1. Logarithms and Logarithm Applications

A. Exponent Laws Before you can learn about logarithms, you need to remember all your exponent laws as these relate directly to logarithms. These laws are:

𝑎𝑛 × 𝑎𝑚 = 𝑎𝑛+𝑚 → When you have the same bases and you are multiplying you add the exponents.

𝑎𝑛 ÷ 𝑎𝑚 = 𝑎𝑛−𝑚 → When you have the same bases and you are dividing, you subtract the exponents. (Note that the base does not change – only the exponents are affected).

(𝑎𝑛𝑏𝑚)𝑝 = 𝑎𝑛 ×𝑝𝑏𝑚 × 𝑝 → when you have an exponent on the outside of a set of brackets with no base (no boyfriend / girlfriend) the exponent is multiplied with all the exponents inside the brackets. The outside exponent is flirty and “gets” with everyone inside the brackets because it doesn’t have a boyfriend /girlfriend.

√𝑎𝑐𝑑= 𝑎

𝑐

𝑑 → the exponent inside the radical is divided by the radical’s power. Another way to remember it is that cat’s live inside the house and dogs live outside the house, the cat is more important than the dog so it goes over the dog.

𝑎0 = 1 → anything to the power of zero equals one (Remember this law – it is very important when we get to exponential graphs).

𝑎−𝑛 = 1

𝑎𝑛 → anything with a negative power goes under one to

make the power positive. Another way to remember it is that if your girlfriend / boyfriend is grumpy (or negative) he / she wants to be alone so you go downstairs (under the one). Remember that the exponent only belongs to its base and does not affect any other

exponent, for example if you have 𝑎−5𝑏2 then only your “a” base has a grumpy boyfriend/ girlfriend so only the “a” base and its exponent will go “downstairs” the “b” base and its exponent will stay

“upstairs” → 𝑏2

𝑎5 . Now the 5 is happy (or positive).

Activity 1.1 1. Simplify the following expressions, leaving all exponents as positive.

a) 𝑎3𝑏2 × 𝑎2𝑏4 b) 𝑎7𝑏3𝑐2

𝑎3𝑏−2𝑐0

c) (𝑎6𝑏−5)2 d) ((𝑎3𝑏4)−1

𝑎−3𝑏7) −1

e) √𝑎6𝑏4

𝑐7𝑑5 ÷ (𝑎2𝑏)−1

𝑐6𝑑5 f) 𝑎𝑐2𝑑2

𝑏4𝑒3 ×

𝑑5𝑒−2

𝑎𝑏2𝑐−2 × 𝑎2𝑏3

𝑐−1𝑑−2𝑒4

g) (𝑎𝑏)−1

(𝑎2𝑏−1 ÷ 𝑎−1𝑏

𝑏2𝑎 h)

𝑎3𝑏𝑐𝑑3

𝑎2𝑏𝑐3𝑑2 ÷ (𝑎2𝑏)−2

√𝑎4𝑏−4 ×

𝑎3𝑏2

𝑐−1𝑑3

2. Solve for x:

a) 𝑥 = 25 b) 3𝑥 = 6. 53

c) 4. 3𝑥 = 2 916 d) 3. 5𝑥 = 1 5

e) 3𝑥 − 3𝑥

4=

729

4 f) 3. 2𝑥 −

5.2𝑥

7 + 9 =

319

7

g) 4𝑥3 = 2 916 h) 5 − 3. 𝑥4 = − 19

i) 3𝑥 = 14 4 𝑎 𝑎

j) 5𝑥+1

4−

3.5𝑥

8= 2 34 3 5 k) 32𝑥+1 + 5. 3𝑥 + 2 = 4

Solutions will follow at the end of the Chapter.

B. Defining logarithms

NOTE:

y = logax, therefore in exponential form: ay = x. In both the log and exponential form, ‘a’ is the base.

Where a base is not written, the base is 10.

