3.3 Exponential & rational equations
Key Takeaways
- The exponent laws combine same-base powers: multiply by adding exponents, divide by subtracting, and raise a power to a power by multiplying.
- An exponential function y = a*b^x has starting value a and base b; b > 1 models growth and 0 < b < 1 models decay.
- Exponential change is repeated multiplication, so values rise or fall far faster than in a linear model.
- Solve rational equations by multiplying through by the least common denominator, but first note excluded values where a denominator equals zero.
- Always check rational-equation answers against the excluded values; a solution that breaks a denominator is extraneous and must be rejected.
Exponential and Rational Equations
This section covers three connected skills: the laws of exponents, exponential growth and decay functions, and rational equations. Each shows up on the TSIA2, and the exponent rules are the foundation for the rest.
Laws of Exponents
The exponent rules let you combine powers of the same base quickly.
| Rule | Formula | Example |
|---|---|---|
| Product | xᵃ · xᵇ = xᵃ⁺ᵇ | x³ · x⁴ = x⁷ |
| Quotient | xᵃ ÷ xᵇ = xᵃ⁻ᵇ | x⁵ ÷ x² = x³ |
| Power of a power | (xᵃ)ᵇ = xᵃᵇ | (x²)³ = x⁶ |
| Zero exponent | x⁰ = 1 | 5⁰ = 1 |
| Negative exponent | x⁻ᵃ = 1 / xᵃ | 2⁻³ = 1/8 |
Example 1. Simplify (x⁵ · x³) ÷ x². Multiply first: x⁵ · x³ = x⁸. Then divide: x⁸ ÷ x² = x⁶.
Example 2. Simplify (2³ · 2⁻¹). Add the exponents: 3 + (−1) = 2, so the result is 2² = 4. A negative exponent lowers the total; it does not make the answer negative.
Keep the base the same when you apply these rules — the product and quotient rules only work when the two powers share a base. When the exponents produce a negative or zero result, fall back on the definitions x⁰ = 1 and x⁻ᵃ = 1/xᵃ to write a clean final form.
Exponential Functions
An exponential function has the form y = a · bˣ, where a is the starting amount (the value when x = 0) and b is the base, or growth factor. When b > 1 the function models growth; when 0 < b < 1 it models decay. Notice the variable is in the exponent, which makes these functions climb or fall much faster than linear functions. To read off the starting value, set x = 0: because b⁰ = 1, y = a · 1 = a, so the constant in front is always the initial amount.
Example 3 (growth). A colony of 50 bacteria doubles every hour, so y = 50 · 2ˣ, where x is hours. After 4 hours: y = 50 · 2⁴ = 50 · 16 = 800 bacteria.
Example 4 (decay). A machine worth $500 loses value by a factor of 0.8 each year, so y = 500 · (0.8)ˣ. After 2 years: y = 500 · (0.8)² = 500 · 0.64 = $320.
Example 5 (reading the model). For y = 100 · 3ˣ, the starting value is 100 (at x = 0), and at x = 2 the value is 100 · 3² = 100 · 9 = 900.
Growth means repeated multiplication, not repeated addition. That is why the bacteria jump from 50 to 100 to 200 to 400 to 800 rather than increasing by the same amount each hour. A linear function would add a fixed number every step; an exponential function multiplies by a fixed factor every step, which is why exponential models eventually overtake any linear one.
Rational Equations
A rational equation contains a variable in a denominator. Two ideas control the whole process:
- Excluded values: any x that makes a denominator zero is not allowed, because division by zero is undefined.
- Clear the denominators by multiplying every term by the least common denominator (LCD), then solve the simpler equation. Because multiplying by the LCD can introduce false answers, you must check every solution against the excluded values; a false answer is called extraneous.
Example 6. Solve (x / 3) + (1 / 2) = 5 / 6. The LCD of 3, 2, and 6 is 6. Multiply every term by 6: 6(x/3) + 6(1/2) = 6(5/6), which gives 2x + 3 = 5. Then 2x = 2, so x = 1. There are no variable denominators here, so no value is excluded and x = 1 is valid.
Example 7. Solve 3 / x = 6 / (x + 2). Excluded values: x ≠ 0 and x ≠ −2. Cross-multiply: 3(x + 2) = 6x, so 3x + 6 = 6x. Then 6 = 3x, giving x = 2. Since 2 is not excluded, the solution is x = 2.
Example 8 (extraneous solution). Solve 1/(x − 1) + 1/(x + 1) = 2/(x² − 1). Because x² − 1 = (x − 1)(x + 1), the excluded values are x ≠ 1 and x ≠ −1, and the LCD is (x − 1)(x + 1). Multiply every term by the LCD: (x + 1) + (x − 1) = 2, which simplifies to 2x = 2, so x = 1. But x = 1 is an excluded value, so it is extraneous. The equation has no solution.
The lesson from Example 8 is why checking is not optional: the algebra produced a number, yet that number breaks the original equation. Always compare each answer to the excluded values before you write it down, and remember that rules for exponents, exponential models, and rational equations all reward careful, step-by-step arithmetic.
A colony of 50 bacteria doubles every hour, modeled by y = 50 * 2^x. How many bacteria are there after 4 hours?
Solve the rational equation 1/(x - 1) + 1/(x + 1) = 2/(x^2 - 1).