AceGamsat - The Mathematics Bible

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The Mathematics Bible Mathematics to Ace the GAMSAT www.AceGAMSAT.com

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Table of Contents PART 1: GAMSAT MATHEMATICS TOPICS

PROPERTIES OF INTEGERS

2

ORDERS OF OPERATIONS

4

BRACKETS DIVISION AND MULTIPLICATION ADDITION AND SUBTRACTION

5 6 7

ALGEBRA

9

SUBSTITUTION INTO ALGEBRAIC EQUATIONS SIMPLIFYING ALGEBRAIC EXPRESSIONS ADDITION AND SUBTRACTION OF ALGEBRAIC FRACTIONS MULTIPLICATION AND DIVISION OF ALGEBRAIC FRACTIONS REMOVING GROUPING SYMBOLS BINOMIAL PRODUCTS SPECIAL PRODUCTS COMPLEX ALGEBRAIC EXPANSIONS

21

COMMON FACTORS FACTORISATION BY GROUPING IN PAIRS DIFFERENCE OF TWO SQUARES

SURDS

21 22 23

25

LOGARITHMS AND EXPONENTS

27

FACTORISATION

BASICS OF EXPONENTS NEGATIVE EXPONENTS FRACTIONAL EXPONENTS CONVERTING BETWEEN EXPONENTS AND LOGARITHMS EVALUATING LOGARITHMS LAWS FOR LOGARITHMS

SCIENTIFIC NOTATION

III

9 11 13 14 16 17 18 20

27 30 31 33 35 37

39

BASIC GEOMETRY

41

RIGHT-ANGLE TRIANGLES SPECIAL RIGHT TRIANGLES CIRCLES

41 42 43

LINEAR GRAPHS

45

PROBABILITY

47

THE PRODUCT RULE THE SUM RULE

47 48

PART 2: ESTIMATION TECHNIQUES FOR THE GAMSAT ESTIMATION TECHNIQUES

50

INTRODUCTION MULTIPLICATION AND DIVISION MULTIPLICATION AND DIVISION (COMPLEX NUMBERS) SCIENTIFIC NOTATION FOR ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION SQUARES AND SQUARE ROOTS FRACTIONS, RATIOS, PERCENTAGES TRIGONOMETRY LOGARITHMS AND NEGATIVE LOGS

64

METRIC SYSTEM PREFIXES SI SYSTEM CONVERSION FACTORS AND CONSTANTS

64 65 66

CHEAT SHEET

IV

50 51 53 54 57 58 60 61

The Mathematics Bible

Part 1: GAMSAT Mathematics Topics As you should be aware fundamental mathematics skills are required for the GAMSAT section 3. Mathematics skills are required in the physics and general chemistry questions. Math can also arise in biology with genetics related questions (probability). Remember, the GAMSAT section III is not directly testing mathematics skills - it is a science and reasoning test. ACER are testing to see if you understand the information and if you know how to apply logic and reasoning through questions. Most importantly they are testing if you know which formulas to use and how to apply them. It is essential that students have a good understanding of all the mathematics-based topics that can arise in the GAMSAT. In part 1 we will examine all the required mathematics-based topics for the GAMSAT. Students should work through every example! In part 2 of this guide we will be examining estimation techniques that can be applied to a range of mathematics-based situations in section III.

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Properties of Integers Students must understand the properties of integers, as these are the foundations of mathematics. 1. A positive number added to another positive number will result in a positive answer. Positive + Positive = Positive For example: 4 + 8 = 12

2. A negative number added to another negative number will result in a negative answer. Negative + Negative = Negative For example: (- 4) + (- 5) = - 9

3. A positive number added to a negative number will result in an answer with the sign of the highest magnitude. Positive + Negative = Sign of number with highest magnitude For example: (- 6) + 2 = - 4. Here we can see that 6 has a greater magnitude than 2, and since it is ‘- 6’ the answer will be negative.

4. A negative number minus a positive number will result in a negative number. Negative - Positive = Negative For example: (- 5) - 10 = - 15

5. A positive number minus a negative number will result in a positive number. Positive - Negative = Positive For example: 2 - (- 6) = 2 + 6 = 8 www.acegamsat.com

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6. A negative number minus a negative number will result in either a negative or a positive number. The answer will have the sign of the number with the highest magnitude. Negative - Negative = Sign of number with highest magnitude For example: (- 5) - (- 7) = (- 5) + 7 = 2 Here we can see that 7 has a greater magnitude than 5, and since it is ‘+ 7’ the answer will be positive.

7. A negative number multiplied by a negative number will result in a positive number. Negative x Negative = Positive For example: (- 6) x (- 2) = 12

8. A positive number multiplied by a negative number will result in a negative number. Positive x Negative = Negative For example: 3 x (- 3) = - 9

9. A positive number divided by a positive number will result in a positive number. Positive ÷ Positive = Positive For example: 10 ÷ 2 = 5

10. A positive number divided by a negative number will result in a negative number. Positive ÷ Negative = Negative For example: 36 ÷ (- 6) = - 6

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Orders of Operations When there is a simple sum that has only two numbers and one operation it is easy to see how to calculate the answer. For example:

3+1=4 5-2=3  4 x 4 = 16 10 ÷ 5 = 2

We simply add, subtract, multiply, or divide to find the answer. When there are several numbers and numerous operations we must follow a set of simple rules so we can determine the order in which to do the problem. For example: 5 + 3 x 3 If we solve this from left to right we get. 5 + 3 = 8 x 3 = 24. This is incorrect! Using the rules we would multiply first, and then add. So the correct answer would be 3 x 3 = 9 + 5 = 14. Here you can see that the rules must be followed or we will end up with an incorrect answer. A useful acronym that students should remember is BODMAS. By following this acronym we will know which order to solve the mathematical problems.

