## Thursday, 28 February 2013

### Lesson Summary (February 27, 2013)

Worksheet A02d
6) log₅(log₃x)=log₁₀₀100
log₅(log₃x)=1
log₃x=5¹
x=3⁵
=243

7) log₅x=a            log√5y=b
express xy² as a power of 5
x=5ª                      y=√5ᵇ
=5^b/2

xy²=5ª x 5^2b/2
=5^a+2b/2
=5^a+b

*Don't cancel your answer even if it seems wrong (sometimes it might be correct) only cancel after realising the mistake and redoing.

Worksheet A02e
9) 6ⁿ + 6^n+1 + 6^+2
= 6ⁿ+6ⁿ6+6ⁿ6²
= 6ⁿ(1+6+36)
= 6ⁿ(43)
∴divisible by 43 for all natural values of n.

11) log₅(5-4x) = log√5(2-x)
= log₅(2-x)/log₅5^1/2
= log₅(2-x)/(1/2)
= log₅(2-x)²

5-4x = (2-x)²
= 4 - 4x + x²
x²=1
x=±1

* Sorry for the late post

## Tuesday, 26 February 2013

### Lesson Summary (February 26th 2013)

Morning Lesson:

We learnt how to solve logarithms using different methods.

Method 1: Covert log form to exponential form.
Method 2: Use change of base formula first.

For example:

Afternoon Lesson:

From the tables we have calculated out, we can conclude that:

- Power:   loga (xp) = p loga x

-  Product: loga (xy) = loga x + loga y

Quotient:

A formula can also be inputed into your graphic calculator to help change base:

Sorry I will input the formula when I get the paper back ><

Otherwise, you can use this log rule to change base too:

- Change of base formula

http://www.mathwords.com/c/change_of_base_formula.htm

Things to take note:

loga (x + y) ≠ loga x + loga y
loga (x – y) ≠ loga x – loga y

## Wednesday, 20 February 2013

### Applications OF Exponential and Logarithm

APPLICATIONS:

OF EXPONENTIAL
GRAPHING EXPONENTIAL

====================================================

OF LOGARITHM

GRAPHING LOGARITHMS

====================================================

KEY RULES OF LOGARITHMS

====================================================

GUIDELINES TO SOLVE LOGARITHMS

### TI84plus : Logarithms - Change of Base

TI84plus : Logarithms - Change of Base

How to use TI84 to Calculate Log of a different Base

How to Programme Change of Base

Solving Quadratic Equation Using TI8plus programme

## Logarithm

### Laws of Logarithm:

1.
a^x = y
is equals to:
log a(y) = x

2.
log10 = lg
log e = l(n), n = natural log.

3.
lg 1 = 0
log10 (1) = 0
10^0 = 1
Because:
loga(1) = 0
a ^0 = 1
where a ≠0.

4.
loga(a) = 1
where a ≠ 0, a >0
log10(10) = 1
because:
log10(10) = 1
10^1 = 10

Note:

5.
log10(-5) = error?
why?

-5 = 10^?

there is no value for the power of 10 to become -5, therefore log a base b = c, where a ≠ a negative integer or  < 0.

6.
loga(b) = log10(b) / log10(a)
= ln(b) / ln(a)

caution!:
lg(b)/lg(a) ≠ b/a (do not cancel!!)

Correct:
lg(b)/ lg(a) = calculate both the top and the bottom.

7.
logc(ab) = logc(a) + logc(b)

8.
logc (a/b) = logc (a) - logc (b)

9.
log10(100) = log10(10)^2
= 2log10(10)
= 2

note:
loga(b)^n = n loga(b)

## Study the video provided for clarity of concept and greater appreciation of TI84plus.

Task 2: Extend your learning by considering the case when the answers are not Real ie. Imaginary

## Thursday, 14 February 2013

### Surds

Important = *

*
(2√3)^2
=2√3 x 2√3
=4√3√3
=4(√3)^2
=4(3)
=12

Conjugate Surds

*
(a+b)(a-b) = a^2 - b^2
(√a+√b)(√a-√b)=(√a)^2-(√b)^2
(a^1/2 + b^1/2) (a^1/2 - b^1/2) = (a^1/2)^2 - (b^1/2)^2

2√3^2=
a) 2√3 √3 (correct)
b)2(√3)^2 (correct)
c)2 . 3^1/2x2 (correct)
d) 4(3) (wrong)
e) √4(3) ^2 (wrong)
f) √4 √9 (wrong)

To convert irrational surds into rational surds

By similar surd
1/√2 x √2(to both numerator and denominator)
= √2 /2

By conjugate surd
(1/ √3 - √2) x √3 + √2 (to both numerator and denominator)
=√3 + √2/ 3-2
=√3 + √2 / 1

## Wednesday, 13 February 2013

### Lesson Summary 13/2/13 Poon Jia Qi

Things to note when plotting:
- Do the whole question on a piece graph paper
- A table of values will be provided, but some values will be missing
- The scale for the X and Y axis may be different, don't assume they are the same (Write the scale on your paper if you have time)  [e.g. -4 ≤ x ≤ 4]

