Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Solve exponential equations using various methods.
• Solve logarithmic equations, checking for valid solutions.
• Apply these functions to solve real-world problems.
• Interpret solutions in the context of the problem.

Lesson summary

Exponential and logarithmic functions are fundamental mathematical tools with wide-ranging applications in various fields, from finance and population growth to radioactive decay and computer science. Understanding these functions empowers you to model and analyze real-world phenomena, predict future trends, and solve complex problems. In South Africa, understanding exponential growth is crucial for analyzing economic trends, tracking the spread of diseases like HIV, understanding compound interest rates, and even modeling population growth for resource planning. This week, we delve deeper into solving equations and applying these functions to real-world problems.

Lesson notes

2.1 Exponential Equations An exponential equation is an equation where the variable appears in the exponent. Solving these equations involves isolating the exponential term and then using logarithms (or sometimes, manipulating to obtain the same base) to solve for the variable.

Method 1: Expressing with the Same Base If we can express both sides of the equation with the same base, we can equate the exponents. If `a^x = a^y`, then `x = y`.

Example 1: Solve `2^(x+1) = 8` Step 1: Express 8 as a power of 2: `8 = 2^3` Step 2: Rewrite the equation: `2^(x+1) = 2^3` Step 3: Equate the exponents: `x + 1 = 3` Step 4: Solve for x: `x = 3 - 1 = 2` Example 2: Solve `9^(x) = 27^(x-1)` Step 1: Express both 9 and 27 as powers of 3: `9 = 3^2` and `27 = 3^3` Step 2: Rewrite the equation: `(3^2)^x = (3^3)^(x-1)` Step 3: Simplify using exponent rules: `3^(2x) = 3^(3x-3)` Step 4: Equate the exponents: `2x = 3x - 3` Step 5: Solve for x: `3 = x` Method 2: Using Logarithms If expressing with the same base is not possible, we can use logarithms. We can take the logarithm of both sides of the equation. It's often convenient to use the natural logarithm (ln) or the common logarithm (log base 10).

Example 3: Solve `3^x = 15` Step 1: Take the logarithm of both sides (using the natural logarithm): `ln(3^x) = ln(15)` Step 2: Use the power rule of logarithms: `x ln(3) = ln(15)` Step 3: Solve for x: `x = ln(15) / ln(3)` Step 4: Approximate the value using a calculator: `x ≈ 2.465` Example 4: Solve `5^(2x-1) = 7^(x)` Step 1: Take the logarithm of both sides (using the common logarithm): `log(5^(2x-1)) = log(7^x)` Step 2: Use the power rule of logarithms: `(2x-1) log(5) = x * log(7)` Step 3: Distribute: `2xlog(5) - log(5) = x*log(7)` Step 4: Rearrange to isolate x: `2xlog(5) - x*log(7) = log(5)` Step 5: Factor out x: `x(2log(5) - log(7)) = log(5)` Step 6: Solve for x: `x = log(5) / (2log(5) - log(7))` Step 7: Approximate the value using a calculator: `x ≈ 1.747` 2.2 Logarithmic Equations A logarithmic equation is an equation where the variable appears within a logarithm. Solving these equations involves isolating the logarithmic term and then using the definition of a logarithm or logarithmic properties to solve for the variable. It is CRUCIAL to check for extraneous solutions because the argument of a logarithm must be positive.

Example 5: Solve `log_2(x + 3) = 4` Step 1: Rewrite the equation in exponential form: `2^4 = x + 3` Step 2: Simplify: `16 = x + 3` Step 3: Solve for x: `x = 16 - 3 = 13` Step 4: Check for extraneous solutions: `log_2(13 + 3) = log_2(16) = 4`. Since 13+3 > 0, the solution is valid.

Example 6: Solve `log(x) + log(x - 3) = 1` (

Note: log here means log base 10)

Step 1: Use the product rule of logarithms: `log(x(x - 3)) = 1` Step 2: Rewrite the equation in exponential form: `10^1 = x(x - 3)` Step 3: Simplify: `10 = x^2 - 3x` Step 4: Rearrange into a quadratic equation: `x^2 - 3x - 10 = 0` Step 5: Factor the quadratic equation: `(x - 5)(x + 2) = 0` Step 6: Solve for x: `x = 5` or `x = -2` Step 7: Check for extraneous solutions: For `x = 5`: `log(5) + log(5 - 3) = log(5) + log(2) = log(10) = 1`. This solution is valid. For `x = -2`: `log(-2)` is undefined. This solution is extraneous.

Therefore, the only solution is `x = 5`.

Example 7: Solve `ln(x+1) - ln(x) = 2` Step 1: Use the quotient rule of logarithms: `ln((x+1)/x) = 2` Step 2: Rewrite in exponential form: `e^2 = (x+1)/x` Step 3: Multiply both sides by x: `xe^2 = x + 1` Step 4: Rearrange to solve for x: `xe^2 - x = 1` Step 5: Factor out x: `x(e^2 - 1) = 1` Step 6: Divide: `x = 1/(e^2 - 1)` Step 7: Check for extraneous solutions: Since e^2 - 1 is positive, x will be positive. x+1 will also be positive, hence the solution is valid.

Step 8: Approximate using a calculator: x ≈ 0.1565 2.3 Applications of Exponential and Logarithmic Functions Exponential and logarithmic functions have many applications in real life.

Compound Interest: The formula for compound interest is `A = P(1 + r/n)^(nt)`, where A is the final amount, P is the principal amount, r is the annual interest rate, n is the number of times interest is compounded per year, and t is the number of years.

Population Growth: The formula for population growth is `P(t) = P_0 e^(kt)`, where `P(t)` is the population at time t, `P_0` is the initial population, k is the growth rate, and t is time.

Radioactive Decay: The formula for radioactive decay is `N(t) = N_0 e^(-λt)`, where `N(t)` is the amount of substance remaining after time t, `N_0` is the initial amount of substance, λ is the decay constant, and t is time. Guided Practice (With Solutions)

Question 1: Solve for x: `4^(x+2) = 64` Solution: Step 1: Express 64 as a power of 4: `64 = 4^3` Step 2: Rewrite the equation: `4^(x+2) = 4^3` Step 3: Equate the exponents: `x + 2 = 3` Step 4: Solve for x: `x = 3 - 2 = 1`

Commentary: This question reinforces the concept of solving exponential equations by expressing both sides with the same base.