The BODMAS acronym stands for: Brackets (the part inside the brackets is always calculated first) Orders (numbers involving square roots or powers) Division Multiplication Addition Subtraction

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Brackets We begin solving an equation by solving what is in the brackets (from left to right).

Example 1: Evaluate 5 x (7 + 3) We will do the sum inside the brackets first (7 + 3), then multiply the answer by 5. = 5 x (7 + 3) = 5 x 10 = 50 If we ignored the brackets and did the sum from left to right we would get. 5 x 7 + 3 = 35 + 3 = 38 (this is incorrect).

Example 2: Evaluate 12 ÷ (4 - 1) We will do the sum inside the brackets first (4 - 1), then divide 12 by the answer. = 12 ÷ (4 - 1) = 12 ÷ 3 =4

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Order Next we will evaluate anything involving a square root or power (these are known as orders).

Example 1: Evaluate 42 + 8 We will evaluate the number containing the power first and then add 8. = 42 + 8 = 16 + 8 = 24

Example 2: Evaluate 122 - 44 We will evaluate the number containing the power first and then subtract 44. = 122 - 44 = 144 - 44 = 100

Division and Multiplication Once we have evaluated parts of the problem involving brackets and orders, we then look at division and multiplication. Multiplication and division are ranked equally, so it does not matter which one you do first. For ease and efficiency work from left to right when evaluating the problem.

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Example 1: Evaluate 5 x 10 ÷ 2 + 1 Multiplication and division are evaluated before the addition. In this case we can just work from left to right. = 5 x 10 ÷ 2 + 1 = 50 ÷ 2 + 1 = 25 + 1 = 26

Example 2: Evaluate 5 + 1 x 6 ÷ 2 Multiplication and division are evaluated before the addition. In this case we can work from left to right, starting at the number one, and leave the addition of 5 to the end. =5+1x6÷2 =5+6÷2 =5+3 =8

Addition and Subtraction

Addition and subtraction are ranked equally, so it does not matter which one you do first. For ease and efficiency work from left to right when evaluating the problem.

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Example 1: Evaluate 5 + 4 + 9 - 1 We can simply work from left to right to evaluate the problem. =5+4+9-1 =9+9-1 = 18 - 1 = 17

Example 2:

Evaluate 5 + 9 - 16 + 2 We can simply work from left to right to evaluate the problem. = 5 + 9 - 16 + 2 = 14 - 16 + 2 =-2+2 =0

Overall, there are 4 levels of rules that we must follow: Brackets, Orders, Division and Multiplication, Addition and Subtraction.

When all operations are the same level for a given problem we can simply work from left to right.

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Algebra We will cover algebra as it is an important skill that will be required in a range of questions in the GAMSAT.

Substitution into algebraic equations a = 2, b = 4 and c = -3.

Example 1:

Evaluate a + bc = a + bc = 2 + 4 x (-3) = 2 + (-12) = 2 - 12 = - 10

Example 2: Evaluate a - c =a-c = 2 - (-3) =2+3 =5

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Example 3: Evaluate (ac - 1) / b = (ac - 1) / b = (2 x -3 - 1) / 4 = (-6 -1) / 4 = -7 / 4

Example 4:

Evaluate c(a-b) = c(a-b) = -3(2 - 4) = -3(-2) =6

Example 5: Evaluate a2 + c2 = a2 + c2 = 22 + (-3) 2 =4+9 = 13

Example 6: Evaluate 3c2 - (3c) 2 + b

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= 3c2 - (3c)2 + b = 3(-3) 2 - (3 x -3) 2 + 4 = 3(9) - (-9) 2 + 4 = 27 - (81) + 4 = 27 - 81 + 4 = 31 - 81 = - 50

Simplifying algebraic expressions Example 1: Simplify 3x + 4x + 10x = 3x + 4x + 10x = 17x

Example 2: Simplify 5y - 2x + 2y - x = 5y - 2x + 2y - x = 7y - 3x

Example 3: Simplify 5ab + 4ba = 5ab + 4ba = 9ab (ab is the same as ba)

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Example 4: Simplify 4a x (-6b) = 4a x (-6b) = - 24ab

Example 5: Simplify (-4a) 2 = (-4a) 2 = -4a x -4a = 16a2

Example 6: Simplify ab ÷ a = ab ÷ a = ab/a = 1ab/a1 = b/1 =b

Example 7:

(10b + 4b) / 7 = (10b + 4b) / 7 = 14b / 7 www.acegamsat.com

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= 214b / 71 = 2b

Example 8: Simplify 4ab ÷ a2b = 4ab ÷ a2b = 4ab / a2b = (4 x a1 x b1) / (a1 x a x b1) =4/a

Addition and Subtraction of Algebraic fractions

When adding and subtracting fractions we find a common denominator and then add/subtract the numerators.

Example 1:

Simplify 7y/2 + y/3 = 7y/2 + y/3 We can see that the lowest common denominator of 2 and 3 is 6. So we multiply the (7y/2) by 3 and the (y/3) by 2. = 21y/6 + 2y/6 = 23y/6

Example 2:

Simplify 5x/2 – x/4 www.acegamsat.com

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= 5x/2 – x/4 = 10x/4 – x/4 = 9x/4

Example 3:

Simplify 3/2y + 7/5y = 3/2y + 7/5y = 15/10y + 14/10y = 29/10y

Example 4: Simplify 4a/3 + 2a/5 = 4a/3 + 2a/5 = 20a/15 + 6a/15 = 26a/15

Multiplication and Division of Algebraic Fractions

When multiplying we cancel and then multiply the numerators and denominators. When dividing we find the reciprocal of the second fraction (turn it upside down) and then multiply.

