Image: http://spectraoflife.files.wordpress.com/2014/01/decaycurve.gif
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. 23 is equal to 2 x 2 x 2. That is:
23= 2 x 2 x 2 = 8 Be careful: some people mistaken this for 2 x 3 which is incorrect!!.
and
24= 2 x 2 x 2 x 2 = 16
25= 2 x 2 x 2 x 2 x 2 = 32
26 = 2 x 2 x 2 x 2 x 2 x 2 = 64
27= 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:
(xa)b = xab
Let us consider this using an example:
(23)2 = 23 x 23 = 26
(Note that 23 x 23 = 23+3 = 26)
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:
104 = 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:
log10 (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:
log2(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:
log10 (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: 103 = 1000
2. Rewrite the following equation in logarithmic form: 43 = 64
3. Rewrite the following equation in exponent form: log5(125) = 3
4. With the aid of a scientific calculator evaluate log10 (2146) to 2 decimal places.
Click here to check your answers
Change the following from exponential to logarithm form
Change the following from logarithmic to exponential form
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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 = log10 You may find it useful to firstly rewrite
the equation as:
log10 = 8.3
Given the relationship between
logarithms and exponents, it follows that:
108.3 =Multiplying
both sides of the equation by [H+] we have:
[H+] x 108.3 = x [H+] Note
that the [H+] terms on
the right hand side of the equation cancel each other out, so we have:
[H+] x 108.3 = 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 108.3 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 ƒ1(x) = 2x is an exponent function.
Value of x | ƒ1(x) = 2x | ƒ2 (x) = 0.5x |
| 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 ƒ1 (x) = 2xis 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 ƒ2 (x) =0.5x is an example of exponential decay (the value of the function decreases by a factor of 0.5).
Note: A word about functional notation: We have used the notation ƒ1 (x) and ƒ2(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: x0 is the initial value of whatever it is that will be growing (or shrinking), r is a constant representing growth (or decay) rate, and xt is the value after t time periods. While the formula's notation may vary between contexts or textbooks (e.g., for the case of compound interest is often denoted 'A' and x0 is often denoted 'P'), the fundamental structure of the equation remains the same.
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)
xt = x0 (1+r)t
x0 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
xt = x0 (1+r)t
xt = 1000 (1+0.05)5
xt = 1000 x 1.055
xt = $1 276.28
Part b)
xt = 1000 (1+0.05)6
xt = $1 340.10
Part c)
xt = 1000 (1+ 0.05)70
xt = $30 426.43
Part d)
xt = 1000 (1+ 0.05)71
xt = $31 947.75
Notice that between the 70th and 71st the size of the investment increased by around $1521 dollars. This is significantly more than the growth of approximately $63 between the 5th and the 6th 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) =2x and the general form of a discrete exponential function may not be immediately apparent. Let us however think about ƒ(x) =2x as follows:
ƒ(x) =2x 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 xt = x0 (1+r)t formula we have:
r = 100% = 1.00
x0 is our starting amount when t =0
Therefore we have:
xt = 1(1+1)t
xt = 2t and using the ƒ(x) notation we have:
f(x) = 2x
The previous section introduced the basic formula for discrete exponential growth as:
xt = x0 (1+r)t
It follows then that the basic formula for discrete exponential decay is:
xt = x0 (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;
xt = 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:
xt = 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 xt = 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:
Solution:
1) As previously calculated
xt = x0 (1+r)t
xt = 1000 (1+0.05)5
xt = 1000 x 1.055
xt = $1 276.28
2) As interest is compounded half-yearly r = 5%/2 = 2.5% = 0.025 and t = 5 x 2 =10
xt = 1000 (1+0.025)10
xt = $1280
3As interest is compounded quarterly r = 5%/4 = 1.25% = 0.0125 and t = 5 x 4 =20
xt = 1000 (1+0.0125)20
xt = $1282
4) As interest is compounded monthly r = 5%/12 = 0.416 ̇ %= 0.00416 ̇ and t = 5 x 12 =60
xt = 1000 (1+0.00416)60
xt = $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:
xt = x0 ert
where x0 is the initial value of the function (i.e., the initial value of whatever is growing or decaying). When r (the rate of growth) is positive we have exponential growth and when r is negative we have exponential decay.
Let us investigate this idea further by considering just the value of er for various values of r. Now the important thing to remember is the e is a number not a variable (recall that e is 2.7182818284590452353602874713527 and goes on forever with no repeating pattern).
Using the ex function on a scientific calculator we obtain the following:
When r = 2 e2 = 7.389 (3 decimal places)
r = 5 e5 = 148.413 (3 dp)
r = 8 e8 = 2 980.958 (3 dp)
Notice how rapidly et 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.
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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:
xt = x0 ert where r < 0 (that is r is negative because we are dealing with decay).
Recall also that xt is the amount of substance remaining after a certain time period and x0 is the initial amount of the substance. Now we are looking for the point where the amount remaining is exactly half of the original amount, and this can be expressed as follows:
You may have recognised that 
is also equivalent to ert so it follows that:
Our aim now is to rearrange this equation to get t by itself because this will be the half-life.
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 ln(x) is the inverse of ex. Hence:
But where did the lne go? Well remember that ln is the same as loge so it follows that
logee = 1 just as log1010 =1, log55 =1, and so on
because
is approximately -0.6931
For clarity we will denote t as t1/2 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).
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