3-Surds, Indices and logarithms

# 3-Surds, Indices and logarithms ## Simplification and operations with surds, Rationalising the denominator and the bounds of an irrational number ### 1. Introduction Welcome to this lesson on surds! Surds are fascinating numbers that appear when we take roots of non-perfect squares. They are irrational numbers, meaning they cannot be expressed as a simple fraction and their decimal representation goes on forever without repeating. In this lesson, we will learn how to simplify surds, perform basic arithmetic operations (addition, subtraction, multiplication, and division) with them, and how to eliminate surds from the denominator of a fraction through a process called rationalisation. Finally, we'll explore how to estimate the value of an irrational number, such as a surd, by determining its bounds. ### 2. Key Concepts #### 2.1 What is a Surd? A **surd** is an irrational number that can be expressed in the form \( \sqrt[n]{a} \), where 'a' is a positive rational number and 'n' is a positive integer. If the root can be simplified to a rational number, it is not a surd. * **Examples of Surds:** \( \sqrt{2} \), \( \sqrt{3} \), \( \sqrt{5} \), \( \sqrt{7} \), \( \sqrt[3]{4} \). * **Examples of numbers that are NOT surds:** \( \sqrt{4} = 2 \), \( \sqrt{9} = 3 \), \( \sqrt[3]{8} = 2 \). These are rational numbers. #### 2.2 Simplification of Surds The goal of simplifying a surd is to express it in the form \( a\sqrt{b} \), where \( b \) has no perfect square factors other than 1. We use the property: \[ \sqrt{ab} = \sqrt{a} \times \sqrt{b} \] **Steps to simplify:** 1. Find the largest perfect square factor of the number under the root sign. 2. Rewrite the surd using the property \( \sqrt{ab} = \sqrt{a} \times \sqrt{b} \). 3. Simplify the perfect square root. #### 2.3 Operations with Surds ##### 2.3.1 Addition and Subtraction of Surds Only **like surds** can be added or subtracted. Like surds are those that have the same number under the root sign after simplification. * Treat like surds similar to how you treat algebraic terms (e.g., \( 2x + 3x = 5x \)). * **Rule:** \( p\sqrt{a} + q\sqrt{a} = (p+q)\sqrt{a} \) * **Rule:** \( p\sqrt{a} - q\sqrt{a} = (p-q)\sqrt{a} \) ##### 2.3.2 Multiplication of Surds To multiply surds, we use the property: \[ \sqrt{a} \times \sqrt{b} = \sqrt{ab} \] * For expressions with coefficients: \( p\sqrt{a} \times q\sqrt{b} = pq\sqrt{ab} \) * **Special Case:** \( \sqrt{a} \times \sqrt{a} = a \) ##### 2.3.3 Division of Surds To divide surds, we use the property: \[ \frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}} \] #### 2.4 Rationalising the Denominator Rationalising the denominator means removing the surd from the denominator of a fraction, making it a rational number. This is done by multiplying both the numerator and the denominator by an appropriate expression. ##### 2.4.1 Monomial Denominators (e.g., \( \frac{a}{\sqrt{b}} \)) * Multiply the numerator and denominator by the surd in the denominator. * **Method:** \( \frac{a}{\sqrt{b}} = \frac{a}{\sqrt{b}} \times \frac{\sqrt{b}}{\sqrt{b}} = \frac{a\sqrt{b}}{b} \) ##### 2.4.2 Binomial Denominators (e.g., \( \frac{a}{b + \sqrt{c}} \) or \( \frac{a}{\sqrt{b} + \sqrt{c}} \)) * Multiply the numerator and denominator by the **conjugate** of the denominator. * The conjugate of \( (x + \sqrt{y}) \) is \( (x - \sqrt{y}) \). * The conjugate of \( (\sqrt{x} + \sqrt{y}) \) is \( (\sqrt{x} - \sqrt{y}) \). * We use the difference of squares identity: \( (A+B)(A-B) = A^2 - B^2 \). * **Method:** \( \frac{a}{b + \sqrt{c}} = \frac{a}{b + \sqrt{c}} \times \frac{b - \sqrt{c}}{b - \sqrt{c}} = \frac{a(b - \sqrt{c})}{b^2 - (\sqrt{c})^2} = \frac{a(b - \sqrt{c})}{b^2 - c} \) #### 2.5 The Bounds of an Irrational Number Finding the bounds of an irrational number means determining two rational numbers (often consecutive integers or numbers with a specific decimal precision) between which the irrational number lies. This helps in estimating its value without a calculator. **Steps:** 1. For a surd \( \sqrt{N} \), find perfect squares that are immediately below and above \( N \). 2. Take the square root of these perfect squares to find the integer bounds. 3. For more precise bounds (e.g., to one decimal place), square numbers with one decimal place until you find values that bound \( N \). ### 3. Vocabulary * **Surd:** An irrational number that is the root of an integer or rational number, such as \( \sqrt{5} \). * **Rational Number:** A number that can be expressed as a fraction \( \frac{p}{q} \) where p and q are integers and \( q \neq 0 \). E.g., \( \frac{1}{2}, 3, -0.75 \). * **Irrational Number:** A number that cannot be expressed as a simple fraction. Its decimal representation is non-terminating and non-repeating. E.g., \( \sqrt{2}, \pi \). * **Like Surds:** Surds that have the same irrational factor after simplification (e.g., \( 2\sqrt{3} \) and \( 5\sqrt{3} \)). * **Conjugate:** For a binomial expression involving a surd, such as \( (a + \sqrt{b}) \), its conjugate is \( (a - \sqrt{b}) \). Used to rationalise denominators. * **Rationalising the Denominator:** The process of converting the denominator of a fraction from an irrational number (a surd) to a rational number. * **Bounds:** The upper and lower limits between which a value lies. ### 4. Examples #### Example 1: Simplification of Surds Simplify \( \sqrt{72} \). **Solution:** \[ \sqrt{72} = \sqrt{36 \times 2} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2} \] #### Example 2: Addition and Subtraction of Surds Simplify \( 3\sqrt{5} + \sqrt{20} - \sqrt{45} \). **Solution:** First, simplify each surd: \( \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \) \( \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \) Now substitute back and combine like surds: \[ 3\sqrt{5} + 2\sqrt{5} - 3\sqrt{5} = (3+2-3)\sqrt{5} = 2\sqrt{5} \] #### Example 3: Multiplication of Surds Simplify \( (2+\sqrt{3})(4-\sqrt{3}) \). **Solution:** Use the distributive property (FOIL): \[ (2+\sqrt{3})(4-\sqrt{3}) = 2(4) - 2(\sqrt{3}) + 4(\sqrt{3}) - (\sqrt{3})(\sqrt{3}) \] \[ = 8 - 2\sqrt{3} + 4\sqrt{3} - 3 \] \[ = (8-3) + (-2+4)\sqrt{3} \] \[ = 5 + 2\sqrt{3} \] #### Example 4: Division of Surds Simplify \( \frac{\sqrt{50}}{\sqrt{2}} \). **Solution:** \[ \frac{\sqrt{50}}{\sqrt{2}} = \sqrt{\frac{50}{2}} = \sqrt{25} = 5 \] #### Example 5: Rationalising a Monomial Denominator Rationalise the denominator of \( \frac{6}{\sqrt{3}} \). **Solution:** Multiply numerator and denominator by \( \sqrt{3} \): \[ \frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \] #### Example 6: Rationalising a Binomial Denominator Rationalise the denominator of \( \frac{5}{3-\sqrt{2}} \). **Solution:** Multiply numerator and denominator by the conjugate of the denominator, which is \( (3+\sqrt{2}) \): \[ \frac{5}{3-\sqrt{2}} = \frac{5}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} \] \[ = \frac{5(3+\sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{15+5\sqrt{2}}{9-2} = \frac{15+5\sqrt{2}}{7} \] #### Example 7: Bounds of an Irrational Number Determine the interval of \( \sqrt{40} \) to one decimal place. **Solution:** First, find integer bounds: \( 6^2 = 36 \) and \( 7^2 = 49 \). Since \( 36 < 40 < 49 \), then \( \sqrt{36} < \sqrt{40} < \sqrt{49} \), so \( 6 < \sqrt{40} < 7 \). Now, for one decimal place: \( 6.1^2 = 37.21 \) \( 6.2^2 = 38.44 \) \( 6.3^2 = 39.69 \) \( 6.4^2 = 40.96 \) Since \( 39.69 < 40 < 40.96 \), then \( \sqrt{39.69} < \sqrt{40} < \sqrt{40.96} \). Therefore, \( 6.3 < \sqrt{40} < 6.4 \). ### 5. Exercises 1. **Simplify the following surds:** a) \( \sqrt{27} \) b) \( \sqrt{125} \) c) \( \sqrt{98} \) 2. **Evaluate the following expressions:** a) \( \sqrt{12} + \sqrt{75} \) b) \( 4\sqrt{3} - \sqrt{48} \) c) \( (3\sqrt{2})^2 \) d) \( (\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2}) \) 3. **Rationalise the denominator of the following:** a) \( \frac{10}{\sqrt{5}} \) b) \( \frac{1}{4+\sqrt{7}} \) c) \( \frac{\sqrt{3}}{\sqrt{6} - \sqrt{2}} \) 4. **Find two consecutive integers between which the following surds lie:** a) \( \sqrt{200} \) b) \( \sqrt{70} \) ### 6. Assignment 1. **Simplify and evaluate:** a) \( (\sqrt{18} - \sqrt{8})^2 \) b) \( \sqrt{5} (\sqrt{45} + \sqrt{80}) \) 2. **Rationalise the denominator and simplify:** a) \( \frac{2\sqrt{3}}{2\sqrt{3}-1} \) b) \( \frac{4}{\sqrt{7}+\sqrt{3}} \) 3. **Given that \( x = 1 + \sqrt{2} \), find the value of \( x^2 - 2x + 1 \).** 4. **Estimate \( \sqrt{7} \) to one decimal place by finding its upper and lower bounds.** ### 7. Summary In this lesson, we've explored the world of surds. We learned that surds are irrational numbers that can be simplified by extracting perfect square factors. Mastering operations with surds is crucial, remembering that only like surds can be added or subtracted, while multiplication and division follow specific rules. We also covered the important technique of rationalising the denominator, which helps in presenting fractions without surds in the denominator, using either the surd itself or its conjugate. Finally, we practiced estimating the value of irrational numbers by finding their bounds, a valuable skill for understanding their magnitude. These skills are fundamental for further studies in algebra and beyond. ## Laws of indices and Simple exponential equations ### Introduction Welcome to this lesson on the Laws of Indices and Simple Exponential Equations! In mathematics, indices (also known as exponents or powers) provide a shorthand way of writing repeated multiplication. For example, \( 2 \times 2 \times 2 \) can be written as \( 2^3 \). Understanding the rules, or 'laws', of indices is crucial for simplifying complex expressions and solving a special type of equation called an exponential equation, where the unknown variable is found in the exponent. This lesson will guide you through the fundamental laws that govern operations with indices and show you how to apply them to solve simple exponential equations. ### Key Concepts #### Indices (Powers) An index (plural: indices) is a small number written above and to the right of a base number. It tells you how many times to multiply the base number by itself. * **Base:** The number being multiplied repeatedly. * **Exponent/Index/Power:** The number that indicates how many times the base is used as a factor. For example, in \( a^n \): * \( a \) is the base. * \( n \) is the exponent (or index or power). #### Laws of Indices These are fundamental rules that simplify calculations involving powers. 