Example 1: Simplify x/2 x 9/2x www.acegamsat.com

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= x/2 x 9/2x = 1x/2 x 9/2x1 = 1/2 x 9/2 = 9/4 =2¼

Example 2: Simplify 8y/7 x 21/4y = 28y1/71 x 321/4y1 = 2/1 x 3/1 = 6/1 =6

Example 3: Simplify ab/6 ÷ b/4 = ab/6 ÷ b/4 = ab/6 x 4/b = ab1/63 x 24/b1 = a/3 x 2/1 = 2a/3

Example 4: Simplify 5xy/4a ÷ 15x/12 = 5xy/4a ÷ 15x/12 www.acegamsat.com

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= 15x1y/14a x 312/315x1 = y/a x 3/3 = y/a x 1 = y/a

Removing grouping symbols

The contents inside the grouping symbols are multiplied by the term outside of the grouping symbols.

Example 1:

Expand and simplify 5(2a + 4b) = 5(2a + 4b)

= 5 x 2a + 5 x 4b = 10a + 20b

Example 2: Expand and simplify b(b - 2) = b(b - 2) = b x b + b x -2 = b2 - 2b

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Example 3: Expand and simplify -(3 - 4a) = - (3 - 4a) = - 1(3 - 4a) = -1 x 3 - 1 x (-4a) = -3 + 4a

Example 4: Expand and simplify a(a + 2) - 5(a + 5) = a(a + 2) - 5(a + 5) = a2 + 2a - 5a - 25 = a2 - 3a – 25

Binomial products

A binomial product is the result of multiplying two binomial expressions. A binomial expression has two terms, for example (4a + 3). Therefore a binomial product would be (4a + 3)(2a + 1). Each term in the first binomial expression multiplies each term in the second binomial expression.

Example 1:

Expand and simplify (a + 3)(a + 5) = (a + 3)(a + 5) = a(a + 5) + 3(a + 5) = a2 + 5a + 3a + 15 = a2 + 8a + 15 www.acegamsat.com

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Example 2: Expand and simplify (2a - b)(a + b) = (2a - b)(a + b) = 2a(a + b) - b(a + b) = 2a2 + 2ab - ba - b2 (remember ab is the same as ba) = 2a2 + ab - b2

Example 3:

Expand and simplify (x - y)(x + y) = (x - y)(x + y) = x(x + y) - y(x + y) = x2 + xy – yx - y2 = x2 - y2

Special products The following special products should be memorised.  (a + b)2 = a2 + 2ab + b2  (a - b)2 = a2 - 2ab + b2  (a - b)(a + b) = a2 - b2

Example 1: Expand and simplify (x + 6)2

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= (x + 6)2 = (x)2 + 2(x)(6) + (6)2 = x2 + 12x + 36

Example 2:

Expand and simplify (2x - 4)2 = (2x - 4)2 = (2x)2 - 2(2x)(4) + (4)2 = 4x2 -16x + 16

Example 3: Expand and simplify (a - 4)(a + 4) = (a - 4)(a + 4) = (a)2 - (4)2 = a2 - 16

Example 4: Expand and simplify (6a + 2)(6a - 2) = (6a)2 - (2)2 = 36a2 - 4

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Complex Algebraic Expansions Example 1: Expand and simplify (x - 4)(x + 4) - (x - 3)2 = (x - 4)(x + 4) - (x - 3)2 = x2 - 16 - (x2 - 6x + 9) = x2 - 16 - x2 + 6x - 9 = 6x - 25

Example 2: Expand and simplify 3(x + 5)2 - (6 - x)(6 + x) = 3(x + 5)2 - (6 - x)(6 + x) = 3(x2 + 10x + 25) - (6 - x)(6 + x) = 3x2 + 30x + 75 - (36 - x2) = 3x2 + 30x + 75 - 36 + x2 = 4x2 + 30x + 39

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Factorisation Common factors Factorisation can be looked at as the opposite of expanding. We look for the largest common factors of the terms in the expression.

Example 1: Factorise 8x - 12 = 8x - 12 = 4(2x - 3)

Example 2:

Factorise 4a - ab = 4a - ab = a(4 - b)

Example 3: Factorise 16xy - 2x = 16xy - 2x = 2x(8y - 1)

Example 4: Factorise 12x2 + 9x2y

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= 12x2 + 9x2y = 3x2(4 +3y)

Example 5: Factorise 5a(a + b) + 7(a + b) = 5a(a + b) + 7(a + b) = (a + b) (5a + 7)

Factorisation by grouping in pairs

Example 1: Factorise xa + ya + xb + yb = xa + ya + xb + yb = a(x + y) + b(x + y) = (x + y)(a + b)

Example 2:

Factorise ab + a2 - b - a = a(b + a) - (b + a) = (b + a)(a - 1)

Example 3: Factorise a2 + bc + ac + ba

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= a2 + bc + ac + ba = a(a + b) + c(a + b) = (a + b)(a + c)

Difference of two squares We can reverse one of the special product rules: a2 - b2 = (a - b)(a + b)

Example 1: Factorise x2 - 16 = x2 - 16 = (x)2 - (4)2 = (x - 4)(x + 4) Note: the answer could also be written as (x + 4)(x - 4).

Example 2: Factorise 81 - 9y2 = 81 - 9y2 = (9)2 – (3y)2 = (9 - 3y)(9 + 3y)

Example 3: Factorise a4 - 1 = a4 - 1

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= (a2)2 - (1)2 = (a2 - 1)(a2 + 1) = (a - 1)(a + 1)(a2 + 1)

Example 4: Factorise x3 - x = x(x2 - 1) = x(x - 1)(x + 1)

Algebraic expression formulae a(b + c) = ab + ac (a + b)(a - b) = a2 - b2 (a + b)(a + b) = (a + b)2 = a2 + 2ab + b2 (a - b)(a - b) = (a - b)2 = a2 - 2ab + b2 (a + b)(c + d) = ac + ad + bc + bd

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Surds Surds are numerical expressions that involve irrational numbers. There are rules for surds that should be learnt as they can be examined in a variety of chemistry and physics questions in the GAMSAT.