Logbasenumber = exponent OR baseexponent = number

Logaa = 1 (a > 0 and a ≠ 1), because a = a

Loga1 = 0 (a > 0 and a ≠ 1), because a = 1

Loga0 is undefined as no number to any given exponent will equal 0.

Restrictions

Logax is defined only for x > 0 and x ≠ 1.

A base cannot be 0 or 1 and must be greater than 0; ie a ≠ 0; a ≠ 1 and a > 0.

By definition, “The logarithm of a number, with a certain base is equal to the exponent to which the base will be raised which is then equal the given number”. These two functions are inverses of one another.

Remember:

Exponent

Base Number

Number Exponent Base

ay = x

Log a x = y

Using the calculator to solve simple logarithms: Use the following keys from the EL-W535HT Scientific Calculator:

This key calculates exponential values. (See Activity 1.2 Question 1)

These keys, when used together, calculate the ‘nth’ root of a given value. (See Activity 1.2 Question 4)

This key calculates the logarithm (of base 10), the exponent to which must be raised to equal the given value. (See Activity 1.4 Question 2, 6 and 9 and Activity 1.7)

These keys, when used together, use the given logarithm to calculate the number (of base 10) to which the exponent is assigned, to calculate the logarithm. (See Activity 1.7 Question 7)

These keys, when used together, calculate the logarithm of a number, x, to a given base, a. (This can be used to solve a logarithm which has any base, where a calculator may be used).

ACTIVITY 1.2:

1. Write the following in logarithm form:

a) 24 = 16 b) 6−3 = 1

216

c) xy = z d) p3 = 343 2. Write the following in exponential form:

a) log𝑎 = 𝑏 b) log41

64= −3

c) log4 1024 = 5 d) log 1000 = 3 3. Simplify the following logarithms, if possible:

a) log8 64 b) log3 3 c) log 100 d) log𝑚 9

e) log12 1 f) log1

4

1

16

g) log3 2 h) log21

32

i) log1

1000 j) log5(−125)

k) log1 3 l) log 0

4. Solve for x, if possible:

a) log3 𝑥 = 5 b) log𝑥 16 = 4 c) log5 125 = 𝑥 d) log1

2

𝑥 = 3

e) log𝑥1

64= 6 f) log9 1 = 𝑥

g) log42 𝑥 = 0 h) log𝑥 = 1 i) 2 log4 32 = 𝑥

C. Laws of Logarithms

NOTE: There are 4 Logarithm Laws Law 1: log𝑛 𝐴𝐵 = log𝑛 𝐴 + log𝑛 𝐵

Law 2: log𝑛𝐴

𝐵= log𝑛 𝐴 − log𝑛 𝐵

Law 3: log𝑛 𝐴𝑚 = log𝑛 𝐴

Law 4: log𝑛 𝐴 = log𝑏 𝐴

log𝑏 𝑛 (change of base law)

Remember that these laws are reversible so they work both forwards and backwards.

Remember your exponent rules as these will help you to remember your logarithm rules.

Logarithms can be used to solve exponential equations.

First write the given equation as a single logarithm.

Convert from logarithm form to exponential form.

Your scientific calculator (the SHARP EL-W535HT) may be used to help solve for x.

Check the roots of the quadratic equation by substituting back into the original equation.

REMEMBER THAT THE LOG OF A NEGATIVE NUMBER IS MEANINGLESS: A>0, n>0, and n1. (these are your restrictions)

Activity 1.3:

1. Write the following as a single logarithm:

a) log 3 – log 4 – log 3 b) 2 log 7 + 7 log 2 c) log4 3 + log2 1 − log4 9 d) log𝑛(𝑥

2 − 1) − log𝑛(𝑥 + 1)

e) log5 3 + 5 f) 1

4log 4 + 4 log

1

4

g) log𝑎 𝑏 + 2 log𝑎 + log𝑎 h) log9 18

log9 2

2. Simplify the following:

a) log 5

log 125 b) log 3 + log 𝑞2 − 2 log 2 + log 𝑞

c) log 4−log2

log 64−log16 d) log25 5 + log7 1 + 3 log2

1

2− log 0.001

3. Expand the following:

a) log 2𝑞𝑟 b) log𝑥𝑦

𝑧

c) log 4𝑎𝑏 d) log(𝑝+𝑞)

(𝑝−𝑞)

4. If log 2 = a, and log 3 = b, determine the following in terms of a and b.

a) log 4 b) log 0,25

c) log 6

log 3 d) log 48

D. Exponential and Logarithmic Equations

Simplify your equation so that you either have a single log on both sides or a single log that is equal to a number.

