6. Indices and logarithm

# 6. Indices and logarithm ## Laws of indices ### Introduction Welcome to the exciting world of indices! In mathematics, indices (also known as exponents or powers) are a way of writing repeated multiplication. Instead of writing \( 2 \times 2 \times 2 \times 2 \times 2 \), we can write \( 2^5 \). This lesson will explore the fundamental rules, or 'laws', that govern how we manipulate and simplify expressions involving indices. Understanding these laws is crucial for further topics in algebra and beyond. ### Key Concepts #### What are Indices? An index (plural: indices) is a small number written to the upper right of another number or variable. It tells us how many times to multiply the base number by itself. * **Base:** The number or variable being multiplied. * **Index / Exponent / Power:** The small number indicating how many times the base is multiplied by itself. For example, in \( 2^5 \): * The base is \( 2 \). * The index (or exponent or power) is \( 5 \). * It means \( 2 \times 2 \times 2 \times 2 \times 2 \). #### Law 1: Multiplication Law When multiplying terms with the same base, we add their indices. **Formula:** \[ a^m \times a^n = a^{m+n} \] **Example:** \( 2^3 \times 2^4 = 2^{3+4} = 2^7 \) (This is \( (2 \times 2 \times 2) \times (2 \times 2 \times 2 \times 2) \), which is \( 2 \) multiplied 7 times.) #### Law 2: Division Law When dividing terms with the same base, we subtract their indices. **Formula:** \[ a^m \div a^n = a^{m-n} \] **Example:** \( 5^6 \div 5^2 = 5^{6-2} = 5^4 \) #### Law 3: Power of a Power Law When raising a power to another power, we multiply the indices. **Formula:** \[ (a^m)^n = a^{mn} \] **Example:** \( (x^3)^2 = x^{3 \times 2} = x^6 \) #### Law 4: Zero Index Law Any non-zero number raised to the power of zero is equal to 1. **Formula:** \[ a^0 = 1 \quad (\text{where } a \neq 0) \] **Example:** \( 7^0 = 1 \) \( (1000)^0 = 1 \) #### Law 5: Negative Index Law A term with a negative index is the reciprocal of the term with a positive index. **Formula:** \[ a^{-n} = \frac{1}{a^n} \] **Example:** \( 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \) \( \frac{1}{x^{-3}} = x^3 \) #### Law 6: Fractional Index Law A fractional index indicates a root. The denominator of the fraction is the root, and the numerator is the power. **Formula:** \[ a^{\frac{1}{n}} = \sqrt[n]{a} \] \[ a^{\frac{m}{n}} = (\sqrt[n]{a})^m = \sqrt[n]{a^m} \] **Examples:** \( 8^{\frac{1}{3}} = \sqrt[3]{8} = 2 \) \( 9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27 \) #### Law 7: Power of a Product Law When a product of bases is raised to a power, each base is raised to that power. **Formula:** \[ (ab)^n = a^n b^n \] **Example:** \( (2x)^3 = 2^3 x^3 = 8x^3 \) #### Law 8: Power of a Quotient Law When a quotient of bases is raised to a power, both the numerator and denominator are raised to that power. **Formula:** \[ \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \quad (\text{where } b \neq 0) \] **Example:** \( \left(\frac{x}{3}\right)^2 = \frac{x^2}{3^2} = \frac{x^2}{9} \) ### Vocabulary * **Index (plural: Indices):** A number or symbol, placed above and to the right of another number or letter (the base), that indicates the number of times the base is to be used as a factor. Also called **exponent** or **power**. * **Base:** The number or variable that is multiplied by itself the number of times indicated by the exponent. * **Reciprocal:** For a number \( x \), its reciprocal is \( \frac{1}{x} \). * **Root:** An operation that is the inverse of raising a number to a power. For example, a square root \( (\sqrt{}) \) is the inverse of squaring a number. ### Examples 1. **Simplify** \( (3x^2y^3) \times (4x^4y) \) * Group terms with similar bases: \( (3 \times 4) \times (x^2 \times x^4) \times (y^3 \times y) \) * Apply Law 1: \( 12 \times x^{2+4} \times y^{3+1} \) * **Result:** \( 12x^6y^4 \) 2. **Evaluate** \( \frac{18a^7b^3}{6a^3b^5} \) * Separate