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UTAS Home › Office of the Pro Vice-Chancellor (Students & Education) › Mathematics Pathways › Pathways to Health Science › Module 8: Logarithms/Growth and Decay

**Image**: http://spectraoflife.files.wordpress.com/2014/01/decaycurve.gif

Click on the link below to take the Pre-Test for Module 8. The Pre-Test is optional but we recommend taking it to test your knowledge of Logarithms/Growth and Decay. There are only 5 questions and it will only take about 10 minutes to complete.

If you receive 80% or greater on the Pre-Test, you have a good knowledge of basic Logarithms/Growth and Decay and can move on to the next module or review the materials in module 8.

If you receive less than an 80%, work your way through the module and then take the quiz at the end to test your knowledge.

Logarithm and exponential relationships describe many areas of science.

The decibel scale for the loudness of sound, the Richter scale of earthquake magnitudes, and the astronomical scale of stellar brightness are all logarithmic scales. Logarithms are also related to pH (a measure of the acidity or alkalinity of a solution) and this will be discussed later in the module in the context of blood pH.

Example

An example of exponential decay relevant to the health area is drug metabolism. For instance Morphine has a half-life of 3 hours. Use this information to think about the following scenario and answer the question.

**Practice example **

16 mg of morphine has been administered to a patient. How long will it take for the amount of morphine in the patient to go down to 2 mg?

** Answer**: Remember that the half-life of morphine is 3 hours. The half-life is the time taken for the amount to reduce to one half of its original amount. So after 3 hours the amount of morphine will have reduced to 8 mg. It would take another 3 hours for it to reduce to 4 mg
and then another 3 hours to reduce to 2 mg. So the total time taken to reduce to 2mg is 9 hours. It may be useful to recall this example later on in the module when we deal with 'half-life' more formally.

The following video clip highlights the prevalence of logarithmic/exponential relationships in the natural world. There are, however a couple of comments to consider in relation to the way the video is presented. The presenter tends to suggest that the advent of the calculator has reduced our 'need' to calculate logarithms by hand now. Perhaps technology affords us the convenience of dealing with logarithmic calculations "at our fingertips" but it does not replace the importance of having a conceptual understanding of logarithms.

Some of the skills and concepts associated with exponents are also covered in the Scientific Notation module. Recall the following notation:

**Image: **www.solving-math-problems.com/image-files/num_exp_base-blue.png

The **exponent** may also be called the index or power. 2^{3} is equal to 2 x 2 x 2. That is:

2^{3}= 2 x 2 x 2 = 8 ** Be careful**: some people mistaken this for 2 x 3 which is incorrect!!.

*and*

2^{4}= 2 x 2 x 2 x 2 = 16

2^{5}= 2 x 2 x 2 x 2 x 2 = 32

2^{6} = 2 x 2 x 2 x 2 x 2 x 2 = 64

2^{7}= 2 x 2 x 2 x 2 x 2 x 2 x 2 = 128

And so on… (you may have already recognised a pattern!!)

The following links to Math Is Fun - Maths Resources provides further discussion of the skills and concepts associated with exponents. The first link focuses on negative exponents and when the exponent is either 0 or 1. The second link deals with negative exponents. Please read through the information provided by both webpages and then complete the review questions (you will find links to each of the questions at the bottom of each of the webpages).

You saw on the previous page the Math Is Fun - Maths Resources webpage, that "anything to the power of zero is 1". You may also find it useful to work through the following reasoning to see why this index law works.

In general,

This formula tells us that any number, except 0, raised to the power zero has a numerical value of 1. this is from: http://www.mathsteacher.com.au/year8/ch07_indices/04_pow/zero.htm

The next weblink deals with
*fractional exponents.* There are two things to point out before you work through the material presented by the link. The first is the fact that:

(x^{a})^{b} = x^{ab}

Let us consider this using an example:

(2^{3})^{2} = 2^{3} x 2^{3} = 2^{6}

(Note that 2^{3} x 2^{3} = 2^{3+3} = 2^{6})

The video also points out that √4 =2 but omits to say that the solution is in fact

√4 = ±2

Think about what number when multiplied by itself gives 4, well one answer is 2 and the other is -2 given that -2 x -2 =4 just as 2 x 2 =4. (Note: don't confuse this with the square root of a negative number as this is an entirely different concept and something that is not dealt with in any of the Health Science Pathways modules).