If the bases of both of those logs are the same then you can “cancel” or “drop” the logs and solve for x from there.

If it is a single log equal to a number remember your basic rule for logarithms → log base number = exponent

Don’t forget that all your logarithm rules still apply and you should learn these off by heart.

Activity 1.4 Solve for x (correct to 2 decimal places, where necessary). 1. log4 𝑥 = log𝑥 4 2. log3𝑥 29 = 6

3. log(𝑥 + 2) = 1

2log(𝑥 + ) 4. 10𝑥 = 35 (Use your calculator)

5. log6 36𝑥 = log 100 6. log 125 − log 25 = 𝑥

7. log(𝑥 + 2) + log(𝑥 − 1) = 2 log 𝑥 8. 4𝑥 = 0.015625 9. (0.5)𝑥 = 0.3 10. 4(6𝑥) = 144

E. Exponential functions and Logarithmic graphs (inverses):

A function occurs when, for each x-value input of an equation, there is only one y-value output. (If there is more than one y-value the graph is not a function)

The two sets of numbers generated from an equation, namely the x- and y-values, can be switched or reversed in order to create an inverse equation or function.

Exponential functions can be written in logarithm form, which is the same as the inverse of the exponential function.

o The equation of an exponential function is y = ax o The equation of the inverse of the above exponential function is

x = ay

o Therefore the logarithm equation is 𝑦 = log𝑎 𝑥

o In logarithm and exponential graphs, the line y = x is the line of symmetry.

o In general, for a logarithmic graph, a>1, a ≠ 1 and x>0. o The Range is given as (−∞; ∞) , and the Domain is given as

(0; ∞). o y = 0 is the asymptote for the graph.

The SHARP EL-W535HT can be used to generate y-values of a function in table format. The Table Mode function is used on the calculator.

To access the Table Mode, press the following keys:

Example:

Given the equation: 𝑦 = log2 𝑥 Step 1: Convert from logarithm form to exponential form. 𝑦 = log2 𝑥 → 2𝑦 = 𝑥 (This is the inverse function) Step 2: Convert from the inverse function to the exponential function 2𝑦 = 𝑥 → 2𝑥 = 𝑦

(To convert from one form to the other, switch the x- and y-values in the exponential function.)

Step 3: To generate y-values using the SHARP EL-W535HT, do the

following: Once the calculator is in Table Mode, you have to enter the

function. Do this by pressing the following keys:

Step 4: Enter the X_Start value, for example -2 (remember to use the

button), then press equals. Step 5: Enter the value of the increment you want the x-values (The X_Step) to increase in, i.e. 1’s or 2’s, etc. Press equals.

Step 6: The following table is displayed on the screen: X ANS Only 3 values are displayed at one time. -2 0.25 The ANS column represents the y-values. -1 0.5 Use the up and down arrow keys to see

0 1 more x- and y-values. The table displays the y-values as decimals.

The conventional table for this equation would look like:

X -2 -1 0 1 2

Y 1 2 4

The table for the inverse values (the logarithmic graph) looks like:

X 1 2 4

Y -2 -1 0 1 2

The graph would look as follows:

y = 2x y = x

x = 2y

For the logarithm graph: The domain is:𝑥 ∈ (0; ∞). The graph never cuts the y-axis. The range is:𝑦 ∈ (−∞; ∞) .