coefficients and variables: \( \left(\frac{18}{6}\right) \times \left(\frac{a^7}{a^3}\right) \times \left(\frac{b^3}{b^5}\right) \) * Apply Law 2: \( 3 \times a^{7-3} \times b^{3-5} \) * Apply Law 5: \( 3a^4b^{-2} = \frac{3a^4}{b^2} \) * **Result:** \( \frac{3a^4}{b^2} \) 3. **Calculate** \( 64^{\frac{2}{3}} \) * Apply Law 6: \( 64^{\frac{2}{3}} = (\sqrt[3]{64})^2 \) * First find the cube root of 64: \( \sqrt[3]{64} = 4 \) (since \( 4 \times 4 \times 4 = 64 \)) * Then square the result: \( 4^2 = 16 \) * **Result:** \( 16 \) 4. **Simplify** \( \left(\frac{m^2n^{-1}}{m^5n^3}\right)^0 \) * Apply Law 4 (anything to the power of zero is 1, as long as the base is not zero): \( \left(\frac{m^2n^{-1}}{m^5n^3}\right)^0 = 1 \) * **Result:** \( 1 \) ### Exercises 1. Simplify the following expressions: a) \( 4^5 \times 4^2 \) b) \( y^8 \div y^3 \) c) \( (p^4)^5 \) d) \( 15^0 \) e) \( 7^{-1} \) 2. Evaluate: a) \( 25^{\frac{1}{2}} \) b) \( 81^{\frac{3}{4}} \) c) \( 100^{-1} \) 3. Simplify: a) \( \frac{x^7y^2}{x^3y^5} \) b) \( (3a^2b^3) \times (2a^{-1}b^4) \) c) \( \left(\frac{2x^3}{y^2}\right)^4 \) ### Assignment 1. Simplify the following expressions, leaving your answers with positive indices only: a) \( (5m^3n^{-2}) \times (2m^{-1}n^5) \) b) \( \frac{(k^2)^3 \times k^4}{k^5} \) c) \( \left(\frac{a^2b^{-3}}{a^{-1}b^2}\right)^{-2} \) 2. Evaluate the following without using a calculator: a) \( 125^{\frac{2}{3}} \) b) \( 16^{-\frac{3}{4}} \) c) \( (0.001)^{\frac{1}{3}} \) 3. If \( x = 2 \) and \( y = -3 \), find the value of \( x^{-2}y^3 \). 4. Show that \( (a^m)^n = a^{mn} \) using an example where \( a=3, m=2, n=3 \). ### Summary Today, we've explored the fundamental Laws of Indices, which provide a powerful toolkit for simplifying expressions involving powers. We covered: * **Multiplication Law:** Add indices when bases are the same (\( a^m \times a^n = a^{m+n} \)). * **Division Law:** Subtract indices when bases are the same (\( a^m \div a^n = a^{m-n} \)). * **Power of a Power Law:** Multiply indices when raising a power to another power (\( (a^m)^n = a^{mn} \)). * **Zero Index Law:** Any non-zero base raised to the power of zero is 1 (\( a^0 = 1 \)). * **Negative Index Law:** A negative index means taking the reciprocal (\( a^{-n} = \frac{1}{a^n} \)). * **Fractional Index Law:** A fractional index indicates a root (\( a^{\frac{m}{n}} = \sqrt[n]{a^m} \)). * **Power of a Product Law:** Apply the power to each factor in a product (\( (ab)^n = a^n b^n \)). * **Power of a Quotient Law:** Apply the power to both numerator and denominator in a quotient (\( \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \)). Mastering these laws will significantly improve your ability to work with algebraic expressions and solve more complex mathematical problems. ## Simple exponential equations ### Introduction Welcome to this lesson on simple exponential equations! In our previous lessons, we learned about indices and how to manipulate expressions involving powers. Today, we're going to take that knowledge a step further by solving equations where the unknown variable is in the exponent. Understanding how to solve these equations is crucial for many areas of mathematics and science. ### Key Concepts #### What are Exponential Equations? An exponential equation is an equation in which the variable (the unknown) appears in the exponent. For example, \( 2^x = 8 \) is an exponential equation because the variable \( x \) is in the exponent. #### The Fundamental Principle for Solving Simple Exponential Equations The core idea behind solving simple exponential equations is to make the bases on both sides of the equation the same. If two exponential expressions with the same base are equal, then their exponents must also be equal. Specifically, if you have an equation like this: \[ a^x = a^y \] where \( a \) is a positive number (and \( a \neq 1 \)), then it must be true that: \[ x = y \] This principle allows us to convert an exponential equation into a simpler linear equation, which we can then solve for the variable. #### Steps to Solve Simple Exponential Equations Follow these steps to solve most simple exponential equations: 1. **Identify a Common Base:** Try to express both sides of the equation with the same base. You might need to use your knowledge of powers and roots. 