Rules Rule 1: √(ab) = √a x √b

For example: Simplify the following, (Note: we look for two factors, the first of which is a perfect square) a) √128 = √64 x √2 = 8√2

b) 5√12 = 5 x √4 x √3 = 5 x 2 x √3 = 10√3

Rule 2: √(a/b) = √a / √b

For example: Simplify the following,

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a) √(36/81) = √36 / √81 = 6/9 = 2/3

Rule 3: (√a)2 = a

For example: Simplify the following:

a) (√9)2 = 32 =9

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Logarithms and Exponents Basics of Exponents

For xn, x is the base, and n is the index/power/exponent. We will refer to ‘n’ as the exponent in the following examples.

Exponent rules: x a x x b = x a+b x a÷ x b = x a-b (xa) b = x ab x 0= 1

Example 1: Simplify y5 x y4 = y5 x y4 = y (5 + 4) = y9

Example 2: Simplify y5 ÷ y = y5 ÷ y = y5 ÷ y1 = y (5 - 1) = y4 www.acegamsat.com

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Example 3: Simplify 35 ÷ 33 = 35 ÷ 33 = 3(5 - 3) = 32 = 9

Example 4:

Simplify 16 x4y ÷ 4 x3 = 4 xy

Example 5: Simplify (a3)5 = (a3) 5 = a3 x 5 = a15

Example 6: Simplify (x1/2)3 = (x1/2)3 = x (1/2 x 3) = x3/2

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Example 7: Simplify (3a4)3 = (3a4)3 = 33 x (a4)3 = 27a12

Example 8: Simplify 5y0 = 5y0 = 5 x y0 =5x1 =5

Example 9: Simplify (4y)0 + 4y0 = (4y) 0 + 4y0 =1+4x1 =1+4 =5

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Negative Exponents Students should understand that the negative power is the reciprocal. x-a = 1 / xa We will go through the following examples:

Example 1: Simplify 5-2 = 5-2 = 1 / 52 = 1 / 25

Example 2: Simplify (1/4) -2 = (1/4) -2 = 1 / (1/4) 2 = 1 / (1/16) = 1 ÷ 1/16 = 1 x 16/1 = 16

Example 3: Simplify 2y-3 www.acegamsat.com

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= 2y-3 = 2 x y-3 = 2 x 1/y3 = 2/y3

Example 4:

Simplify (2x2)-3 = (2x2)-3 = 1/(2x2)3 = 1/(8x6)

Example 5: Simplify x4/x9 = x4/x9 = x4

÷ x9

= x-5 = 1/x5

Fractional Exponents The fractional power rule should be memorised so numbers that have fractional powers can be solved without the use of a calculator.

The rule is as follows: x1/a = a√x So, x1/2 = √x, x1/3 = 3√x So, the law can be written as xa/b = a√xb www.acegamsat.com

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Example 1: Simplify 41/2 = 41/2 = √4 =2

Example 2: Simplify 81/3 = 81/3 = 3√8 =2

Example 3: Simplify 163/2 = 163/2 Tip: take the root first and then the power to keep number small as possible = (161/2)3 = (√16)3 = 43 = 64

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Converting between Exponents and Logarithms If you understand that a log is another way to write an exponent, it will help when you are required to work through various log problems in the GAMSAT. These problems usually arise with pH and half-life related questions. Definition of a log function: The logarithmic function with a base b, where b is not equal to one and b > 0 is represented by logb. It is defined by y = logb x, only if by = x In the GAMSAT you may need to convert an equation written in log form to exponential form and vice versa. When you are converting from log form to exponential form, b is your base, Y is your exponent, and X is what the exponential expression is equal to.

Example 1:

Express the logarithmic equation 3 = log4 64 exponentially. We will use the above definition: y = logb x, only if by = x. We can see that the b in the definition relates to the 4 in the problem, so the base is 4. The log is set to equal to exponent, so the exponent is 3. Now that leaves 64 as the result the exponential expression is set to. Now we get 3 = log4 64  43 = 64 We have now written the log form in exponential form. By looking at the exponential form we can see it is a true statement.

Example 2: Express the logarithmic equation log9 81 = y exponentially. We will use the definition - y = logb x, only if by = x. We can quickly see that the base is 9, since b in out definition relates to 9 in the problem. Remember that the log is set to equal the exponent, so the exponent is y. The exponent expression is set to equal 81.

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Now we get log9 81 = y  9y = 81. We have now written the logarithmic form in exponential form.

Example 3: Express the exponential equation: 7-2 = 1/49 in logarithmic form. Now we are going in the opposite direction – from exponential to logarithmic form. We will use the definition - y = logb x, only if by = x. We can see that the b in the definition relates to 7 in the problem. We can quickly see that the exponent is -2. The value of the exponential expression is set to 1/49 (represented by x in the log function). Now we get 7-2 = 1/49  log7 1/49 = -2 We have now written the exponential form in logarithmic form.

Example 4: Express the exponential equation √49 = x in logarithmic form. We will use the definition - y = logb x, only if by = x. First we need to rewrite the original problem using exponents. √49 = x (49)1/2 = x We can see that the b in the definition relates to 49 in the problem. We can quickly see that the exponent is ½. The value of the exponential expression is set to equal x. This goes inside the log function. √49 = x (49)1/2 = x ½ = log49 x www.acegamsat.com

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Evaluating Logarithms

You should now be comfortable with the definition of logs and how they are related to exponents. We will now examine examples that involve evaluating logs. The following 3 simple steps should be used when evaluating logs. 1. Set the log equal to x. 2. Now use the definition of logs (y = logb x, only if by = x.) to write the equation in exponential form. 3. Then find x.

Example 1: Evaluate the expression log5 125. The thought behind this is we are trying to find the power that we would need to raise 5 to so we can get 125. 1. Set the log equal to x. log5 125 = x 2. Write the equation in exponential form. 5x = 125 Now we think what power can we raise 5 to so we can get 125? The answer is 3 x=3

Example 2: Evaluate the expression log6 1. The thought behind this is we are trying to find the power that we would need to raise 6 to so we can get 1.