Determine the value of ‘a’ in the equation 𝑦 = log𝑎 𝑥, if the graph

passes through the point : (4; 1

3)

Step 1: Substitute the values from the point, into the equation:

1

3= log𝑎 4

Step 2: Change to exponential form and solve for ‘a’:

𝑎1

3 = 4

(𝑎1

3)3

= 43

∴ 𝑎 = 64 Activity 1.5 1. Write the following exponential equations in logarithm form:

a) 34 = 1 b) (1

2)3=

1

8

c) 0.001 = 10−3 d) 102 = 100 2. Write the following logarithm equations in exponential form:

a) log4 256 = 4 b) log21

32= −5

c) log 1 = 0 d) log𝑥 𝑎 = 𝑦

Notes for graphs:

In the general equation,𝑦 = log𝑎 𝑥, the x-intercept is (1;0) and for

(the inverse), the y-intercept is (0;1).

If there is a coefficient, b, for 𝑦 = 𝑎𝑥, the y-intercept is affected. The

coefficient value, b, will always be the y-intercept. Coefficients with

the same value; but with opposite signs, namely being positive or

negative, will mirror one another about the x-axis.

If there is a constant term in an equation, i.e., the 𝑓(𝑥) = (1

3) 3𝑥 − 1

the (-1) represents the translation of the points of the graph

𝑓(𝑥) = (1

3) 3𝑥 , one unit down, including the y-intercept. So if the

y-intercept is (1

3), it will now be moved down one unit to become

(−2

3). The asymptote will move down one unit as well, therefore it will

now be y = -1.

Generally, the constant translates the graph up or down by the number

of units represented by the constant. The constant also determines the

asymptote of the graph.

Activity 1.6

1. y x = 2 g f x 0 B -1 A (2;-1)

The diagram shows the curves of 𝑓(𝑥) = (1

3)𝑥

and 𝑔(𝑥). 𝑔(𝑥) is a

quadratic equation with a turning point at A (2;-1), and passes through the origin.

a) Show that the equation of 𝑔(𝑥) is 𝑔(𝑥) = 1

4(𝑥 − 2)2 − 1.

b) Write down the coordinates of B c) Is g(x) a one-to-one function or a one-to-many function? Give a

reason for your answer. d) Write down the equation of f -1 in the form y = …. e) Determine, from the graph, for which values of x is g(x) > f(x).

2. y

f

P (1;3)

Q (0;1) g

x 0

a) Explain why the coordinates of Q are (0; 1). b) Write down the equation of f(x). c) Write down the equation of f -1 in the form y = … d) Is f -1 a function? Give a reason for your answer.

e) Determine the equation of h(x), the graph obtained when f(x) is reflected about the y-axis.

f) Determine the value of ‘a’ in g(x) which is a hyperbola (the asymptotes are the axes).

g) Determine the equation of m(x), the graph obtained when g(x) is moved 2 units to the right and 1 unit up.

h) Calculate the intercepts of m(x) with the x- and y-axis.

3. y

h

Q

A (-1;4

1 )

x

0 a) Calculate the value of ‘a’ in the equation h(x) = ax

b) Write down the equation of the inverse equation, h -1, in the form y = ….

c) Draw a sketch of h -1. Show on the graph the coordinates of two points that lie on this graph.

d) Read off from your graph the values for x for which log4 𝑥 > −1.

4. y x 0 1 16 f -2 The above sketch is the graph of 𝑓(𝑥) = − log4 𝑥 a) Write down the domain and range of f b) Write down the equation f -1, in the form of y = …. c) Write down the asymptote of f -1 . d) Explain, by using the graph of f, how you would sketch the

graphs with the following equations:

i) 𝑔(𝑥) = log4 𝑥 ii) ℎ(𝑥) = 4−𝑥 − 3

e) Use the graph of f to solve for x, where log4 𝑥 < 1

2

F. Applications of Logarithms Notes:

In Financial Mathematics, we use four different formulae, namely simple increase and decrease, and compound increase and decrease. These should all be familiar to you as you have learnt them from grade 9.