1. **Multiplication Law:** When multiplying powers with the same base, add the exponents. \[ a^m \times a^n = a^{m+n} \] 2. **Division Law:** When dividing powers with the same base, subtract the exponents. \[ a^m \div a^n = a^{m-n} \] 3. **Power of a Power Law:** When raising a power to another power, multiply the exponents. \[ (a^m)^n = a^{mn} \] 4. **Power of a Product Law:** When raising a product to a power, raise each factor to that power. \[ (ab)^n = a^n b^n \] 5. **Power of a Quotient Law:** When raising a quotient to a power, raise both the numerator and denominator to that power. \[ (\frac{a}{b})^n = \frac{a^n}{b^n} \] 6. **Zero Index Law:** Any non-zero number raised to the power of zero is equal to 1. \[ a^0 = 1 \quad (\text{where } a \neq 0) \] 7. **Negative Index Law:** A base raised to a negative exponent is equal to its reciprocal raised to the positive exponent. \[ a^{-n} = \frac{1}{a^n} \] 8. **Fractional Index Law:** A base raised to a fractional exponent can be expressed as a root. * \( a^{\frac{1}{n}} = \sqrt[n]{a} \) * \( a^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m} \) #### Simple Exponential Equations An exponential equation is an equation where the variable appears in the exponent. Simple exponential equations can often be solved by expressing both sides of the equation with the same base. **Principle:** If \( a^x = a^y \) and \( a > 0 \), \( a \neq 1 \), then \( x = y \). **Steps to solve simple exponential equations:** 1. Express both sides of the equation with the same base. 2. Equate the exponents. 3. Solve the resulting linear equation for the variable. ### Vocabulary * **Index (plural: Indices):** A number indicating the power to which a base number is raised. * **Exponent:** Another term for index. * **Base:** The number that is multiplied by itself according to the exponent. * **Power:** The entire expression \( a^n \), or simply another term for exponent/index. * **Exponential Equation:** An equation in which the variable occurs in the exponent. * **Reciprocal:** For a number \( x \), its reciprocal is \( \frac{1}{x} \). * **Root:** The inverse operation of raising a number to a power (e.g., square root, cube root). ### Examples #### Simplifying Expressions using Laws of Indices 1. **Simplify:** \( 2^3 \times 2^5 \) * **Solution:** \( 2^{3+5} = 2^8 \) 2. **Simplify:** \( \frac{x^7}{x^3} \) * **Solution:** \( x^{7-3} = x^4 \) 3. **Simplify:** \( (3x^2)^3 \) * **Solution:** \( 3^3 \times (x^2)^3 = 27x^{2 \times 3} = 27x^6 \) 4. **Evaluate:** \( 5^0 \) * **Solution:** \( 1 \) 5. **Evaluate:** \( 4^{-2} \) * **Solution:** \( \frac{1}{4^2} = \frac{1}{16} \) 6. **Evaluate:** \( 8^{\frac{2}{3}} \) * **Solution:** \( (\sqrt[3]{8})^2 = (2)^2 = 4 \) #### Solving Simple Exponential Equations 1. **Solve for \( x \):** \( 2^x = 16 \) * **Solution:** Express 16 as a power of 2: \( 16 = 2^4 \). So, \( 2^x = 2^4 \). Equating the exponents, \( x = 4 \). 