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1. Set the log equal to x log6 1 = x 2. Write the equation in exponential form. 6x = 1 Now we think what power can we raise 6 to so we can get 1? We know that any number raised to a power of zero will get one. The answer is 0. x = 0.

Example 3: Evaluate the expression log5 5. 1. Set the log equal to x log5 5 = x 2. Write the equation in exponential form. 5x = 5 Now we think what power can we raise 5 to so we can get 5? The answer is 1. x=1

Example 4: Evaluate the expression log3 √3. 1. Set the log equal to x log3 √3 = x 2. Write the equation in exponential form. 3x = √3

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Now we think what power can we raise 3 to so we can get √3? The answer is 1/2. x=½

Laws for Logarithms Logarithm product rule: logb(x y) = logb(x) + logb(y)

For example: log2 3 x 6 = log2 3 + log2 6 log10 4 + log10 2 = log10 8

Logarithm quotient rule: logb(x / y) = logb(x) - logb(y)

For example: log10 (22/5) = log10 22 - log10 5 log10 48 - log10 12 = log10 (48/12) = log10 4

Logarithm power rule: logb(x y) = y logb(x)

For example: log4 52 = 2 log4 5 1/2 log6 9 = log6 91/2 = log6 3 (remember 161/2 = √16)

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Logarithm of the base: logb(b) = 1

For example: log2 2 = 1 log10 10 = 1

Logarithm with x equal to 1:

For example: logb 1 = 0

For example: log2 1 = 0 log10 1 = 0

10 power log rule: 10log10M = M 10log105 = 5

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Scientific Notation Very large, or very small numbers are written as a product of a number between 1 and 10, and a power of ten.

Example 1:

Express 145.4 in scientific notation = 145.4 We move the decimal place two places to the left so we get a number between 1 and 10. = 1.454 x 102 The number of decimal places moved by the decimal point equals the power of 10. = 1.454 x 102

Example 2: Express 1 000 000 in scientific notation = 1 000 000 Move decimal place 6 places to the left and make 6 the power of 10. = 1 x 106

Example 3: Express 0.006 in scientific notation = 0.006

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We need to move decimal places 3 places to the right to get a number between 1 and 10 (6). If we move the decimal place three places to the right this is the same as multiplying by 1000. We must then divide by 1000 so the value is not changed. = 6 ÷ 1000 = 6 x 1/1000 = 6 x 10-3 (We will learn a shortcut method for this in Part 2)

Example 4: Express 3.2445 x 103 in decimal notation = 3.2445 x 103 Since 10 is to the power of 3, we will move the decimal place 3 places to the right. This is the same as multiplying by 1000. = 3244.5

Example 5:

Express 6 x 10-4 in decimal notation = 6 x 10-4 Since the 10 is to the power of - 4 , we will move the decimal place 4 places to the left. This is the same as multiplying by 1/10000. = 0.0006

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Basic Geometry Right-angle triangles We name each of the sides of a right-angle triangle as follows:

Note: the hypotenuse is the longest side. The opposite and adjacent sides are named with respect to θ.

Students should know the following: sin θ = opposite / hypotenuse (SOH) cos θ = adjacent / hypotenuse (CAH) tan θ = opposite / adjacent (TOA) Remember: ‘SOH CAH TOA’

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Example 1: Find the value of h in meters. Note: tan 60 = 1.73

tan 60 = opp/adj tan 60 = h / 5 1.73 = h / 5 h = 1.73 x 5 h = 8.65 cm h = 8.65 / 100 m h = 0.0865 m

Special Right Triangles

Since calculators are prohibited in the GAMSAT exam, it is crucial that students memorise the following special right triangles. These triangles contain the standard values of the trigonometric functions.

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Gradient = vertical rise / horizontal run Positive gradient (slope ascends from left to right) Negative gradient (slope descends from left to right)

Circles Students have previously been required to determine the cross-sectional area of a tube. The cross-sectional area of a tube is a circle. So we can use the formula for a circle to determine the cross-sectional area. Cross-sectional area = π.r2 Since calculators are prohibited in the GAMSAT students need to memorise that π is approximately equal to 3.14, and r = the radius. Some GAMSAT questions may ask to determine the circumference of the tube. This is the same as the circumference of a circle so we can use the following formula. Circumference of circle = 2π.r

Example 1: Find the cross-sectional area of a tube with diameter of 8 cm. Cross-sectional area = π.r2 Diameter = 2 x radius www.acegamsat.com

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Diameter = 8 cm, therefore radius = 4 cm = 3.14 x 42 = 3.14 x 16 = 50.24 cm2

Example 2:

Find the circumference of the same tube with a diameter of 8 cm. Circumference of circle = 2π.r Diameter = 2 x radius, therefore 2r = 8 Circumference = π x 2r

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Linear Graphs Equations that are in the form y = mx + b are termed linear equations. When graphed these form a linear graph. The value of y is where the linear graph intersects the y-axis and is called intercept b. The constant m is the slope of the line.

Let’s look at the above graph with equation y = 2x + 3. It cuts the y-axis at 3 and the x-axis at -1.5. Since we have the equation we can quickly see that the slope (m) = 2. If we did not have the equation, we could use two selected points on the graph to calculate the slope as follows.

The gradient between any two points: (x1, y1) and (x2, y2) are any two points, on the straight line. Gradient = rise / run = difference of y co-ordinates / difference of x co-ordinates = (y2 - y1) / (x2 - x1)

We will choose the given points on the graph above (-1.5, 0) and (0, 3). = (y2 - y1) / (x2 - x1)

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= (0 - 3) / (-1.5 - 0) = - 3 / - 1.5 =2

So, even without the given equation we are able to find the slope. A positive slope ascends as it extends to the right (graph above). A negative slope descends as it extends to the right. A horizontal line will have a slope of zero.

