Apply exponential and logarithmic functions to compound interest, half-life, and growth models. Grade 12 Math 30-1 - exam essential.
Lesson 3.8 of Exponents & Logarithms in Math 30-1 — Alberta curriculum lessons.
There are two types of equations:
Exponential Equation. Variable in the exponent: 5^x = 150
Logarithmic Equation. Variable inside the log: (x) - (20) = (0.2)
Each type has different solving methods — that's where the three cases come in. Cases 1 and 2 are for exponential equations. Case 3 is for logarithmic equations.
Your job. Identify which case before you start solving.
How to solve. Step 1: Isolate the exponential.
Step 2: Can both sides be written with the same base? YES → Case 1. NO → Case 2.
When to use. Both sides are powers of the same number (2, 3, 5, 10, etc.)
Steps. Change both sides to same base → Set exponents equal → Solve
Example. 9^(x-2) = (1)/(243) becomes 3^(2(x-2)) = 3^(-5), so 2(x-2) = -5
When to use. One side isn't a clean power (like 5^x = 150)
Method A — Direct conversion. c^x = N becomes x = _c(N)
Method B — Take log of both sides. Apply log → Use Power Law → Solve. Better when variables are on both sides.
Step 1: Identify the common base. Both 9 and 243 are powers of 3.
9 = 3^2, so 9^(x-2) = (3^2)^(x-2) = 3^(2(x-2)) = 3^(2x-4)
243 = 3^5, so (1)/(243) = 3^(-5)
Step 2: Rewrite the equation. 3^(2x-4) = 3^(-5)
Step 3: Set exponents equal. 2x - 4 = -5
2x = -1
x = -(1)/(2)
✓ Verify. 9^(-(1)/(2)-2) = 9^(-(5)/(2)) = (3^2)^(-(5)/(2)) = 3^(-5) = (1)/(243) ✓
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