2. **Solve for \( x \):** \( 3^{2x-1} = 27 \) * **Solution:** Express 27 as a power of 3: \( 27 = 3^3 \). So, \( 3^{2x-1} = 3^3 \). Equating the exponents: \( 2x-1 = 3 \) \( 2x = 4 \) \( x = 2 \) 3. **Solve for \( x \):** \( 5^x = \frac{1}{25} \) * **Solution:** Express \( \frac{1}{25} \) as a power of 5: \( \frac{1}{25} = \frac{1}{5^2} = 5^{-2} \). So, \( 5^x = 5^{-2} \). Equating the exponents, \( x = -2 \). 4. **Solve for \( x \):** \( 4^{x+1} = 8^{x-1} \) * **Solution:** Express both 4 and 8 as powers of the same base (base 2). \( 4 = 2^2 \) and \( 8 = 2^3 \). Substitute these into the equation: \( (2^2)^{x+1} = (2^3)^{x-1} \). Apply the power of a power law: \( 2^{2(x+1)} = 2^{3(x-1)} \). \( 2^{2x+2} = 2^{3x-3} \). Equate the exponents: \( 2x+2 = 3x-3 \) \( 2+3 = 3x-2x \) \( 5 = x \) Therefore, \( x = 5 \). ### Exercises 1. **Simplify the following expressions, leaving your answers in index form where appropriate:** a) \( y^6 \times y^4 \) b) \( m^{10} \div m^3 \) c) \( (p^5)^2 \) d) \( (2xy^3)^4 \) e) \( (\frac{a^3}{b^2})^2 \) 2. **Evaluate the following:** a) \( 7^0 \) b) \( 9^{-1} \) c) \( 16^{\frac{1}{4}} \) d) \( 27^{\frac{2}{3}} \) 3. **Solve for \( x \) in the following exponential equations:** a) \( 5^x = 125 \) b) \( 2^{x-3} = 32 \) c) \( 7^{2x} = \frac{1}{49} \) d) \( 9^x = 3^{x+4} \) ### Assignment 1. **Simplify:** \( \frac{(a^3b^{-2})^2}{a^2b^{-5}} \) 2. **Evaluate:** \( 81^{-\frac{3}{4}} \) 3. **Solve for \( x \):** \( 6^{x+2} = 216^{x-1} \) 4. **Solve for \( x \):** \( 2^{x+1} + 2^x = 48 \) *(Hint: Factor out the common term)* 5. **Solve for \( x \):** \( (\frac{1}{4})^x = 8^{x-2} \) ### Summary In this lesson, we explored the foundational Laws of Indices, which provide a systematic way to work with powers. We learned about the rules for multiplication, division, power of a power, zero index, negative index, and fractional indices. These laws are essential for simplifying algebraic expressions. Furthermore, we applied these laws to solve simple exponential equations by transforming them so that both sides have the same base, allowing us to equate their exponents and solve for the unknown variable. Mastering these concepts is a crucial step for more advanced topics in algebra and calculus. ## Laws of logarithms and Simple logarithmic equations ### Introduction Welcome to this lesson on the Laws of Logarithms and Simple Logarithmic Equations! In previous lessons, we learned about indices and the basic definition of a logarithm. Recall that a logarithm is essentially the inverse operation of exponentiation. If \( a^x = N \), then \( x \) is the logarithm of \( N \) to the base \( a \), written as \( x = \log_a N \). In this lesson, we will explore the fundamental laws that govern logarithms. These laws are incredibly useful for simplifying complex logarithmic expressions and, more importantly, for solving equations that involve logarithms. Mastering these laws will open up new ways to tackle problems in mathematics and science. ### Key Concepts #### 1. Definition of a Logarithm (Recap) As a reminder, the relationship between exponential and logarithmic form is: If \( a^x = N \), where \( a > 0 \), \( a \neq 1 \), and \( N > 0 \), then \( \log_a N = x \). #### 2. Laws of Logarithms There are several key