Logarithm Expressions and Equations Revision Notes for HSC SSCE Mathematics Advanced

Logarithm Expressions and Equations

Introduction

After studying this topic, you will be able to:

Apply logarithm laws to simplify complex logarithmic expressions
Solve logarithmic equations and verify that solutions are valid
Solve exponential equations by taking the logarithm of both sides
Use digital tools to evaluate logarithmic expressions and determine if they are rational or irrational

Simplifying and evaluating logarithmic expressions

When we work with logarithmic expressions, we can often simplify them by applying logarithm laws. These laws allow us to combine or rewrite multiple logarithm terms to produce either a single logarithm or a simpler form.

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Fundamental Logarithm Laws for Simplification

The three key logarithm laws you must know are:

Product law: $\log_a(xy) = \log_a(x) + \log_a(y)$
Quotient law: $\log_a\left(\frac{x}{y}\right) = \log_a(x) - \log_a(y)$
Power law: $\log_a(x^n) = n \log_a(x)$

Remember: You can only combine logarithms with the same base!

Once we have simplified an expression, we can use digital tools such as calculators to evaluate it and obtain either a rational or an irrational value.

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Understanding Rational and Irrational Logarithms

Rational logarithms evaluate to a rational number (a number that can be expressed as a fraction). For example, $\log_{10}(100) = 2$ is rational because $2$ can be written as $\frac{2}{1}$ .

Irrational logarithms evaluate to an irrational number (a number that cannot be expressed as a simple fraction and has non-repeating, non-terminating decimals). For example, $\log_{10}(15)$ is typically irrational and must be approximated using technology.

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Worked Example 1: Simplifying and Evaluating Logarithmic Expressions

Let's simplify then evaluate each logarithmic expression, rounded to four decimal places, and state whether the evaluated value is rational or irrational.

Part a: $\log_5(25) + \log_5(4) - \log_5(2)$

Strategy:

Since all the logarithms have the same base (base 5), we can use the product law and quotient law to combine terms.

Working:

\log_5(25) + \log_5(4) - \log_5(2) = \log_5(25 \times 4) - \log_5(2)

Apply product law

= \log_5\left(\frac{100}{2}\right)

Apply quotient law

= \log_5(50)

Evaluate inside the brackets

= \log_5(5^2 \times 10)

Rewrite 50 as $5^2 \times 10$

= \log_5(5^2) + \log_5(10)

Apply product law

= 2 + \log_5(10)

Apply power law: $\log_5(5^2) = 2$

\approx 3.4307

Evaluate and round using technology

Since $\log_5(50)$ cannot be simplified to a rational number, it is irrational.

Part b: $3 \log_2(5) - \log_2(25)$

Strategy:

Since the logarithms have the same base, we can use the power law to rewrite the first term and the quotient law to combine terms.

Working:

3 \log_2(5) - \log_2(25) = \log_2(5^3) - \log_2(25)

Apply power law

= \log_2(125) - \log_2(25)

Simplify: $5^3 = 125$

= \log_2\left(\frac{125}{25}\right)

Apply quotient law

= \log_2(5)

Evaluate inside the brackets

\approx 2.3220

Evaluate and round using technology

Since $\log_2(5)$ cannot be simplified to a rational number, it is irrational.

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Key Takeaways: Simplifying Logarithmic Expressions

Logarithmic expressions are simplified using product, quotient, and power laws. Digital tools evaluate results, yielding rational values (e.g., $\log_{10}(100) = 2$ ) or irrational values (e.g., $\log_5(50)$ ).

Solving logarithmic equations

Logarithmic equations can be solved by applying logarithm laws and properties. The key is to ensure solutions satisfy domain restrictions. Remember that for logarithms, we need $a > 0$ , $a \neq 1$ , and all arguments must be positive.

To solve logarithmic equations, we typically follow these steps:

Use logarithm laws to combine logarithms or simplify the equation
Convert to exponential form using the relationship: $\log_a(x) = y \iff a^y = x$
Solve for the variable
Check that all arguments in the original equation are positive (domain check)
Verify the solution by substituting back into the original equation

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Critical Domain Restrictions

When solving logarithmic equations, all arguments must be positive. This means:

For $\log_a(x)$ , we must have $x > 0$
The base must satisfy $a > 0$ and $a \neq 1$
Any solution that makes an argument negative or zero must be rejected

Always perform a domain check before finalizing your answer!

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Worked Example 2: Solving a Simple Logarithmic Equation

Solve $\log_2(x + 3) = 3$ .

Strategy:

Convert to exponential form using $\log_a(x) = y \iff a^y = x$ and check the domain.

Working:

\log_2(x + 3) = 3

Write the equation

x + 3 = 2^3

Convert to exponential form

x + 3 = 8

Evaluate $2^3 = 8$

x = 5

Subtract $3$ from both sides

Checking the domain:

We need $x + 3 > 0$ . When $x = 5$ , we have $x + 3 = 5 + 3 = 8 > 0$ , so the argument is valid.