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Probability Probability will commonly arise in genetics-related questions. We will examine the key points of probability by examining traits of pea plants using Mendelian crosses. The probability of an event is calculated by dividing the number of times the event occurs by the total number of opportunities for the event to occur.

The Product Rule The pea-plant characteristics studied by Mendel were transmitted as discrete units from the parent to offspring. Different characteristics were transmitted independently of one another and could be considered in separate probability analyses. For example, a cross between a plant with green, wrinkled seeds and a plant with yellow, round seeds produced offspring with a 3:1 ratio of green: yellow seeds and a 3:1 ratio of round: wrinkled seeds. Since the transmission of these characteristics is independent, the product rule can be applied. The product rule states that the probability of two independent events occurring together can be calculated by multiplying their individual probabilities of each even occurring alone. So, what is the probability of the above pea plants producing green, wrinkled seeds? Probability of green is ¾ Probability of wrinkled is ¼ Therefore, probability of producing green, wrinkled seeds are, 3/4 x 1/4 = 3/16.

We can also examine another example. What is the probability of flipping a coin and landing on heads, and rolling a 6-sided die and obtaining a 3. Heads on coin = ½ Rolling a 3 = 1/6 Therefore, probability of landing on heads and rolling a 3 is, www.acegamsat.com

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1/2 x 1/6 = 1/12

The Sum Rule

The sum rule states that the probability of the occurrence of one event or the other, of two mutually exclusive events, is the sum of their individual probabilities. The word ‘or’ indicates that the sum rule is to be used. For example, a 6-sided die is rolled. What is the probability of rolling a 2 or a 5? These events are mutually exclusive because they cannot happen at the same time. To find the probability we will determine the probability of each event and then add the probabilities together. The probability of rolling a 2 is 1/6. The probability of rolling a 5 is 1/6. Therefore, the probability of rolling a 2 or a 5 is, = 1/6 + 1/6 = 2/6 = 1/3 The sum rule can be applied to show the probability of having just one dominant trait in the F2 generation of a dihybrid cross.

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Part 2: Estimation Techniques for the GAMSAT

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Estimation Techniques Introduction Estimation techniques can be used to increase speed, whilst still maintaining accuracy when solving problems in the GAMSAT. However, just like any skill, consistent practice is required to maximise speed and efficiency. Some answer choices will have numbers that are not very close to each other, so if you can do the correct math and get something close to the answer, you can quickly rule out the incorrect choices without long complicated equations and significant figures. Example: How many moles of Br- ions are found in 4.95 L of a 0.095 molar solution? You should quickly notice that the moles of Br- ions are approximately equal to 0.5 moles. Using a calculator this value is 0.47025, which is close enough to 0.5. If this question were in the GAMSAT the answers would be far enough from each other so that estimations can be accurately made. If this question was examined in the GAMSAT it would be too time consuming to work out the exact value using cross multiplication methods. We will now examine the steps taken: Firstly, we can see that 4.95 is very close to 5, so we can just round this to 5. Next, 0.095 is very close to 0.1, so we can just round to 0.1. Now we have 5 x 0.1 ≈ 0.5 mol In some cases there may be two solutions that are very close to each other. For example, let’s say that we have two close multiple-choice answers of 0.47 mol and 0.53 mol. We know that 4.95 and 0.095 were both rounded up to give our answer of 5. Now we know that our answer of 0.5 is slightly rounded up from the correct answer. Here we can quickly see that 0.47 < 0.5, so that must be the correct answer. Applying these techniques to GAMSAT practice-questions is the fastest way to improve!

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Multiplication and Division We will now examine a shortcut used when multiplying and dividing numbers by factors of 10. Sometimes you will need to round a number to the nearest 10 to use this shortcut. Using this method is easier and faster than doing it the long way.

Multiplying: Example: Find the value of 3 x 10? For this question we can simply just add a zero to get the answer of 30. What we are really doing is moving the invisible decimal place after the three one space to the right. Let’s see what happens when there is a visible decimal? Example: What is the value of 4.5 x 10?

In this case there is a visible decimal, so we are not able to just add a zero. Instead we need to shift the decimal one space to the right and we get 45. Example: What is the value of 16 x 100?

Now we are dealing with 100, which has two zeros. We will shift the decimal over to the right twice (two zeros) and we get 1600. Part of a question in the GAMSAT may require conversion between different units.

Example: Students may be required to convert 4.1 kg to g as part of a larger question. This can easily be converted mentally. We know that there are 1000 g per kg so we can straight away create a mental equation of 4.1 x 1000. There are three zeros so we will shift the decimal in the equation three places to the right to get 4100. When done in your head this should take less than a few seconds.

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Dividing: If multiplying by 10 means to move the decimal one place to the right, then dividing by 10 means to move the decimal one place to the left.

Example: What is the value of 20 ÷ 10?

We simple move the invisible decimal (after the 30) one place to the left. We then get the answer of 2.

Example: What is the value of 850 ÷ 100?

There are two decimals in 100 so we move the invisible decimal two places to the left. We get the answer of 8.5.

Now what if we are starting with a decimal?

Example: What if we are required to convert 0.044 kg to g. In other words, what is the value of 0.044 ÷ 1000?

There are three zeros in a thousand, so we will move the decimal three places to the left. We get the answer of 0.000044. This answer will usually be written in scientific notation – 4.4 x 105. Now, let’s see some examples that are not as simple.

Example: What is the value of 4.16 x 9.75?

We can round 9.75 to 10, and then just move the decimal in 4.16 one space to the right. We get the approximate answer of 41.6, which is very close to the actual answer of 40.46.

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Multiplication and Division (Complex Numbers) We will now examine more advanced multiplication and division when are given multiple digits or decimal spaces in the equation. The shortcut of moving the decimal places left and right can be further used in more complex questions.