laws that help us manipulate and simplify logarithmic expressions: ##### a. Product Law The logarithm of a product of two numbers is the sum of their logarithms. If \( M \) and \( N \) are positive real numbers and \( a \) is a positive real number not equal to 1, then: \[ \log_a (MN) = \log_a M + \log_a N \] ##### b. Quotient Law The logarithm of a quotient of two numbers is the difference of their logarithms. If \( M \) and \( N \) are positive real numbers and \( a \) is a positive real number not equal to 1, then: \[ \log_a \left(\frac{M}{N}\right) = \log_a M - \log_a N \] ##### c. Power Law The logarithm of a number raised to a power is the product of the power and the logarithm of the number. If \( M \) is a positive real number, \( P \) is any real number, and \( a \) is a positive real number not equal to 1, then: \[ \log_a M^P = P \log_a M \] ##### d. Change of Base Law This law allows us to change the base of a logarithm from one base to another. If \( A, B, C \) are positive real numbers and \( B \neq 1, C \neq 1 \), then: \[ \log_B A = \frac{\log_C A}{\log_C B} \] A common application is changing to base 10 or natural logarithm (base \( e \)). A special case of the change of base law is: \[ \log_B A = \frac{1}{\log_A B} \] ##### e. Special Logarithmic Identities These are important results derived from the definition of logarithms: * \( \log_a a = 1 \) (The logarithm of the base to itself is always 1, because \( a^1 = a \)) * \( \log_a 1 = 0 \) (The logarithm of 1 to any base is always 0, because \( a^0 = 1 \)) * \( \log_a a^x = x \) (The logarithm of \( a^x \) to base \( a \) is \( x \)) * \( a^{\log_a x} = x \) (The base \( a \) raised to the power of \( \log_a x \) is \( x \)) #### 3. Solving Simple Logarithmic Equations A logarithmic equation is an equation that involves the logarithm of a variable. To solve them, we often use the laws of logarithms to simplify the equation and then convert it to an exponential equation or equate the arguments. ##### a. Type 1: \( \log_a x = k \) If you have an equation in the form \( \log_a x = k \), convert it directly to its exponential form: \[ x = a^k \] ##### b. Type 2: \( \log_a x = \log_a y \) If you have an equation where logarithms with the same base are equal, then their arguments must be equal: \[ x = y \] ##### c. General Strategy 1. **Isolate the logarithmic term(s):** Rearrange the equation so that all logarithmic terms are on one side, or one logarithmic term is isolated. If there are multiple logarithmic terms, use the laws of logarithms to combine them into a single logarithm. Any constant terms should be moved to the other side. Remember that a constant \( k \) can be written as \( \log_a a^k \). 2. **Convert to exponential form or equate arguments:** Once the equation is in the form \( \log_a X = k \) or \( \log_a X = \log_a Y \), proceed as described in Type 1 or Type 2. 3. **Solve for the variable:** Solve the resulting algebraic equation. 4. **Check for extraneous solutions:** This is a CRUCIAL step. The argument of a logarithm must always be positive. If a solution makes any original logarithmic term's argument non-positive (zero or negative), it is an extraneous solution and must be discarded. ### Vocabulary * **Logarithm:** The power to which a base must be raised to produce a given number. * **Base:** The number that is raised to a power in an exponential expression, or the number that remains constant in a logarithmic expression.