Therefore, $x = 5$ .

Verify:

$\log_2(5 + 3) = \log_2(8) = \log_2(2^3) = 3$ ✓

This confirms our solution.

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Worked Example 3: Solving an Equation with Multiple Logarithms

Solve $\log_3(2x - 1) + \log_3(x + 1) = 2$ .

Strategy:

Since the logarithms have the same base, use the product law to combine logarithms, convert to exponential form, and check the domain.

Working:

\log_3(2x - 1) + \log_3(x + 1) = 2

Write the equation

\log_3[(2x - 1)(x + 1)] = 2

Apply product law

\log_3(2x^2 + x - 1) = 2

Expand $(2x - 1)(x + 1)$

2x^2 + x - 1 = 3^2

Convert to exponential form

2x^2 + x - 1 = 9

Evaluate $3^2 = 9$

2x^2 + x - 10 = 0

Subtract $9$ from both sides

(2x + 5)(x - 2) = 0

Factorise

x = -\frac{5}{2},\ 2

Solve for $x$

Checking domain:

For $x = 2$ : $2x - 1 = 3 > 0$ and $x + 1 = 3 > 0$ ✓ Valid

For $x = -\frac{5}{2}$ : $2x - 1 = -6 < 0$ ✗ Invalid (argument cannot be negative)

Therefore, $x = 2$ .

Verify:

$\log_3(2 \times 2 - 1) + \log_3(2 + 1) = \log_3(3) + \log_3(3) = 1 + 1 = 2$ ✓

This confirms the solution.

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Key Takeaways: Solving Logarithmic Equations

Logarithmic equations are solved using logarithm laws and properties, converting to exponential form or combining terms. Solutions must satisfy domain restrictions (all arguments must be positive), and digital tools can verify results.

Solving exponential equations with logarithms

Exponential equations of the form $a^x = b$ , where $a > 0$ , $a \neq 1$ , and $b > 0$ , can be solved using logarithms if rewriting with the same base is not feasible.

The method involves taking the logarithm (base $10$ or base $e$ ) of both sides and applying the power law. This gives us:

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Standard Method for Solving Exponential Equations

a^x = b

\log_c(a^x) = \log_c(b)

Take logarithm of both sides (base $c$ )

x \log_c(a) = \log_c(b)

Apply power law

x = \frac{\log_c(b)}{\log_c(a)}

Divide both sides by $\log_c(a)$

Here, $c$ is the logarithm base, typically $10$ (common logarithm) or $e$ (natural logarithm).

Digital tools can evaluate $x$ , which may be irrational. For example, solving $a^x = b$ might give a value like $\frac{\ln(7)}{\ln(2)}$ . We can verify solutions by substituting back into the original equation.

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Worked Example 4: Solving Exponential Equations

Solve for $x$ , rounded to four decimal places using technology.

Part a: $4^x = 15$

Strategy:

Take the natural logarithm of both sides and use the power law.

Working:

4^x = 15

Write the equation

\ln(4^x) = \ln(15)

Take natural logarithm of both sides

x \ln(4) = \ln(15)

Apply power law

x = \frac{\ln(15)}{\ln(4)}

Divide both sides by $\ln(4)$

x \approx 1.9534

Evaluate using technology

Verify:

$4^{1.9534} \approx 14.9996 \approx 15$ ✓

This confirms the solution.

Part b: $2^{3x-1} = 10$

Strategy:

Take the common logarithm of both sides, apply the power law, and solve for $x$ .

Working:

2^{3x-1} = 10

Write the equation

\log(2^{3x-1}) = \log(10)

Take common logarithm of both sides

(3x - 1)\log(2) = 1

Apply power law and note that $\log(10) = 1$

3x - 1 = \frac{1}{\log(2)}

Divide both sides by $\log(2)$

3x = \frac{1}{\log(2)} + 1

Add $1$ to both sides

x = \frac{\frac{1}{\log(2)} + 1}{3}

Divide both sides by $3$

x \approx 1.4408

Evaluate using technology

Verify:

$2^{3 \times 1.4408 - 1} = 2^{3.3224} \approx 9.998 \approx 10$ ✓

This confirms the solution.

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Remember: Key Points for Logarithms and Exponential Equations

Logarithm laws are essential for simplification: Use the product law to add logarithms, the quotient law to subtract them, and the power law to bring down exponents.
Always check domain restrictions: When solving logarithmic equations, ensure all arguments are positive. Solutions that make arguments negative or zero must be rejected.
Convert to exponential form when needed: The relationship $\log_a(x) = y \iff a^y = x$ is crucial for solving logarithmic equations.
Use logarithms to solve exponential equations: When the equation $a^x = b$ cannot be solved by rewriting with the same base, take the logarithm of both sides and apply the power law to isolate $x$ .
Verify all solutions: Substitute your answer back into the original equation to confirm it works, and use technology to check decimal approximations.

Logarithm Expressions and Equations (HSC SSCE Mathematics Advanced): Revision Notes