- Draw in PENCIL
- Plot your points, be consistent and use crosses (x)
- Write your equation after you're done

Possible questions:

1. Find the turning point, using the graph
- Characteristics
• Tangent = 0
• Minimum or maximum point
• Use the line of symmetry to find out the turning point

2. Tangent
- Draw the tangent line on the graph
- Find the gradient of the line, using the formula (y1-y2)/(x1-x2)

3. Intersection of two equations
- Plot the other equation (Normally a linear graph)
- Find the intersection points of the two equations and write them down

4. Co-ordinates
- Represent in the form (x,y)

Simplify -32
-32 = -1 * 3 * 3
= -(3)2
-32 ≠ (-3)2
= (-3) * (-3)

Simplify 2 √ab
2 √(ab) ≠ 2√(ab)
= √a * √b
= a½ * b½

2 √(ab) = 2 * √a * √b
= 2 * a½ * b½

Simplify 2 √(ab) * 3 √(ab)
2 √(ab) * 3 √(ab) = 2 * √a * √b * 3 * √a * √b
= 6 (√a)2 (√b)2
= 6ab

Good luck for the upcoming level test and study hard ^^
p.s. Sorry for the late post

I
Q5

Farrell Nah (08)

Question 2)

## Wednesday, 6 February 2013

### Lesson Summary 7/2/13 Shaun Ng

Level Test

Elementary math

Duration: 45 mins
30 marks

Topics:
1. Algebraic manipulation
• Factorisation
• Expansion
1. Algebraic Fraction
3  Indices (no surds + logs) (quiz c)
1. Quadratic Function/graph plotting  (U shaped and n shaped)

Duration: 45 mins
30 marks

Topics:
1. Polynomials
2. Remainder Factor Theorem
3. Cubic expression/equation
4. Partial fraction
(All Jumbled up)
(Start doing Exam Prep Questions)

Question for fun

Given that the roots of a quadratic function is -3 and 1. Find the function if x=-3 (x+3)=0

1. Case 1
the coefficient of x^2 is 2     (3)

1. Case 2
y intercept is -6.              (3)

Sketch the above function(s), showing clearly the x and y intercepts, and the turning point. State also the nature of turning point. (2)

y = f(x) = ax^2+bx+c
= A(x+d)(x+e)
= A(x_3)(x-1)
= A(x^2+2x-3)
=Ax^2+2Ax-3A

1. f(x) = 2(x+3)(x-1)
= 2x^2+4x+6

1. f(x) = A(x+3)(x-6)
x=0, f(0) = A(3)(-1) = -6
A = 2
Hence, f(x) = 2(x+3)(x-1)
= 2x^2+4x-6

Nature of turning point = Minimum tp
Turning point = Two x intercepts added/2
Sub x value into answer for a.]]]

### Lesson Summary 6th Feb - Wednesday ; Kaelan

Practice 1:

1 = A(x+2) + B(x+1)
1 = (A+B)x + 2A + B

Compare Co-efficients. A+B = 0
2A+B=1
Therefore:  A = 1; B = -1

1             1
------   -   ------
(x+1)      (x+2)

b) Simple:

9x+9                 9(x+1)                 9
--------------- =  -----------------  =  --------
(x+1)(x-2)          (x+1)(x-2)          (x-2)

c) 3x+5 = A(x+2)(x+3) + B(x+1)(x+3) + C(x+1)(x+2)

Sub x = -1

2 = 2A
A = 1

Sub x = -2, -1 = -B, B = 1
Sub x = -3, -4 = 2C, C = -2

Therefore :    1                      1                    2
-------------   +  -------------   -  ------------
(x+1)               (x+2)              (x+3)

Practice 2 :

1 = Ax(x-1) + B(x-1) + Cx^2 OR        1=(Ax+B)(x-1)+Cx^2
1 = (A+C)x^2 + (B-A)x -B OR     x=1, C=1
Compare Coefficients OR       x=0, B=-1
B=-1 OR        x=-1, A = -1
A =-1
C = 1

A question :

x^2
------------------
9-x^2

Long divide to give :  -1 + 9/9-x^2
= -1  + 9/(3+x)(3-x)
9 = A(3-x) + B(3+x)
x=3 , B=1.5
x=-3, A=1.5

-1 + 3/(6-2x) + 3/(6+2x)

NOTE: ALWAYS FACTORISE, A NON FACTORISED POLYNOMIAL WILL NOT WORK. have fun in a loop.

## Tuesday, 5 February 2013

### Partial Fractions - Jia Qi and Carissa

Definition:
Partial fractions are used to simplify a complex fraction into two or more fractions. This is so that there will be less confusion when dealing with algebraic fractions.

Proper Rational Expression:
The degree of the top is less than the degree of the bottom

Improper Expression:
The degree of the top is greater than, or equal to the degree of the bottom.
Use polynomial long division first. The remainder will be the proper fraction, use the proper fraction to solve the question.