Example: A boy kicks a stone weighing 69 grams giving it an initial acceleration of 215 m/s2. With how much force did the elephant hit the stone? This would be very easy to solve with a calculator, however without a calculator it could be time consuming. We will now break down the question to make it easier to solve. When solving this question, we need to look beyond the story and find the numbers you are going to calculate. Here we are given: 69 g and 215 m/s2. We will be using the formula: F = ma. This looks simple, but remember force has units of kg/m/s2 . First we must convert the grams to kilograms. 69 g ÷ 1000 = 0.069 kg (moving decimal three places to the left) Now using the formula: F = ma = 0.069 kg x 215 m/s2 0.069 x 215 looks complicated, so we will round to make it simple. We can round 0.069 to 0.070 and 215 to 200. Now we have a simplified question: 0.070 x 200 For some students, this still may be difficult to do in their heads. So now we can do the following: Take the first number (0.070) and multiply it by 100 and take the second number (200) and divide by 100. 0.070 x 100 = 7 200 ÷ 100 = 2 www.acegamsat.com

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Now we have a much more simple question that can easily be solved mentally: 7 x 2 = 14 N Using a calculator, the exact answer was 14.835, which on the GAMSAT is close enough to the options that will be given. Our goal here is to get rid of the confusing decimals, which make an equation difficult to mentally solve. Let’s looks at another example:

Example: Find the value of 0.05 x 400.

Here we can multiply the 0.05 x 100, and then divide the 400 by 100. We end up with: 5 x 4 = 20 The decimal place has been removed and the answer is very straight forward. If a similar question arises in the GAMSAT just mentally move the decimal to the right for the small number and to the left for the large number by the same number of spaces.

Scientific Notation for Addition, Subtraction, Multiplication and Division When it comes to adding, and subtracting exponents, as long as the power is the same you can just add or subtract the coefficient and keep the power.

Example: 5.5 x 10-4 + 3.5 x 10-4

Both numbers in the question have ‘x 10-4’, so all we have to do is add 5.5 and 3.5. 5.5 + 3.5 = 9 So the answer is 9.0 x 10-4 When the powers are not the same, but they are close enough, we want to turn them into the same powers.

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Example: 4 x 10-4 + 3 x 10-5

The powers are different so we cannot simply just add 4 and 3 together like in the previous example. We will have to either make both powers ‘-4’ or ‘-5’. We can make 4 x 10-4 into the suitable form by: Dividing the 10-4 by 10, and multiplying the 4 by 10. This will not change the value of the whole number, and will result in 40 x 10-5 Now we have both numbers with the power of -5 and we can simply add them together 43 x 10-5 With typical scientific notation, the first number should be between 1 and 10. So we can multiply 10-5 by 10, and divide 43 by 10 to get 4.3 x 10-4 A faster way is to remember that when you decrease the first number (moving the decimal place to the left) the exponent value is increased.

Multiplication When it comes to multiplication, simply multiply the numbers and add the exponents.

Example: Find the value of 2 x 106 x 3 x 103

We multiply 2 and 3 to get 6, and add 6 and 3 to get 9. We end up with 6 x 109

Division When it comes to division, simply divide the numbers and subtract the exponents.

Example: Find the value of 4 x 108 ÷ 8 x 104 www.acegamsat.com

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4/8 is the same as 1/2, which is the same as 0.5. We subtract 4 from 8 and get an exponent of 4. We are left with 0.5 x 104 This is not proper scientific notation so we will move the decimal one place to the right. Remember that moving the decimal one place to the right decreases the exponent by the value of 1. We are left with 5 x 103 This can commonly arise in many general chemistry GAMSAT questions.

Example: Find the SO42- concentration when 5.8 x 10-7 moles of H2SO4 are added to 4.1 x 10-3 L of H2O.

We want to solve for molarity = M = mol ÷ L 5.8 x 10-7 ÷ 5.1 x 10-3 Looking at the exponents we have -7 and -3. We will now subtract them to determine the exponent. -7 - (-3) = -4. Therefore, we have 10-4. Now we have 5.8, which is approximately 6, and we have 5.1, which is approximately 5. We can say that we have 6/5 x 10-4. This answer is not appropriate so we need to make it simpler. 6/5 is the same as 5/5 + 1/5 1 + 1/5 We know that 1/5 is 0.2, so the number is 1.2. So, the final answer is 1.2 x 10-4. By typing the original numbers into the calculator we get 1.41 x 10-4, which on the GAMSAT is close enough to be able to choose the correct answer.

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Squares and Square Roots Knowing the shortcuts for square and square roots is essential when racing against time in the GAMSAT.

Example: Find the hydronium ion concentration in 0.05 molar solution of acetic acid given an acid dissociation constant of 1.8 x 10-5.

The concept behind this question is discussed in ‘The Chemistry Bible’, however we will now just focus on the math. We have a 0.05 M solution and a Ka of 1.8 x 10-5. Ka = [H+][A-]/[HA] = x2/[HA] Now we have Ka = x2/[HA] We will multiply both sides by [HA] [HA] x Ka = x2 We will square root each side to isolate x √ ([HA] x Ka) = √x2 √ ([HA] x Ka) = x [HA] is the initial molarity = 0.05, and Ka of 1.8 x 10-5, so we get x = √ ((1.8 x 10-5) x 0.05) We can multiple 0.05 by 100 and then divide 1.8 x 10-5 by 100. Remember when dividing we drop the value of the exponent once for every factor of 10. So for 100 it will decrease by two. We now get x = √(5 x 1.8 x 10-7) 1.8 is approximately equal to 2, so we get x = √(5 x 2 x 10-7) x = √(10 x 10-7) By dividing the 10 on the left by 10, and multiplying 10-7 by 10, we get x = √(1.0 x 10-6) www.acegamsat.com

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Take the square root of each number. The square root of 1 is 1, and the square root of 10-6 is found by dividing the power by 2. So the square root of 10-6 is 10-3. x = 10-3. Therefore the concentration of the hydronium ion [H2O+] = 10-3. Students should know the rule that √X = X1/2.