\( (a \) in \( \log_a N \) ). * **Argument (of a logarithm):** The number inside the logarithm, whose logarithm is being taken.\( (N \) in \( \log_a N \) ). It must always be positive. * **Common Logarithm:** A logarithm with base 10, often written as \( \log x \) (without the base explicitly stated). * **Natural Logarithm:** A logarithm with base \( e \) (Euler's number, approximately 2.71828), written as \( \ln x \). * **Logarithmic Equation:** An equation that contains one or more logarithms involving a variable. * **Extraneous Solution:** A solution to an equation that arises during the solving process but does not satisfy the original equation (in this context, usually because it violates the domain of the logarithm). ### Examples #### Example 1: Applying the Laws of Logarithms Simplify the following expressions: **a) Express \( \log_2 8 + \log_2 4 \) as a single logarithm.** Using the Product Law: \( \log_2 8 + \log_2 4 = \log_2 (8 \times 4) = \log_2 32 \) **b) Express \( \log_3 27 - \log_3 9 \) as a single logarithm.** Using the Quotient Law: \( \log_3 27 - \log_3 9 = \log_3 \left(\frac{27}{9}\right) = \log_3 3 \) **c) Express \( 2 \log 5 \) as a single logarithm.** Using the Power Law: \( 2 \log 5 = \log 5^2 = \log 25 \) **d) Expand \( \log_a (x^2 y^3) \).** Using Product and Power Laws: \( \log_a (x^2 y^3) = \log_a x^2 + \log_a y^3 = 2 \log_a x + 3 \log_a y \) **e) Evaluate \( \log_5 25 \).** We need to find \( x \) such that \( 5^x = 25 \). Since \( 5^2 = 25 \), then \( \log_5 25 = 2 \). Alternatively, using the power law: \( \log_5 5^2 = 2 \log_5 5 = 2(1) = 2 \). #### Example 2: Solving Simple Logarithmic Equations (Type 1) **a) Solve for \( x \): \( \log_3 x = 4 \)** Convert to exponential form: \( x = 3^4 \) \( x = 81 \) *Check*: \( \log_3 81 = 4 \) is true, and \( 81 > 0 \). So, \( x=81 \) is the solution. **b) Solve for \( x \): \( \log_5 (2x - 1) = 2 \)** Convert to exponential form: \( 2x - 1 = 5^2 \) \( 2x - 1 = 25 \) \( 2x = 26 \) \( x = 13 \) *Check*: When \( x = 13 \), \( 2x-1 = 2(13)-1 = 26-1 = 25 \). Since \( 25 > 0 \), \( x=13 \) is a valid solution. #### Example 3: Solving Simple Logarithmic Equations (Type 2) **a) Solve for \( x \): \( \log_7 (x+3) = \log_7 8 \)** Since the bases are the same, equate the arguments: \( x + 3 = 8 \) \( x = 5 \) *Check*: When \( x = 5 \), \( x+3 = 8 \). Since \( 8 > 0 \), \( x=5 \) is a valid solution. **b) Solve for \( x \): \( \log (3x-2) = \log (x+4) \)** (Note: \( \log \) without a base usually implies base 10). Equate the arguments: \( 3x - 2 = x + 4 \) \( 2x = 6 \) \( x = 3 \) *Check*: When \( x = 3 \), \( 3x-2 = 3(3)-2 = 7 \) and \( x+4 = 3+4 = 7 \). Both arguments are \( 7 > 0 \). So, \( x=3 \) is a valid solution. #### Example 4: Solving Logarithmic Equations using Laws **a) Solve for \( x \): \( \log_2 (x+2) + \log_2 x = 3 \)** Apply the Product Law to combine the logarithms: \( \log_2 ( (x+2)x ) = 3 \) \( \log_2 (x^2 + 2x) = 3 \) Convert to exponential form: \( x^2 + 2x = 2^3 \) \( x^2 + 2x = 8 \) \( x^2 + 2x - 8 = 0 \) Factor the quadratic equation: \( (x+4)(x-2) = 0 \) Possible solutions are \( x = -4 \) or \( x = 2 \). *Check for extraneous solutions:* For \( x = -4 \): \( \log_2 (x+2) \) becomes \( \log_2 (-4+2) = \log_2 (-2) \). The argument is negative, which is undefined. So, \( x=-4 \) is an extraneous solution. For \( x = 2 \): \( \log_2 (x+2) = \log_2 (2+2) = \log_2 4 \) (valid) \( \log_2 x = \log_2 2 \) (valid) Both original arguments are positive. So, \( x=2 \) is the only valid solution. **b) Solve for \( x \): \( 2 \log_3 x - \log_3 4 = \log_3 (x-1) \)** Apply the Power Law on the first term: \( \log_3 x^2 - \log_3 4 = \log_3 (x-1) \) Apply the Quotient Law on the left side: \( \log_3 \left(\frac{x^2}{4}\right) = \log_3 (x-1) \) Equate the arguments: \( \frac{x^2}{4} = x-1 \) Multiply by 4: \( x^2 = 4(x-1) \) \( x^2 = 4x - 4 \) \( x^2 - 4x + 4 = 0 \) Factor the quadratic equation: \( (x-2)^2 = 0 \) Possible solution is \( x = 2 \). *Check for extraneous solutions:* For \( x = 2 \): \( \log_3 x \) becomes \( \log_3 2 \) (valid) \( \log_3 (x-1) \) becomes \( \log_3 (2-1) = \log_3 1 \) (valid) All original arguments are positive. So, \( x=2 \) is the valid solution. ### Exercises 1. **Express as a single logarithm:** a) \( \log_5 100 - \log_5 4 \) b) \( \log_b x + \log_b y - \log_b z \) c) \( 3 \log_2 A + \log_2 B \) d) \( \frac{1}{2} \log_3 36 - 2 \log_3 2 \) 2. **Expand the following logarithms:** a) \( \log_a (pqr^2) \) b) \( \log \left(\frac{x^3}{y^2}\right) \) c) \( \log_5 \sqrt{xy} \) 3. **Solve for \( x \) in each of the following equations. Remember to check for extraneous solutions:** a) \( \log_4 x = 3 \) b) \( \log_2 (x+5) = 4 \) c) \( \log_6 (3x-1) = \log_6 5 \) d) \( \log x + \log (x-3) = 1 \) (Hint: Base is 10) e) \( \log_3 (x+4) - \log_3 (x-2) = 2 \) f) \( 2 \log x = \log (2x+3) \) g) \( \log_2 (x^2 - x) = \log_2 6 \) ### Assignment 1. **Simplify and evaluate (where possible):** a) \( \log_7 49 + \log_7 1 \) b) \( 5 \log_2 2 - \log_2 16 \) c) If \( \log_a 2 = 0.301 \) and \( \log_a 3 = 0.477 \), find \( \log_a 6 \) and \( \log_a 9 \). 2. **Solve the following logarithmic equations for \( x \):** a) \( \log_5 (x-1) + \log_5 (x+3) = 1 \) b) \( 2 \log_4 (x+1) = \log_4 (11x - 1) \) c) \( \log_2 (x^2 - 9) - \log_2 (x-3) = 3 \) d) \( \log_3 (2x-5) - \log_3 (x-3) = 0 \) e) \( \log_x 81 = 2 \) (Hint: solve for the base \( x \)) 3. **Challenge Question:** Given \( \log_{10} 2 = 0.3010 \) and \( \log_{10} 3 = 0.4771 \), without using a calculator, evaluate \( \log_{10} 12 \). ### Summary In this lesson, we delved into the powerful **Laws of Logarithms**, which are essential tools for manipulating and simplifying logarithmic expressions. We covered: * **Product Law:** \( \log_a (MN) = \log_a M + \log_a N \) * **Quotient Law:** \( \log_a \left(\frac{M}{N}\right) = \log_a M - \log_a N \) * **Power Law:** \( \log_a M^P = P \log_a M \) * **Change of Base Law:** \( \log_B A = \frac{\log_C A}{\log_C B} \) * **Special Identities:** \( \log_a a = 1 \), \( \log_a 1 = 0 \), \( \log_a a^x = x \), \( a^{\log_a x} = x \) We also learned how to apply these laws to solve **Simple Logarithmic Equations**. The general strategy involves combining logarithmic terms, converting the equation to an exponential form or equating arguments, and crucially, **checking for extraneous solutions** to ensure that the arguments of all logarithms in the original equation remain positive. Consistent practice with these laws and solving techniques will build your confidence and proficiency in handling logarithmic problems.