In many applications however, the only permissible solution is +2, but this should not mean that we overlook the second solution from a mathematical viewpoint.

Look at the following resource from the Math Is Fun - Maths Resources on Fractional Exponents.

The **exponent** of a number says how many times to use the number in a multiplication.

Example 1:

10^{4} = 10 x 10 x 10 x 10 = 10 000

The **logarithm **tells us what exponent we need to raise a number to in order to obtain another number.

Example 2:

log_{10} (10 000) = 4

We may interpret this as "using 10 as a base, what is the logarithm of 10 000"? The answer is 4. In other words the logarithm gives you the exponent as its answer.

Example 3:

log_{2}(32) = 5

We may interpret this as "using 2 as the base, what is the logarithm of 32"? The answer is 5.

It is important to note that logarithms are not always integers.

Example 4:

log_{10} (3162) ≈ 3.5

It is convenient to use a scientific or CAS calculator to evaluate logarithms, particularly those that cannot be solved easily by inspection such as those in examples 2 and 3. The following video clip deals with some notation associated with logarithms, including how to interpret logarithmic
functions on a scientific calculator. The presenter focuses on logarithms to base 10 and the natural logarithm to base *e.* The "natural" logarithm will feature in a later section of this module.

The following web link to Math Is Fun - Maths Resources provides further discussion of some of the key skills and concepts associated with logarithms including Please click on the link below, work through the material provided and then complete the ten questions (you will find links to each of the questions at the bottom of the webpage).

Please click on the following link from the Khan Academy for some further practice with evaluating logarithms.

**https://www.khanacademy.org/math/algebra2/logarithms-tutorial/logarithm_basics/e/logarithms_1.5**

*All Khan Academy content is available for free at* *www.khanacaemy.org*

Practice Task 1

The following practice examples are designed to assist you to become fluent with and to understand the relationship between exponentials and logarithms. Many health science contexts (e.g., the drug metabolism example used previously) are underpinned by logarithmic and exponential relationships. It is therefore important to be familiar and to not be intimidated with the procedures involved in solving logarithmic and exponential problems.

1. Rewrite the following equation in
logarithmic form: 10^{3} = 1000

2. Rewrite the following equation in
logarithmic form: 4^{3} = 64

3. Rewrite the following equation in
exponent form: log_{5}(125)
= 3

4. With the aid of a scientific
calculator evaluate log_{10 } (2146) to 2 decimal places.

Click **here **to check your answers

Change the following from exponential to logarithm form

- 2
^{5 }= 32 - 5
^{4}= 625 - 10
^{2}= 100

Change the following from logarithmic to exponential form

- log
_{7}(49 = 2 - log
_{5}(^{1}/_{25}) = -2 - log
_{10}10 000 = 4

Click **here **to check your answers

**Blood pH**

Health care workers monitor several patient vitals such as respiration, heart rate, and blood pressure. They also monitor blood pH levels.

The kidneys and lungs maintain the proper balance of acids and bases, in the body.

If the body correctly maintains this balance then the blood pH range for an adult should be between 7.35 and 7.45. For infants and children (under16 yrs.), the blood pH levels should be between 7.35 and 7.42.

You may have noticed that the range of blood pH levels from dangerously low to normal to dangerously high seems quite small.

Example

The difference between an extreme low (7.2) to the low normal (7.35) is only 0.15. Similarly the difference between a high normal (7.45) in adults and an extreme high (7.5) in adults is only 0.05. While these differences seem very small, the implications for the patients are extremely serious.

Before we proceed with some calculations involving blood pH we will first talk about pH itself.

Swedish chemist S.P. L. Sorensen developed the pH system in the early 1900's. He defined pH as the logarithm (base 10) of the reciprocal (e.g., the reciprocal of 2 is ^{1}/_{2} ) of the concentration of the hydrogen ions (H^{+} ions) measured in moles per litre, in a solution. To unpack this definition, let us consider that when an acid is added to water, it releases hydrogen ions and increases the concentration
of hydrogen ions in the resulting solution. The resulting solution is said to have a pH lower than 7 (see the pH scale below).