2. **Equate the Exponents:** Once both sides have the same base, set the exponents equal to each other. 3. **Solve the Linear Equation:** Solve the resulting linear equation for the unknown variable. #### Working with Different Bases and Fractions * **Converting to a Common Base:** Sometimes, the bases might not look the same initially, but they can be expressed as powers of a common number. For example, \( 9 \) can be written as \( 3^2 \), and \( 16 \) can be written as \( 2^4 \) (or \( 4^2 \)). * **Handling Fractions:** Remember that a fraction like \( \frac{1}{a^n} \) can be written as \( a^{-n} \). This is very useful when one side of the equation is a fraction. * **Handling Roots:** Remember that \( \sqrt[n]{a} = a^{1/n} \). For example, \( \sqrt{a} = a^{1/2} \). ### Vocabulary * **Exponential Equation:** An equation where the unknown variable is in the exponent. * **Base:** The number that is raised to a power (e.g., in \( 2^x \), 2 is the base). * **Exponent (or Index/Power):** The number that indicates how many times the base is multiplied by itself (e.g., in \( 2^x \), \( x \) is the exponent). * **Common Base:** A base that can be used to express all numbers in an exponential equation. ### Examples Let's work through some examples to see how these steps are applied. #### Example 1: Solve for \( x \) in \( 2^x = 16 \). **Solution:** 1. Express both sides with the same base. We know that \( 16 = 2^4 \). So, the equation becomes \( 2^x = 2^4 \). 2. Equate the exponents: \( x = 4 \) #### Example 2: Solve for \( x \) in \( 3^{x+1} = 27 \). **Solution:** 1. Express both sides with the same base. We know that \( 27 = 3^3 \). So, the equation becomes \( 3^{x+1} = 3^3 \). 2. Equate the exponents: \( x+1 = 3 \) 3. Solve the linear equation: \( x = 3 - 1 \) \( x = 2 \) #### Example 3: Solve for \( x \) in \( 5^{2x-1} = \frac{1}{25} \). **Solution:** 1. Express both sides with the same base. We know that \( 25 = 5^2 \), so \( \frac{1}{25} = 5^{-2} \). So, the equation becomes \( 5^{2x-1} = 5^{-2} \). 2. Equate the exponents: \( 2x-1 = -2 \) 3. Solve the linear equation: \( 2x = -2 + 1 \) \( 2x = -1 \) \( x = -\frac{1}{2} \) #### Example 4: Solve for \( x \) in \( 8^x = 4 \). **Solution:** 1. Express both sides with a common base. Both 8 and 4 can be expressed as powers of 2. \( 8 = 2^3 \) and \( 4 = 2^2 \). So, the equation becomes \( (2^3)^x = 2^2 \). Using the power rule \( (a^m)^n = a^{mn} \), we get \( 2^{3x} = 2^2 \). 2. Equate the exponents: \( 3x = 2 \) 3. Solve the linear equation: \( x = \frac{2}{3} \) #### Example 5: Solve for \( x \) in \( \left(\frac{1}{2}\right)^{x-1} = 8 \). **Solution:** 1. Express both sides with a common base. \( \frac{1}{2} = 2^{-1} \) and \( 8 = 2^3 \). So, the equation becomes \( (2^{-1})^{x-1} = 2^3 \). Using the power rule, we get \( 2^{-(x-1)} = 2^3 \). This simplifies to \( 2^{-x+1} = 2^3 \). 2. Equate the exponents: \( -x+1 = 3 \) 3. Solve the linear equation: \( -x = 3 - 1 \) \( -x = 2 \) \( x = -2 \) ### Exercises Solve for the unknown variable in each of the following exponential equations: 1. \( 4^x = 64 \) 2. \( 5^{x-2} = 125 \) 3. \( 2^{3x} = \frac{1}{8} \) 4. \( 9^{2x} = 3^4 \) 5. \( 6^{x+3} = 1 \) (Hint: Any non-zero number raised to the power of 0 is 1) ### Assignment Continue practicing your skills with these problems. Show all your steps clearly. 