The formulae are as follows: 𝐴 = 𝑃(1 + 𝑛 ) Simple increase

𝐴 = 𝑃(1 − 𝑛 ) Simple decrease 𝐴 = 𝑃(1 + )𝑛 Compound increase 𝐴 = 𝑃(1 − )𝑛 Compound decrease

P = Principal amount A = Final amount i = interest rate n = period (Easy to remember – PAIN)

Logarithms can be used in the calculations to find the value of ‘n’, an exponent, by making ‘n’ the subject of the formula.

Activity 1.7 1. Mrs Nkosi, a teacher in Limpopo, invests R6500 in a bank account

which pays 10.5% interest per annum, compounded quarterly. She now has R18500 in her account. How long ago, rounded off to the nearest year, did she invest the original amount into the account?

2. Determine the time in years, for an amount of money to be doubled if

the interest rate is 13.24% per annum, compounded every 4 months. 3. R2250 is invested at 12% per annum, compound interest. After how

many years will the investment be worth R25000? 4. A photocopier valued at R28000 depreciates at a rate of 16% per

annum on the reducing-balance method. After how many years will its value be R16000?

𝐴 = 𝑃(𝑛 + 𝑖)𝑛

1 500 = 6 500 1 + 0.105

4 4𝑥

3

13= (

21

00)4𝑥

∴ 𝑥 = 10.09 ≈ 10 𝑦𝑒𝑎𝑟𝑠.

Remember that when doing these questions always write down all of your given

information – for example from question 1.

P = ?

A =?

i = ?

n = ?

Now fill in what you can from the story:

eg. P = R6 500

A = R18 500

i = 10.5% ÷ 4 (compounded quarterly) (don’t forget to change your percentage

into a decimal)

n = x × 4.

Now you are ready to substitute into your equation.

log821

800

37

13= 4𝑥 (Now you can put this log straight into your SHARP EL-W535HT –

To get to the log press then for fractions press the button.

Your answer should be: 4x = 40.367

5. The De Bruyn family purchase a car to the value of R380000. It depreciates in value on the reducing-balance method, by 11.4% per annum. Determine how long, to the nearest year, it will take for their car to be worth R200000.

6. Devon inherits R150000 and decides to invest it in a savings account

earning interest at 8.5% per annum compounded monthly. Calculate how many years it will take for his investment to be worth R300000. Give your answer to the nearest year.

7. Mary needs R400000 to start her own business. According to her

calculations, if she makes monthly payments of R6500, she will be able to repay the loan plus the interest in 7 years. If her bank offers her a loan with interest compounded monthly, what annual interest rate are they offering her (correct to one decimal place)?

Examples of Logs in the Real world:

Often logs are used in the “real world” to calculate the growth of populations such as bacteria or people. These would be considered application questions.

Growth rate of bacteria: The rate of the increase in the number of bacteria cells is proportional to the number of bacteria present at that time.

Formula: Rate of increase of bacteria cells = x number of cells. (𝜇 is the growth rate constant, its unit is hr)

To calculate the growth rate constant, we can use a formula using logs: log10𝑁 − log10𝑁0(N = Number of cells at time t) (N = Number of cells at time 0) or 𝜇 = [(log10𝑁 − log10𝑁0) × 2.303] ÷ ( − 0) (t = time taken)

(t0 = time is 0) Time interval

By measuring the increase in the number of bacteria cells during a certain time period, the growth rate constant can be calculated.

Calculate the growth rate, (𝜇),of a number of bacteria cells over a given time period.

Information: t0= 1.5 hours t = 8.5 hours log10𝑁0 = 1.92 log10𝑁 = .53

Information and diagram taken from wvlc.uwaterloo.ca

Solution: 𝜇 = [(log10𝑁 − log10𝑁0) × 2.303] ÷ ( − 0) = [(8.53-1.92) x 2.303] ÷ (8.5-1.5)

= (6.61 x 2.303) ÷ 7

= 15.22 ÷ 7

= 2.18hr-1