Workings:

Different cases will result in different partial fraction forms:

### Lesson Summary 5th February

Homework :
Linoit
Comment on the posts on the math blog about the error
Prepare a thing plastic red file for filing .
Corrections

Remember :
Rules of index
Thanks to Justin

For Comparison
In order to rid the denominator of the surd/square root  , we rationalize it .
Rationalizing is the process whereby we multiply the denominator by another fraction with the numerator and denominator like it . We make use of the identities such as : (a+b)^2 = a^2 + 2ab + b^2 ,
(a-b)^2 = a^2-2ab+b^2 or (a+b)(a-b) = a^2-b^2

Recall : Prime factorization
It is a way of breaking down a compound number into  prime factors

## Monday, 4 February 2013

### Lesson Summary ~(•O•)~ 4/2/13

RECALL:
~(•o•)~ Partial Fractions! Read through the summaries posted by yourself/others
~(•o•)~ Level Tests are on Week 7 & 8, starting mugging.

HOMEWORK:
~(•o•)~ Check what you haven't done on the homework spreadsheet.
~(•o•)~ AceLearning for those who did badly on the various quizzes and tests.
~(•o•)~ Assignment 01 and 02 are now due on Friday~!

REVISE:
~(•o•)~ The 9 laws of indices: (here's a helpful table)

Please remember to fully revise all laws as they will be used in Indices and Surds.

LESSON:
~(•o•)~ Indices and Surds: Forms

Index Form: When there is an index for the base. (Eg. 2^6)
Surd Form: When the base has a squareroot with it (√5)
Radical Form: When the base has an fractional index (2^(2/123))

~(•o•)~ Indices and Surds: Rationalization
Rationalization is the act of eliminating the squareroot (√x) from the denominator.

BACK TO BASICS– "  1/2 VS 1/3"
How do we determine which is bigger?
We change make them have an equal denominator. Thus, the LCM between 2 and 3 is 6. Therefore, we multiply 1/2 with 3/3 (Note how this is 1 whole), and 1/3 with 2/2. Eventually, we get "3/6 > 2/6"

ON TO SURDS! – "1/2 VS 1/√2"

How do we determine which is bigger?
Once again, we change them to make them have an equal denominator. In this case, we change the denominator to 2. As the first fraction already has a denominator of 2, we will work with 1/√2.
2 divided by √2 is √2. Thus, we multiply 1/√2 by √2/√2 (1 whole), and get √2/2.
Therefore, we would get "1/2 < √2/2"

Rationalizing with a^2 - b^2 – "1/(3 - 2) VS 1/(√3 - √2)"
To rid of the surd denominator, we use the 'a^2  b^2 = (a + b)(a - b)' concept.
In this scenario, our denominator is (√3 - √2) which is our (a - b). Thus, to rid of it, we multiply it by (√3 + √2) which represents our (a + b).
Thus, we multiply 1/(√3 - √2) by (√3 + √2)/(√3 - √2) <-- which is one whole!
This gives us (√3 + √2)/(3 - 2) = (√3 + √2)/1
Therefore, "1 < (√3 + √2)"

### TASK FOR 5th February 2013

Period 1

1.  Self Assessment ( please refer to Quiz C [10mins]
- complete the corrections
- identify the errors committed be it (i) conceptual or (ii) carelessness

2. Collaborative Effort - via Maths Blog  -  Linoit exercise  [20mins]

- S3-05   complete the portion on Indices
- S3-09   complete the portion on Surds

3. Peer Assessment - via Maths Blog activity

INDICES & SURDS - The Mistake Fragment part 1
- post the responses as comments

4. Self Assessment - via Maths Blog activity
- INDICES & SURDS - The Mistake Fragment part 2
================================================================Period 2

5. Diagnostic Test 2 - Indices, Surds and Logarithms
- do be done on foolscap papers
- scribe of day to collect and past to teacher

### INDICES & SURDS - The Mistake Fragment 2

By Mr Johari

Identify the mistakes shown below and post your corrections.
For the correction identify the correct rules.

Find the algebra mistake:
1.
2.
3.

4.      Find the algebra mistake:

5.     Find the algebra mistake:

### INDICES & SURDS - The Mistake Fragment part 1

BY MR JOHARI

Context
Below shown the solution posted by the secondary 3 students during the 4th February 2013 Maths Quiz C.
There are errors in the solution posted.

Focus
Error Analysis and Peer Evaluation
Individual Effort

You are required to post the following as a comment:

(i) Nature of error - conceptual or carelessness (be very specific in your description)
example - careless error due to arithmetic (addition)
example - conceptual error due to error in (a + b)^2 = a^2 + b^2.

(2) Correction to the error
Ensure that your solution is concise and correct.

## Sunday, 3 February 2013

### INDICES, SURDS & LOGARITHMS

BY MR JOHARI This is an Interclass Collaboration on INDICES and SURDS S3-05 focuses on INDICES S3-09 focuses on SURDS TASK 1. Identify as many laws of Indices as possible. 2. Include condition, where possible such as a > 0. 3. Include also examples on how the rules are used.
CLICK

Indices and Surds to access Linoit