1. \( 7^{2x+1} = 49 \) 2. \( \left(\frac{1}{3}\right)^x = 27 \) 3. \( 16^{x-1} = 8^{x+2} \) 4. \( 100^x = 0.01 \) 5. \( 2^{x+2} \cdot 4^x = 32 \) (Hint: Use index laws to combine terms with the same base) ### Summary In this lesson, we learned how to solve simple exponential equations. The key takeaway is the fundamental principle: if \( a^x = a^y \) (where \( a > 0 \) and \( a \neq 1 \)), then \( x = y \). This allows us to equate the exponents once we have expressed both sides of the equation with the same base. Remember to utilize your knowledge of index laws, including those for fractions and roots, to find a common base and simplify equations effectively. Keep practicing, and you'll master solving these equations in no time! ## Laws of logarithms ### Introduction Welcome to this lesson on the Laws of Logarithms! In previous lessons, we learned about indices and the basic definition of a logarithm. Remember that a logarithm is simply the inverse operation to exponentiation. Just as we have rules (laws) for working with indices, we also have fundamental rules for manipulating and simplifying logarithmic expressions. These laws are incredibly useful for solving complex equations, simplifying expressions, and understanding various phenomena in science and engineering. By the end of this lesson, you will be able to confidently apply these laws to solve problems involving logarithms. ### Key Concepts #### Review: What is a Logarithm? Before diving into the laws, let's quickly recap the definition of a logarithm: If \( b^x = N \), then \( \log_b N = x \). * **b** is the **base** (must be positive and not equal to 1). * **N** is the **argument** (must be positive). * **x** is the **exponent** or the **logarithm**. #### The Laws of Logarithms There are three main laws of logarithms that you must know. These laws are derived directly from the laws of indices. ##### 1. The Product Law This law states that the logarithm of a product of two numbers is the sum of the logarithms of the individual numbers. **Law:** \[ \log_b (MN) = \log_b M + \log_b N \] **Conditions:** \( M > 0 \), \( N > 0 \), \( b > 0 \), \( b \ne 1 \). **Explanation:** Imagine you're multiplying two numbers, \( M \) and \( N \). If you express them in terms of a common base, say \( b^x = M \) and \( b^y = N \), then \( MN = b^x \cdot b^y = b^{x+y} \). Taking the logarithm base \( b \) of both sides gives \( \log_b (MN) = x+y \). Since \( x = \log_b M \) and \( y = \log_b N \), we get \( \log_b (MN) = \log_b M + \log_b N \). ##### 2. The Quotient Law This law states that the logarithm of a quotient of two numbers is the difference of the logarithms of the individual numbers. **Law:** \[ \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \] **Conditions:** \( M > 0 \), \( N > 0 \), \( b > 0 \), \( b \ne 1 \). **Explanation:** Similar to the product law, if \( M = b^x \) and \( N = b^y \), then \( \frac{M}{N} = \frac{b^x}{b^y} = b^{x-y} \). Taking the logarithm base \( b \) gives \( \log_b \left(\frac{M}{N}\right) = x-y \). Substituting \( x \) and \( y \) gives \( \log_b M - \log_b N \). ##### 3. The Power Law This law states that the logarithm of a number raised to a power is the power multiplied by the logarithm of the number. **Law:** \[ \log_b M^p = p \log_b M \] **Conditions:** \( M > 0 \), \( b > 0 \), \( b \ne 1 \), and \( p \) is any real number. **Explanation:** If \( M = b^x \), then \( M^p = (b^x)^p = b^{xp} \). Taking the logarithm base \( b \) of both sides gives \( \log_b M^p = xp \). Since \( x = \log_b M \), we get \( \log_b M^p = p \log_b M \). #### Special Properties of Logarithms * **Logarithm of 1:** The logarithm of 1 to any valid base is always 0. \[ \log_b 1 = 0 \] **Reason:** Because \( b^0 = 1 \) for any non-zero base \( b \). * **Logarithm of the Base:** The logarithm of a number to its own base is always 1. \[ \log_b b = 1 \] **Reason:** Because \( b^1 = b \) for any base \( b \). ### 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 (e.g., 'b' in \( b^x = N \) or \( \log_b N \)). * **Argument:** The number inside the logarithm (e.g., 'N' in \( \log_b N \)). * **Product Law:** The law stating \( \log_b (MN) = \log_b M + \log_b N \). * **Quotient Law:** The law stating \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \). * **Power Law:** The law stating \( \log_b M^p = p \log_b M \). ### Examples Let's apply these laws with some examples. **Example 1: Using the Product Law** Simplify \( \log_2 (8 \times 4) \). **Solution:** Using the Product Law, \( \log_b (MN) = \log_b M + \log_b N \). \( \log_2 (8 \times 4) = \log_2 8 + \log_2 4 \) We know that \( 2^3 = 8 \) and \( 2^2 = 4 \). So, \( \log_2 8 = 3 \) and \( \log_2 4 = 2 \). Therefore, \( \log_2 (8 \times 4) = 3 + 2 = 5 \). *(Check: \( 8 \times 4 = 32 \). \( \log_2 32 = 5 \) because \( 2^5 = 32 \). The answer is correct.)* **Example 2: Using the Quotient Law** Simplify \( \log_3 \left(\frac{81}{9}\right) \). **Solution:** Using the Quotient Law, \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \). \( \log_3 \left(\frac{81}{9}\right) = \log_3 81 - \log_3 9 \) We know that \( 3^4 = 81 \) and \( 3^2 = 9 \). So, \( \log_3 81 = 4 \) and \( \log_3 9 = 2 \). Therefore, \( \log_3 \left(\frac{81}{9}\right) = 4 - 2 = 2 \). *(Check: \( \frac{81}{9} = 9 \). \( \log_3 9 = 2 \) because \( 3^2 = 9 \). The answer is correct.)* **Example 3: Using the Power Law** Simplify \( \log_5 25^3 \). **Solution:** Using the Power Law, \( \log_b M^p = p \log_b M \). \( \log_5 25^3 = 3 \log_5 25 \) We know that \( 5^2 = 25 \), so \( \log_5 25 = 2 \). Therefore, \( 3 \log_5 25 = 3 \times 2 = 6 \). *(Check: \( 25^3 = (5^2)^3 = 5^6 \). So \( \log_5 25^3 = \log_5 5^6 = 6 \). The answer is correct.)* **Example 4: Combining Laws** Express \( \log a + \log b - \log c \) as a single logarithm. **Solution:** First, apply the Product Law to the addition: \( \log a + \log b = \log (ab) \) Then, apply the Quotient Law to the subtraction: \( \log (ab) - \log c = \log \left(\frac{ab}{c}\right) \) So, \( \log a + \log b - \log c = \log \left(\frac{ab}{c}\right) \). **Example 5: Evaluating with Special Properties** Evaluate \( \log_{10} 1000 + \log_7 1 - \log_5 5 \). **Solution:** * \( \log_{10} 1000 = 3 \) (since \( 10^3 = 1000 \)) * \( \log_7 1 = 0 \) (using the property \( \log_b 1 = 0 \)) * \( \log_5 5 = 1 \) (using the property \( \log_b b = 1 \)) So, \( 3 + 0 - 1 = 2 \). ### Exercises 1. Evaluate the following without using a calculator: a) \( \log_4 (16 \times 64) \) b) \( \log_2 \left(\frac{32}{8}\right) \) c) \( \log_{10} 100^5 \) d) \( \log_6 6 + \log_3 1 \) 2. Express each of the following as a single logarithm: a) \( \log_2 3 + \log_2 5 \) b) \( \log_5 20 - \log_5 4 \) c) \( 4 \log_{10} 2 \) d) \( 2 \log a + 3 \log b - \log c \) 3. Given that \( \log_x 2 = 0.301 \) and \( \log_x 3 = 0.477 \), find the value of: a) \( \log_x 6 \) b) \( \log_x 9 \) c) \( \log_x \left(\frac{3}{2}\right) \) ### Assignment 1. Simplify the following expressions: a) \( \log_3 27 + \log_3 9 - \log_3 81 \) b) \( \frac{1}{2} \log_2 16 - \log_2 \sqrt{2} \) c) \( \log_b (x^2 y^3) \) 2. Express \( \log_5 \left(\frac{x^3 y^2}{z}\right) \) in terms of \( \log_5 x \), \( \log_5 y \), and \( \log_5 z \). 3. If \( \log_a M = 2 \) and \( \log_a N = 3 \), evaluate \( \log_a \left(\frac{M^2}{N}\right) \). ### Summary In this lesson, we explored the three fundamental laws of logarithms and their special properties, which are essential for manipulating and solving logarithmic expressions. We learned that: * **Product Law:** \( \log_b (MN) = \log_b M + \log_b N \) * **Quotient Law:** \( \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \) * **Power Law:** \( \log_b M^p = p \log_b M \) We also reviewed that \( \log_b 1 = 0 \) and \( \log_b b = 1 \). These laws allow us to expand and condense logarithmic expressions, making complex calculations more manageable. Practicing these laws will build your confidence and proficiency in solving a wide range of mathematical problems. ## simple logarithmic equations ### Introduction Welcome to this