured in years.

$f\left(t\right)={A}_{0}{e}^{-0.0000000087t}$

The formula for radioactive decay is important in
radiocarbon datingwhich is used to calculate the approximate date a plant or creature 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 1% fault for plants or animals that died within the final threescore,000 years.

Carbon-fourteen is a radioactive isotope of carbon that has a half-life of 5,730 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 12 and is not radioactive. Scientists have adamant the ratio of carbon-xiv to carbon-12 in the air for the terminal 60,000 years using tree rings and other organic samples of known dates—although the ratio has changed slightly over the centuries.

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Equally long as a constitute or animal is alive, the ratio of the two isotopes of carbon in its trunk is shut to the ratio in the atmosphere. When information technology dies, the carbon-14 in its body decays and is not replaced. By comparing the ratio of carbon-xiv to carbon-12 in a decaying sample to the known ratio in the atmosphere, the date the found or brute died can be approximated.

Since the half-life of carbon-fourteen is v,730 years, the formula for the corporeality of carbon-fourteen remaining later on
t years is

$A\approx {A}_{0}{e}^{\left(\frac{\mathrm{ln}\left(0.five\right)}{5730}\right)t}$

where

• $A$ is the amount of carbon-xiv remaining
• ${A}_{0}$ is the corporeality of carbon-14 when the plant or animal began decaying.

This formula is derived as follows:

$\brainstorm{array}{l}\text{ }A={A}_{0}{east}^{kt}\hfill & \text{The continuous growth formula}.\hfill \\ \text{ }0.v{A}_{0}={A}_{0}{e}^{k\cdot 5730}\hfill & \text{Substitute the half-life for }t\text{ and }0.5{A}_{0}\text{ for }f\left(t\right).\hfill \\ \text{ }0.five={east}^{5730k}\hfill & \text{Divide both sides by }{A}_{0}.\hfill \\ \mathrm{ln}\left(0.5\correct)=5730k\hfill & \text{Take the natural log of both sides}.\hfill \\ \text{ }g=\frac{\mathrm{ln}\left(0.v\right)}{5730}\hfill & \text{Divide both sides past the coefficient of }grand.\hfill \\ \text{ }A={A}_{0}{due east}^{\left(\frac{\mathrm{ln}\left(0.5\right)}{5730}\correct)t}\hfill & \text{Substitute for }r\text{ in the continuous growth formula}.\hfill \end{assortment}$

To find the age of an object we solve this equation for
t:

$t=\frac{\mathrm{ln}\left(\frac{A}{{A}_{0}}\correct)}{-0.000121}$

Out of necessity, nosotros fail here the many details that a scientist takes into consideration when doing carbon-14 dating, and nosotros only expect at the bones formula. The ratio of carbon-14 to carbon-12 in the atmosphere is approximately 0.0000000001%. Let
r be the ratio of carbon-14 to carbon-12 in the organic artifact or fossil to be dated adamant by a method chosen liquid scintillation. From the equation $A\approx {A}_{0}{e}^{-0.000121t}$ nosotros know the ratio of the pct of carbon-14 in the object nosotros are dating to the pct of carbon-14 in the temper is $r=\frac{A}{{A}_{0}}\approx {eastward}^{-0.000121t}$. Nosotros solve this equation for
t, to get

$t=\frac{\mathrm{ln}\left(r\right)}{-0.000121}$

### How To: Given the percentage of carbon-14 in an object, determine its age

1. Express the given percentage of carbon-xiv as an equivalent decimal
r.
2. Substitute for
rin the equation $t=\frac{\mathrm{ln}\left(r\correct)}{-0.000121}$ and solve for the age,
t.

### Example: Finding the Age of a Bone

A bone fragment is found that contains xx% of its original carbon-14. To the nearest year, how old is the bone?

### Endeavor Information technology

Cesium-137 has a half-life of about xxx years. If we begin with 200 mg of cesium-137, will it accept more or less than 230 years until only ane milligram remains?

less than 230 years; 229.3157 to exist exact

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