Lesson Note on MATHEMATICS for SS3 MS-WORD

This Free Lesson Note on MATHEMATICS for SS3  Note was pulled from our book ( Lesson Note on MATHEMATICS for SS3 MS-WORD ); Compiled to serve as reference material to help teachers draw out their lesson plan easier, saving you valuable time to focus on the core job of teaching.

The Lesson notes are based on the current NERDC curriculum (UBE compliant)[/su_note]

This Free Lesson Note on MATHEMATICS for SS3 Lesson Note Covers The Following Topics

1. THEORY OF LOGARITHM
2. SURDS: MEANING OF RATIONAL AND IRRATIONAL NUMBERS LEADING TO THE DEFINITION OF SURDS
3. APPLICATION OF SURDS TO TRIGONOMETRICAL RATIOS
4. MATRICES AND DETERMINANTS
5. LINEAR AND QUADRATIC EQUATION
6. WORD PROBLEMS LEADING TO ONE LINEAR-ONE QUADRATIC EQUATION; SURFACE AREAS AND VOLUME OF SPHERES AND HEMISPHERICAL SHAPE
7. LATITUDE AND LONGITUDE: IDENTIFICATION OF NORTH AND SOUTH POLESTHEORY OF LOGARITHM

 

WEEK 1

Topic: THEORY OF LOGARITHM

Laws of Logarithms

1. Product law

Log_a(MN) = log_a M + log_a N

e.g. log₂(2 x 4) = log₂ + log₂ 4

 

2. Quotient law

Log_a (M/N) =  log_a M – log_a N

e.g. log₁₀(1000/10) = log₁₀1000 – log₁₀10

 

3. Power law

Log_a M^p = P log_a M

e.g. log₂2³ = 3 log₂2

4a. log_a 1 = 0 where a ¹ 1

4b. log_aa = 1 e.g. log₁₀10 = 1, log₂2 = 1

 

5. Fractional Power

Log_aMx/y = x log_aM/y e.g. log₁₀100^3/2

6. Root Law

In a special case of 5 when x = 1

Log_aM1/x = log_a ^yÖM = log_aM/y e.g. log₂ 16^1/4 = log₂ ⁴Ö16 = log₂16/4

 

7. Change of base

log_aM = 1/log_m^a e.g. log₂8 = log₈2

 

Calculations based on the application of the basic rule

Simplify without using table

1.  Log₂ ³Ö8/log₂4

Log₂ 8^1/3/log₂2²

1/3 log₂2³/2log₂2

Log₂2/2log₂2 = ½

2.  3 log₁₀10²+ log₁₀ 10^-3

(2)3 log₁₀10 + -3 log₁₀10

6 log₁₀10 + (-3) log₁₀10

6 + (-3) = 3

3.  ½ log₅25 – log₅0.2

= log₅25^1/2 – log₅1/5

= log₅Ö25 – log₅1/5

= log₅5 –log₅1/5

= log₅ (5/1/5)

= log₅ 25 = log₅5² = 2log₅5 = 2

 

Application of Logarithm in Solving Problem Involving Calculations

APPLICATIONS OF EXPONENTIAL AN LOGARITHMIC FUNCTIONS

To solve an exponential or logarithmic word problem, convert the narrative to an equation and solve the equation.

In this section, we will review population problems. We will also discuss why the base of e is used so often with population problems.

 

Example 1: Suppose that you are observing the behavior of cell duplication in a lab. In one experiment, you started with one cell and the cells doubled every minute. Write an equation with base 2 to determine the number (population) of cells after one hour.

 

Solution and Explanations:

First record your observations by making a table with two columns: one column for the time and one column for the number of cells. The number of cells (size of population) depends on the time. If you were to graph your findings, the points would be formed by (specific time, number of cells at the specific time). For example, at t = 0, there is 1 cell, and the corresponding point is (0, 1). At t = 1, there are 2 cells, and the corresponding point is (1, 2). At t = 2, there are 4 cells, and the corresponding point is (2, 4). At t = 3, there are 8 cells, and the corresponding point is (3, 8).

It appears that the relationship between the two parts of the point is exponential. At time 0, the number of cells is 1 or 2⁰ = 1. After 1 minute, when t = 1, there are two cells or 2¹ = 2. After 2 minutes, when t = 2, there are 4 cells or 2² = 4.

Therefore, one formula to estimate the number of cells (size of population) after t minutes is the equation (model)

f (t) = 2^t.

 

Determine the number of cells after one hour:

Convert one hour to minutes. ^. = 60 min

Substitute 60 for t in the equation. f (t) = 2^t:

f (60) = 2⁶⁰ = 1.15×10¹⁸

 

Example 2: Determine how long it would take the population (number of cells) to reach 100,000 cells.

 

Solution and explanation:

In this example, you know the number of cells at the beginning of the experiment (1) and at the end of the experiment (100,000), but you do not know the time. Substitute 100,000 for f(t) in the equation f (t) = 2^t:

100, 000 = 2^t

Take the natural logarithm of both sides:

ln(100, 000) = ln(2^t)

Simplify the right side of the equation using the third rule of logarithms:

ln(100, 000) = t ln(2)

Divide both sides by ln(2):

t = = 16.60964 min

It would take 16.6 minutes, rounded, for the population (number of cells) to reach 100,000.

 

 

Example 3: Write an equation with base 5 that is equivalent to the equation

f (t) = 2^t.

 

Solution and Explanation:

Let’s start with a generic exponential equation with base 5:

f (t) = a ^. 5^bt.

The f(t) represents the size of the population at time t, the t represents the time, and the a and b represent adjusters when we change the base. The value of a is the number of cells (size of population) at the beginning of the study, and the value of b is the relative growth rate based on a base of 5. We need to find the values of a and b.

 

We know that the population is 1 at time 0, so insert these numbers in the equation

f (t) = a ^. 5^bt.

We have

1 = a ^. 5^{b . 0} = a ^. 5⁰ = a ^. 1 = a.

 

We now know that the value of a in the adjusted equation is 1.

Rewrite the equation

f (t) = a ^. 5^bt

with a = 1.

f (t) = 1 ^. 5^bt

 

which in turn can be rewritten as

f (t) = 5^{b . t}

 

We know that the population after 1 minute is 2 cells, so insert these numbers in the equation

f (t) = 5^{b . t}

to obtain

2 = 5^{b . 1}.

 

Solve for b by taking the natural logarithm of both sides of the equation 2 = 5^b.

ln(2) = ln(5^b).

 

Simplify the right side of the equation using the third rule of logarithms:

ln(2) = b ln(5).

 

Divide both sides of the equation by ln(5) and simplify:

b =In(2)/In(5) = 0.43067658075,

rounded to 0.4307.

 

Insert this value of b in the equation

f (t) = 5^{b . t},

and the equation is simplified to

f (t) = 5^0.4307t

 

We know that the population is 8 after 3 seconds, so use these values to check the validity of the above equation. Substitute 3 for t in the right side of the above equation. If the answer is 8, or close to 8 because we rounded, then the model (equation) is correct. 5^0.4307(3) = 8.00906, rounded to 8.

We know from the original equation that after 4 seconds, the population is 16. Let’s do a second check. 5^0.4307(4) = 16.02415, rounded to 16 cells. The check would be closer had we rounded b to more decimals.

ASSESSMENT

• Suppose that you are observing the behavior of cell duplication in a lab. In one experiment, you started with one cell and the cells tripled every minute. Write an equation with base 2 to determine the number (population) of cells after one hour.

 

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