lesson on simple logarithmic equations! In earlier lessons, you learned about logarithms and their properties. Just as you solve algebraic equations involving variables, you can also solve equations where the variable is inside a logarithm. These are called logarithmic equations. Understanding how to solve them is a key skill that builds on your knowledge of indices and logarithms, and it has applications in various fields like science, engineering, and finance. ### Key Concepts #### What is a Logarithmic Equation? A **logarithmic equation** is an equation that involves the logarithm of a variable. For example, \( \log_2 x = 3 \) is a simple logarithmic equation where \( x \) is the unknown. #### Revisiting Logarithm Properties Before we solve equations, let's quickly recall the fundamental properties of logarithms that will be crucial in simplifying and solving these equations: * **Product Rule:** The logarithm of a product is the sum of the logarithms. \[ \log_b (MN) = \log_b M + \log_b N \] * **Quotient Rule:** The logarithm of a quotient is the difference of the logarithms. \[ \log_b \left(\frac{M}{N}\right) = \log_b M - \log_b N \] * **Power Rule:** The logarithm of a number raised to a power is the power times the logarithm of the number. \[ \log_b M^p = p \log_b M \] * **Inverse Property (Converting to Index Form):** The most important property for solving simple logarithmic equations is the relationship between logarithmic and exponential forms. If \( \log_b x = y \), then \( x = b^y \). * **One-to-One Property:** If two logarithms with the same base are equal, then their arguments must be equal. If \( \log_b M = \log_b N \), then \( M = N \). * **Special Logarithms:** * \( \log_b 1 = 0 \) (The logarithm of 1 to any base is 0). * \( \log_b b = 1 \) (The logarithm of a number to its own base is 1). #### Strategies for Solving Simple Logarithmic Equations There are several strategies, often used in combination, to solve logarithmic equations: ##### Method 1: Converting to Index Form This method is used when you can isolate a single logarithm on one side of the equation and a constant on the other side. **Steps:** 1. Isolate the logarithmic term on one side of the equation. 2. Convert the equation from logarithmic form to its equivalent exponential (index) form. 3. Solve the resulting algebraic equation for the variable. 4. **Important:** Always check your solution(s) in the original equation to ensure that the argument of the logarithm is positive. Logarithms are only defined for positive numbers. ##### Method 2: Using the One-to-One Property This method is used when you have a single logarithm with the same base on both sides of the equation. **Steps:** 1. Use logarithm properties (Product, Quotient, Power Rules) to express both sides of the equation as a single logarithm with the same base. 2. Apply the one-to-one property: If \( \log_b M = \log_b N \), then set their arguments equal: \( M = N \). 3. Solve the resulting algebraic equation for the variable. 4. **Important:** Check your solution(s) in the original equation to ensure that the argument(s) of the logarithms are positive. ##### Method 3: Using Logarithm Properties to Simplify Often, an equation might have multiple logarithmic terms. You will need to use the Product, Quotient, or Power Rules to combine these into a single logarithm before applying Method 1 or Method 2. ### Vocabulary * **Logarithm:** The power to which a base must be raised to produce a given number. If \( b^y = x \), then \( oken{log}_b x = y \). * **Base:** The number that is raised to a power in an exponential expression, or the subscript in a logarithm (e.g., \( b \) in \( \log_b x \)). * **Argument (or Antilogarithm):** The number or expression inside the logarithm (e.g., \( x \) in \( \log_b x \)). The argument of a logarithm must always be positive. * **Logarithmic Equation:** An equation that contains one or more logarithms involving a variable. ### Examples #### Example 1: Converting to Index Form Solve: \( \log_3 x = 2 \) **Solution:** 1. The logarithm is already isolated. 