**Image**: www.abundanthealthcenter.com/blog/wp-content/uploads/2011/04/PH-Scale.jpg

Symbolically we express pH as:

**pH = log **where** [H ^{+}]** represents the concentration of Hydrogen ions.

Note also that the **reciprocal **of** [H ^{+}]** is

**Example 1**

If the concentration of hydrogen ions in a solution is 0.0078 moles per litre, what is the corresponding pH of the solution? Use your scientific calculator to complete this calculation

pH = log

pH = log

pH = log (128.2)

Note: on your calculator you would enter the following 1 ÷ 0.0078 to yield 128.2 to 2 decimal places.

pH = 2.1

**Example 2**

If the pH of a solution is 8.3, what is the concentration of hydrogen ions in moles per litre.

Solution:

pH = log

8.3 = log Now remember that we are dealing with log to base 10 here so:

8.3 = log_{10} You may find it useful to firstly rewrite
the equation as:

log_{10} = 8.3
Given the relationship between
logarithms and exponents, it follows that:

10^{8.3} =Multiplying
both sides of the equation by [H^{+}] we have:

[H^{+}] x 10^{8.3} = x [H^{+}] Note
that the [H^{+}] terms on
the right hand side of the equation cancel each other out, so we have:

[H^{+}] x 10^{8.3} = 1 We want [H^{+}] by itself
(because we are finding the concentration of the hydrogen ions) so we divide
both sides of the equation by 10^{8.3} which
gives us:

To may evaluate using your scientific calculator:

1 ÷ 10 ^ 8.3 = 5.01 x 10^{-9} (you may
wish to refer to or revisit the Scientific Notation module)

[H^{+}] = 5.01 x 10^{-9} moles per
litre.

Note: that this can also be written as 0.00000000501 mol/L.

**Example 3**

The blood for one adult patient is
showing hydrogen ion concentration of 7.94 x 10^{-8} moles per
litre.

Determine the pH for this patient's blood

Solution:

pH = log (12594458) evaluate using a scientific calculator

pH = 7.1

Note: that this patient's blood pH is outside of the normal blood range for adults which is 7.35 to 7.45.

Growth and Decay

**Exponential growth** (or **exponential
decay** if the growth rate is negative) is modelled by a mathematical relationship (function) with a variable exponent.

As the value of the variable changes, the value of the function increases (or decreases) in proportion to its current value.

Example

The function **ƒ _{1}(x) = 2^{x}**

As the value of the exponent *x *increases by 1 the value of the function doubles, or in other words grows by 100% of its existing value (see the table below).

For the function **ƒ**_{2} (*x) = 0.5*^{x} , each time the value of x increases by 1, the value of the function halves, or decreases by 50% (see the table below):

## Value of x |
_{1}(x)
= 2^{x} |
_{2} (x) = 0.5^{x} |

0 | 1 | 1 |

1 | 2 | 0.5 |

2 | 4 | 0.25 |

3 | 8 | 0.125 |

4 | 16 | 0.0625 |

5 | 32 | 0.03125 |

6 | 64 | 0.015625 |

7 | 128 | 0.0078125 |

8 | 256 | 0.00390625 |

- | - | - |

20 | 1048576 | 0.0000009536743164 |

In the above examples we saw that ** ƒ_{1} (x) = 2^{x}**is an example of an exponential growth function (the function grows by a constant factor of 2 in other words it doubles after each growth period) and

Note: A word about functional notation: We have used the notation ƒ_{1} (*x*) and ƒ_{2}(*x*) to denote two different functions. We could have just as easily used something like *f*(x) and, *g*(*x*) or similar.

*The discrete case*

The basic formula for discrete exponential growth is:

Where: ** x_{0}** is the initial value of whatever it is that
will be growing (or shrinking),

We call this **discrete **exponential growth (as distinct from continuous
exponential growth) because all possible values of** t** are some distance
apart from each other.

Example

You may be familiar with the context of compound interest where interest may be added say half yearly, quarterly, monthly, fortnightly etc. Before we look at the concept of exponential growth in the context of health science, it is useful to use the common example of compound interest to demonstrate the way discrete exponential growth works.

At the beginning of the year, you deposit $1000 into a bank account, with an annual interest rate of 5%. Assuming no other deposits or withdrawals are made and the interest rate stays constant, what will be the value of the account after;

a) 5 years if interest is compounded annually (i.e., interest is added at the end of each year).

b) 6 years if interest is compounded annually.

