Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Define and identify exponential and logarithmic functions.
• Solve exponential and logarithmic equations.
• Sketch graphs of these functions and their transformations.
• Apply these functions to solve real-world problems.
• Use logarithmic laws to simplify expressions.

Lesson summary

Exponential and logarithmic functions are powerful mathematical tools that describe situations involving growth and decay. Understanding these functions allows us to model and analyze various phenomena we encounter in everyday life, from population growth to compound interest and radioactive decay. In the South African context, these functions are crucial for understanding financial growth (investments, loans), predicting the spread of diseases, and even modeling resource depletion.

Lesson notes

2.1 Exponential Functions An exponential function is defined as `f(x) = a^x`, where 'a' is a positive constant (a > 0) and 'a' is not equal to

1. The variable 'x' is the exponent. The constant 'a' is called the base.

Base: The base determines the rate of growth or decay. If `a > 1`, the function represents exponential growth. If `0 0).

Asymptote: The x-axis (y = 0) is a horizontal asymptote. As x approaches negative infinity, `f(x)` approaches

0. Y-intercept: The graph always passes through the point (0, 1) since `a^0 = 1`.

Increasing/Decreasing: If `a > 1`, the function is increasing. If `0 0, a ≠ 1).

Domain: The domain of `y = logₐ(x)` is all positive real numbers (x > 0).

Range: The range of `y = logₐ(x)` is all real numbers (y ∈ ℝ).

Asymptote: The y-axis (x = 0) is a vertical asymptote.

X-intercept: The graph always passes through the point (1, 0) since `logₐ(1) = 0`.

Increasing/Decreasing: If `a > 1`, the function is increasing. If `0 x = logₐ(y)` `a^(logₐ(x)) = x` `logₐ(a^x) = x` Common Logarithms: When the base is 10, it's called the common logarithm and written as `log(x)` or `log₁₀(x)`.

Natural Logarithms: When the base is e (Euler's number, approximately 2.718), it's called the natural logarithm and written as `ln(x)` or `logₑ(x)`. 2.3 Logarithmic Laws These laws are essential for simplifying logarithmic expressions and solving equations: Product Rule: `logₐ(mn) = logₐ(m) + logₐ(n)` Quotient Rule: `logₐ(m/n) = logₐ(m) - logₐ(n)` Power Rule: `logₐ(m^p) = p * logₐ(m)` Change of Base Rule: `logₐ(b) = logₓ(b) / logₓ(a)` (This is useful for evaluating logarithms with bases not available on a calculator. You can change the base to 10 or e).

Example 3: Simplifying using Logarithmic Laws Simplify: `log₂(8x) - log₂(4)` Solution: Apply the quotient rule: `log₂(8x/4)` Simplify the fraction: `log₂(2x)` Apply the product rule: `log₂(2) + log₂(x)` Since `log₂(2) = 1`: `1 + log₂(x)` 2.4 Solving Exponential Equations To solve exponential equations, we aim to isolate the exponential term and then take the logarithm of both sides using a convenient base (usually base 10 or base e).

Example 4: Solving Exponential Equation Solve for x: `3^(x+1) = 27` Solution: Express both sides with the same base: `3^(x+1) = 3^3` Equate the exponents: `x + 1 = 3` Solve for x: `x = 3 - 1 = 2` Example 5: Solving Exponential Equation with Logarithms Solve for x: `5^x = 12` Solution: Take the logarithm of both sides (base 10): `log(5^x) = log(12)` Apply the power rule: `x * log(5) = log(12)` Solve for x: `x = log(12) / log(5)` Using a calculator: `x ≈ 1.544` 2.5 Solving Logarithmic Equations To solve logarithmic equations, we aim to isolate the logarithmic term and then rewrite the equation in exponential form. We must also check for extraneous solutions (solutions that do not satisfy the original equation because they result in taking the logarithm of a negative number or zero).

Example 6: Solving Logarithmic Equation Solve for x: `log₂(x + 3) = 4` Solution: Rewrite in exponential form: `x + 3 = 2^4` Simplify: `x + 3 = 16` Solve for x: `x = 16 - 3 = 13` Check: `log₂(13 + 3) = log₂(16) = 4` (Solution is valid)

Example 7: Solving Logarithmic Equation with Multiple Logarithms Solve for x: `log(x) + log(x - 3) = 1` Solution: Apply the product rule: `log(x(x - 3)) = 1` Rewrite in exponential form (base 10): `x(x - 3) = 10^1` Expand: `x^2 - 3x = 10` Rearrange into a quadratic equation: `x^2 - 3x - 10 = 0` Factorize: `(x - 5)(x + 2) = 0` Solve for x: `x = 5` or `x = -2` Check: For `x = 5`: `log(5) + log(5 - 3) = log(5) + log(2) = log(10) = 1` (Solution is valid) For `x = -2`: `log(-2)` is undefined (Solution is extraneous) Therefore, the only solution is `x = 5`. Guided Practice (With Solutions)

Question 1: Sketch the graph of `f(x) = 3^x` and identify its domain, range, and y-intercept.

Solution: To sketch the graph, we can plot a few points: x = -1: f(x) = 3^(-1) = 1/3 x = 0: f(x) = 3^0 = 1 x = 1: f(x) = 3^1 = 3 x = 2: f(x) = 3^2 = 9 The graph will be an increasing exponential curve passing through (0, 1). The x-axis (y=0) is a horizontal asymptote.

Domain: All real numbers (x ∈ ℝ)

Range: y > 0 Y-intercept: (0, 1)

Question 2: Solve for x: `2^(x - 1) = 16` Solution: Express both sides with the same base: `2^(x - 1) = 2^4` Equate the exponents: `x - 1 = 4` Solve for x: `x = 4 + 1 = 5` Question 3: Simplify: `log₅(25) + log₂(8) - log₃(3)` Solution: `log₅(25) = log₅(5²) = 2` `log₂(8) = log₂(2³) = 3` `log₃(3) = 1` Therefore, `log₅(25) + log₂(8) - log₃(3) = 2 + 3 - 1 = 4` Question 4: Solve for x: `log₃(x) = 2` Solution: Rewrite in exponential form: `x = 3²` Solve for x: `x = 9` Independent Practice (Questions Only) Sketch the graph of `y = (1/3)^x` and identify its domain, range, and y-intercept.

Solve for x: `4^(2x + 1) = 64` Solve for x: `7^x = 49√7` Simplify: `2log(5) + log(4)` Solve for x: `log₂(x + 1) + log₂(x - 1) = 3` The population of a town is modeled by the equation P(t) = 1000 * (1.05)^t, where t is the number of years since 2023.