2. Convert to index form: \( x = 3^2 \) 3. Solve for \( x \): \( x = 9 \) 4. Check: \( \log_3 9 = 2 \) (since \( 3^2 = 9 \)). The argument \( 9 \) is positive. So, \( x=9 \) is the correct solution. #### Example 2: Converting to Index Form with a Complex Argument Solve: \( \log_2 (x+1) = 3 \) **Solution:** 1. The logarithm is isolated. 2. Convert to index form: \( x+1 = 2^3 \) 3. Solve for \( x \): \( x+1 = 8 \implies x = 7 \) 4. Check: \( \log_2 (7+1) = \log_2 8 = 3 \) (since \( 2^3 = 8 \)). The argument \( 8 \) is positive. So, \( x=7 \) is the correct solution. #### Example 3: Using the One-to-One Property Solve: \( \log_5 (2x-1) = \log_5 9 \) **Solution:** 1. Both sides have a single logarithm with the same base (base 5). 2. Apply the one-to-one property: \( 2x-1 = 9 \) 3. Solve for \( x \): \( 2x = 10 \implies x = 5 \) 4. Check: \( \log_5 (2(5)-1) = \log_5 (10-1) = \log_5 9 \). The argument \( 9 \) is positive. So, \( x=5 \) is the correct solution. #### Example 4: Using Logarithm Properties to Simplify First (Product Rule) Solve: \( \log_4 x + \log_4 (x-3) = 1 \) **Solution:** 1. Use the Product Rule to combine the logarithms on the left side: \( \log_4 [x(x-3)] = 1 \) \( \log_4 (x^2 - 3x) = 1 \) 2. Convert to index form: \( x^2 - 3x = 4^1 \) \( x^2 - 3x = 4 \) 3. Rearrange into a quadratic equation and solve: \( x^2 - 3x - 4 = 0 \) \( (x-4)(x+1) = 0 \) This gives two potential solutions: \( x=4 \) or \( x=-1 \). 4. Check solutions: * For \( x=4 \): \( \log_4 4 + \log_4 (4-3) = \log_4 4 + \log_4 1 = 1 + 0 = 1 \). Both arguments (4 and 1) are positive. So, \( x=4 \) is a valid solution. * For \( x=-1 \): \( \log_4 (-1) \) is undefined because the argument must be positive. Therefore, \( x=-1 \) is not a valid solution (it's an extraneous solution). The only solution is \( x=4 \). #### Example 5: Using Logarithm Properties to Simplify First (Quotient Rule) Solve: \( \log_2 (x+3) - \log_2 (x-1) = \log_2 4 \) **Solution:** 1. Use the Quotient Rule to combine the logarithms on the left side: \( \log_2 \left(\frac{x+3}{x-1}\right) = \log_2 4 \) 2. Apply the one-to-one property: \( \frac{x+3}{x-1} = 4 \) 3. Solve the resulting algebraic equation: \( x+3 = 4(x-1) \) \( x+3 = 4x - 4 \) \( 7 = 3x \) \( x = \frac{7}{3} \) 4. Check solution: * For \( x = \frac{7}{3} \): Argument 1: \( x+3 = \frac{7}{3} + 3 = \frac{7}{3} + \frac{9}{3} = \frac{16}{3} \) (positive). Argument 2: \( x-1 = \frac{7}{3} - 1 = \frac{7}{3} - \frac{3}{3} = \frac{4}{3} \) (positive). Since both arguments are positive, \( x = \frac{7}{3} \) is a valid solution. ### Exercises Solve each of the following logarithmic equations. Remember to check your answers! 1. \( \log_4 x = 3 \) 2. \( \log_7 (2x+5) = 2 \) 3. \( \log_3 (x+6) = \log_3 (2x) \) 4. \( \log_6 x + \log_6 2 = 2 \) 5. \( \log_5 (x+2) - \log_5 x = 1 \) ### Assignment Continue practicing at home by solving these equations: 1. \( \log_{10} x = -1 \) 2. \( \log_8 (3x-1) = \frac{2}{3} \) 3. \( 2 \log_2 x = \log_2 (x+2) \) 4. \( \log_2 (x^2 - 4) - \log_2 (x-2) = 3 \) ### Summary In this lesson, we explored how to solve simple logarithmic equations. We reviewed the key properties of logarithms including the Product, Quotient, and Power Rules, as well as the crucial inverse property for converting between logarithmic and index forms, and the one-to-one property. We learned two primary methods for solving: converting the logarithm to its index form, and using the one-to-one property when both sides of the equation contain a single logarithm with the same base. We also practiced combining multiple logarithmic terms using the properties before solving. Most importantly, we learned the necessity of checking our solutions to ensure that the arguments of the logarithms in the original equation are always positive.