Fractions, ratios, percentages You may come across a question as of follows in the GAMSAT.

For example: Find the molarity of a solution made by dissolving 35 g of NaBr in 500 ml of water.

The science portion of this question will be overviewed in ‘The General Chemistry Bible’, but for now we will just focus on the math. We are trying to find molarity (M) M = moles / L We must first solve for moles and then liters, which results in many different fractions. To solve for moles we set up the dimensional analysis in a form of ‘given x ratio’. Where 35 g is the given, and we multiply this by the ratio (1 mol NaBr )/ (Molar mass) Molar mass of NaBr = 22.99 + 79.9 = 23 + 80 = 103 (Note – the molar mass values of each atom will be given in the GAMSAT) Now we use ‘given x ratio’ 35 g x (1 mol NaBr) / (103 g) The grams cancel out and we are left with 35/103 Now we will find the liters, We have 500 ml x 1 L / 1000ml Cancel out the zeros and we get 5/10 = ½ Now we have (35/103)/(1/2) www.acegamsat.com

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When dividing by a fraction we simply multiply the top by the reciprocal of the bottom. 35/103 x 2 = ? 35/103 is approximately 1/3 because 33/100 is 1/3. In the GAMSAT we are allowed to round if it is close enough, so we will make 35/103 equal to 1/3. Now we have 1/3 x 2 = 2/3 You should know that 2/3 is the same as 0.67. So, we get 0.67 M By typing the exact values into a calculator, we get 0.68 M, which on the GAMSAT is close enough. TIP: Fractions are harder to put into longer calculations so try to turn them into decimals or whole numbers!

When it comes to percentages we want to treat these in a similar way to fractions.

For example: Find the percent ‘s’ character in a molecules that has Sp3 hybridisation.

When finding a percentage we solve for (part / whole) x 100 Where the whole is going to be the 100 value. The ‘s’ is a part, and we only have 1s and we have 3p. We therefore have a total of 4 parts. So, Part / whole = ¼ = 0.25. Now we will multiply 0.25 by 100 %. Since 100 has two zeros we will just move the decimal place two places to the right. We get the answer of 25%.

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Trigonometry Once again, it is expected that students know how to use sin cos and tan with a right-angled triangle. In the GAMSAT some questions may require you to find the value of some sin ratios that produce fractions. For example: sin 60 will equal √3/2. This may be hard to write in decimal form in the exam and can be very time consuming. The following table of Sin and Cos functions can be memorised to help increase speed and efficiency of approximations in the GAMSAT.

0 (degrees)

30

45

60

90

180

Sin

0

0.5

0.7

0.9

1

0

Cos

1

0.9

0.7

0.5

0

-1

Now, let’s apply this to a problem that may arise in the GAMSAT. A man pulls on a 4 kg toy car with a force of 9 N. Find the initial acceleration of the car if the force is applied at 30 degrees to the ground. The physics part of this question will be covered in ‘The physics Bible’ F=ma We want to find acceleration, so a=F/m We want to find the force that the car is being pulled in the x direction so we can find the acceleration. In other words, we want to find the component of the direction because the car is being pulled in the x direction.

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Force in the x direction = Fx, Fx = F x cos 30 Looking at the table above we get cos 30 as 0.9 Fx = 9 N x 0.9 We get 9 x 0.9. Since 9 x 9 is 81, we can just write 81 and move the decimal place one place to the left to get the answer of 8.1. Therefore, force in x-direction is 8.1 To find acceleration in x direction use a = F/m Round 8.1 to 8, mass is 4 kg a = 8/4 a = 2 m/s2

Logarithms and Negative Logs Students must have a basic understanding of logarithms to solve the pH and rate-law problems in the GAMSAT. Again, here are the log rules:

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Rule name

Rule

Logarithm product rule

logb(x ∙ y) = logb(x) + logb(y)

Logarithm quotient rule logb(x / y) = logb(x) - logb(y) Logarithm power rule

logb(x y) = y ∙ logb(x)

Logarithm of the base

logb(b) = 1

10 power log rule

10log10M = M

Logarithms commonly arise in the GAMSAT in pH related questions.

For example: Find the pH of a 4 x 10-4 molar solution of KOH.

Note: log 4 = 0.602 The chemistry part of this question will be covered in ‘The General Chemistry Bible’. First we will find the concentration of OH-, then the pOH, and then the pH. [OH-]  pOH  pH We know that the [OH-] is 4 x 10-4 To find the pOH we use the following equation: pOH = -log [OH-] = - log [4 x 10-4] - Logarithm product rule = - log [4 x 10-4] = - log 4 - log 10-4 = - log 4 + 4 (log 10) = - log 4 + 4(1) Logarithm of the base = 4 - log 4

It is given in the question that log 4 = 0.602

= 4 - 0.602 = 3.398 Therefore, pOH = 3.398 www.acegamsat.com

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Now we use the following equation to find pH: pOH + pH = 14 pH = 14 - pOH pH = 14 - 3.398 pH = 10.602 = 10.6

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Cheat Sheet Metric System Prefixes

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SI System

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Conversion Factors and Constants 1000 g = 1 kg

1 mL = 1 cc (cubic centimetre

1000 mL = 1 L

1000 L = 1 m

760 torr = 760 mm Hg = 1 atm

1 atm ≃ 105 Pa

Avogadro’s number ≃ 6. x 1023 p/mole

6.022 x 1023 amu = 1 gram

Density of H2O = 1g/mL

Density of H2O =1000 kg/m3

Charge on an electron = -1.6 x 10-19 C

1 eV = 1.6 x 10-19 C

1 angstrom (1Å) = 1010 m

Acc. due to gravity = 9.8 m/s

°C = Kelvin degrees - 273

Speed of light (vacuum) = 3 x 108 m/s

√2 = 1.4

√3 = 1.7

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