*c) *70 years if
interest is compounded annually *(this length of time may be a little
unrealistic but here we are interested in the behaviour of the function for
larger values of t.*

*d) *71 years if
interest is compounded annually

*Solution:*

Part a)

**x _{t }= x_{0} (1+r)^{t} **

** x_{0}** is $1000 (your initial deposit)

*r* is the rate of growth (interest)
per time period and as interest is added annually each time period is 1 year so
*r* =5% = 0.05

*t* is the number of time periods
which in this case is 5

**x _{t }= x_{0} (1+r)^{t} **

*x*_{t }= 1000 (1+0.05)^{5}

*x*_{t } = 1000 x 1.05^{5}

*x*_{t } =** $1 276.28**

Part b)

*x*_{t } = 1000 (1+0.05)^{6}

*x*_{t } **= $1 340.10**

Part c)

*x*_{t } = 1000 (1+ 0.05)^{70}

*x*_{t } = **$30 426.43**

Part d)

*x*_{t } = 1000 (1+ 0.05)^{71}

*x*_{t } =** $31 947.75**

*Notice that between the
70 ^{th} and 71^{st} the size of the investment increased by
around $1521 dollars. This is significantly more than the growth of
approximately $63 between the 5^{th} and the 6^{th} year. The
purpose of these calculations was to demonstrate that difference between two
consecutive years becomes larger (up to a point) as t becomes large.*

At first glance the
connection between ** ƒ(x) =2^{x} ** and the general form of a discrete exponential
function may not be immediately apparent. Let us however think about

* ƒ(x) =2^{x} *We
saw previously that this function grows by a constant factor of 2 in other
words it doubles after each growth period. We can also think of doubling as
100% growth. That is in one time period the function grows from 4 to 8 and in
the next time period it grows from 8 to 16 and so on.

In other words, the rate of growth per time period is 100% (i.e., 1.00 expressed as a decimal).

Hence
using our ** x_{t }= x_{0} (1+r)^{t}** formula we have:

*r* = 100% = 1.00

** x_{0}** is our starting amount when

Therefore we have:

*x _{t} *= 1(1+1)

x_{t} = 2^{t} and using the ƒ(*x*) notation we have:

*f(x) = 2 ^{x}*

The previous section introduced the basic formula for discrete exponential growth as:

*x _{t }= x_{0} (1+r)^{t}*

It follows then that the basic formula for discrete exponential decay is:

**x _{t }= x_{0} (1-r)^{t}** There is a minus sign instead of a plus sign because we are dealing with a negative rate of growth.

*Elimination of a drug from the body*

When people take medicine, the drug is metabolised and eliminated at a certain rate. Suppose the initial amount of a drug in the body is 200 mg and is eliminated at a rate of 30% per hour. We can model this situation as follows;

x_{t} = 200 (1 -0.3)^{t} Notice that 30% has been entered as a decimal (i.e.,0.3)

An equivalent representation of this equation is:

x_{t} = 200 (0.7)^{t}

Note: that for exponential decay the base will always be between 0 and 1 (in this example the base is 0.7). Compare this situation with the case of exponential growth where the base is always greater than 1.

Using the formula x_{t} = 200 (0.7)^{t} to calculate the amount of drug left in the
body after 2 hours.

Click **here **to check your answers.

So far we have discussed discrete (or non-continuous) exponential growth and decay. Let us return for a moment to our compound interest example from the previous section. At the beginning of the year, you deposit $1000 into a bank account, with an annual interest rate of 5%. Assuming no other deposits or withdrawals are made and the interest rate stays constant, this time we will investigate the way the value of the account changes after 5 years if interest is added:

- annually?
- half-yearly?
- Quarterly?
- Monthly

Solution:

1) As previously calculated**x _{t }= x_{0} (1+r)^{t}**

*x*_{t }= 1000 (1+0.05)^{5}

*x*_{t } = 1000 x 1.05^{5}

*x*_{t } =** $1 276.28**

2) As interest is compounded half-yearly r = ^{5%}/_{2} = 2.5% = 0.025 and *t *= 5 x 2 =10

*x*_{t }= 1000 (1+0.025)^{10}

*x*_{t } =** $1280**

3As interest is compounded quarterly r = ^{5%}/_{4 }= 1.25% = 0.0125 and t = 5 x 4 =20

*x*_{t }= 1000 (1+0.0125)^{2}^{0}

*x*_{t} =** $1282**

4) As interest is compounded monthly r = ^{5%}/_{12} = 0.416 ̇ %= 0.00416 ̇ and t = 5 x 12 =60

*x*_{t }= 1000 (1+0.00416)^{6}^{0}

*x*_{t }= **$1283.36**

Notice that as the distance between two consecutive time periods becomes less (i.e., interest is compounded more often), the value of the investment increases. If interest was compounded weekly then the value of the investment would be higher again and even higher if interest was compounded daily, hourly, and so on. Imagine the distance between two consecutive time periods as being infinitesimally small. This intuition is at the core of the concept of continuous exponential growth.

The following explanation from Purple Math touches on a very important number that arises in the development of exponential functions, and that is the "natural" exponential.

Ignoring the principal, the interest rate, and the number of years by setting all these variables equal to "1", and looking only at the influence of the number of compoundings, we get:

how often compounded | computation |

yearly | |

semi-annually | |

quarterly | |

monthly | |

weekly | |

daily | |

hourly | |

every minute | |

every second |

http://www.purplemath.com/modules/expofcns5.htm

As you can see, the computed value keeps getting larger and larger, the more often you compound. But the growth is slowing down; as the number of compounding increases, the computed value appears to be approaching some fixed value.

You might think that the value of the compound-interest formula is getting closer and closer to a number that starts out "2.71828". And you'd be right; the number we're approaching is called "e". This phenomenon is relevant to any situation involving continuous growth or decay- it does not have to involve money!

*The general formula used for continuous exponential growth is:**x _{t} = x_{0} e^{rt}*

where

Let us investigate this idea further by considering just the value of *e*^{r} for various values of ** r**. Now the important thing to remember is the

Using the ** e^{x}** function on a scientific calculator we obtain the following:

When ** r** = 2

r = 5 e^{5} = 148.413 (3 dp)

r = 8 e^{8} = 2 980.958 (3 dp)

Notice how rapidly e^{t} increases as r increases.

When e is raised to an increasingly negative power the function decreases.

r = -2 Therefore |

r = -5 Therefore |

r = -8 Therefore |

Certain bacteria, given favourable growth conditions, grow continuously at a rate of 4.6% a day.

Find the bacterial population after thirty-six hours, if the initial population was 250 bacteria.

Click **here** to check your answer

**Worked Example 2**

*Half-life decay*

Half-life is
defined as the amount of time needed for a system undergoing exponential decay
to decrease to half of its initial value. The term 'half-life' is usually used
in **describing** the characteristics of radioactive elements and testing
pharmaceutical substances.

The half-life of the element
carbon-14 is about 5730 years. This means that it takes about 5730 years for
Carbon-14 to decay to half its original amount. It is important to understand
that half-life does **not **depend on the initial amount of the substance.
Rather, it depends only on the **rate **of decay.

Recall that the general formula for continuous exponential decay is:

* x_{t} = x_{0} e^{rt }*where

Recall also that *x**_{t}* is the amount of substance remaining after a
certain time period and

You may have recognised that is also equivalent to ** e^{rt}** so it follows that:

Our aim now is to rearrange this equation to get

Taking the natural log of both sides of the equation releases the exponents* rt*. Recall that logs 'undo' exponents and exponents 'undo' logs (they are inverses of each other). So

But where did the ln*e* go? Well remember that ln is the same as **log _{e}** so it follows that

log= 1 just as_{e}elog=1,_{10}10log=1, and so on_{5}5

becauseis approximately -0.6931

For clarity we will denote **t** as t_{1/2} to make it clear that it is the half-life we are interested in.

Therefore we can use the formula to calculate the half-life of any system given the value of r (i.e., the rate of decay).

- A penicillin solution containing 300 units/ml has a half-life of 8 days in plasma. What will the concentration be in 7 days?
- A penicillin solution has a half-life of 6 days. How long will it take for the concentration to drop to 70% of the initial concentration?

Click **here** to check your answers

Click on the link below to take the online self-assessed quiz.

There are 10 mathematics questions on the Quiz and they are about the information in this module.

To pass this quiz, you will need to get a mark of 80%. Feedback will be provided for both correct and incorrect answers at the end of the Quiz. If you answer questions incorrectly, then it is strongly recommended that you review the sections of the modules to review those topics. You will be able to re-take the quiz if needed.

Make sure to enter your name and email address in the quiz so your results can be mailed to you for your records. You may need to show your results to your university.

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2 May, 2018

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