Exponential Word Problem - Algebra-1

1. Fundamental Concepts

Definition: Exponential word problems are questions that solve practical growth or decay problems based on exponential function models. Their core is to use the exponential function \(y = a \cdot b^x\) to describe how quantities change over time (or other variables).
Core Elements:
- a: The initial quantity (the quantity at the starting moment in the problem);
- b: The growth/decay factor (b > 1 indicates growth, and 0 < b < 1 indicates decay);
- x: Time (or the number of changes, attention should be paid to matching the unit with the period);
- y: The final quantity after x time.

2. Key Concepts

Characteristics of Exponential Growth/Decay: The change in quantity is proportional to the current quantity, rather than a fixed increment (different from linear relationships). For example, descriptions such as "doubling", "halving", "growing/decaying at a certain percentage" all belong to exponential changes.
Period Matching: If the problem involves periodic changes such as "doubling every 3 years" or "decaying every half a year", the time unit needs to be unified into the period unit. For instance, 30 years contain 10 periods of 3 years (\(30 \div 3 = 10\)).
Comparison with Linear Relationships: Linear relationships are "uniform growth" (e.g., increasing by a fixed number each month), while exponential relationships are "accelerated/decelerated growth" (e.g., growing proportionally). In the long run, exponential growth (b > 1) usually far exceeds linear growth.

3. Examples

1. Easy Level

Problem: A bacterial population initially has 200 individuals, and its quantity becomes 1.5 times the original every hour. What is the population size after 4 hours?

Establishing the function formula: The general form of an exponential function is \(y = a \cdot b^x\), where:
- \(a = 200\) (initial number of bacteria);
- \(b = 1.5\) (growth factor per hour);
- x is time (in hours);
- Therefore, the function formula is: \(y = 200 \cdot 1.5^x\).
Substituting data for calculation: When \(x = 4\), \(y = 200 \cdot 1.5^4 = 200 \cdot 5.0625 = 1012.5 \approx 1013\) (individuals).

2. Medium Level

Problem: A car initially costs 300,000 yuan, and its value depreciates (decreases) by 15% every year. What is the value of the car after 5 years?

Establishing the function formula: The general form of an exponential function is \(y = a \cdot b^x\), where:
- \(a = 300000\) (initial value, in yuan);
- \(b = 1 - 15\% = 0.85\) (depreciation factor per year);
- x is time (in years);
- Therefore, the function formula is: \(y = 300000 \cdot 0.85^x\).
Substituting data for calculation: When \(x = 5\), \(y = 300000 \cdot 0.85^5 \approx 300000 \cdot 0.4437 = 133110\) (yuan).

3. Hard Level

Problem: There are initially 1000 trees. In a 30-year plan, what is the final number of trees under the scheme where "the number doubles every 3 years"?

Establishing the function formula: The general form of an exponential function is \(y = a \cdot b^x\), where:
- \(a = 1000\) (initial number of trees);
- \(b = 2\) (growth factor every 3 years);
- x is the number of 3-year cycles (30 years contain \(30 \div 3 = 10\) cycles);
- Therefore, the function formula is: \(y = 1000 \cdot 2^x\).
Substituting data for calculation: When \(x = 10\), \(y = 1000 \cdot 2^{10} = 1000 \cdot 1024 = 1024000\) (trees).

Problem: Two investment plans: Plan A has a fixed annual income of 500 yuan; Plan B starts with 1000 yuan and has an annual income growth of 10%. Which plan will have a higher total income after 10 years? Analysis:

Plan A (linear): Total income = \(500 \times 10 = 5000\) yuan, final total amount = \(1000 + 5000 = 6000\) yuan.
Plan B (exponential): Initial quantity \(a = 1000\), growth factor \(b = 1 + 10\% = 1.1\), final total amount after 10 years \(y = 1000 \cdot 1.1^{10} \approx 1000 \times 2.5937 = 2593.7\) yuan.
Conclusion: Plan A has a higher income (Note: Here, due to the low initial amount, linear growth may be better in the short term, but exponential growth is better in the long term).

4. Problem-Solving Techniques

Determine the model type: Judge whether it is an exponential relationship (such as "doubling", "percentage growth/decay", "changing to k times the original every n years/months") or a linear relationship (such as "increasing by a fixed number every month") through key words.
Extract key parameters: Find the initial quantity a, growth/decay factor b, the unit and period of the time variable x, and ensure the units are unified (e.g., convert years to the number of periods).
Establish the function expression: Substitute into the exponential function formula \(y = a \cdot b^x\) and clarify the actual meaning of each parameter.
Calculate and verify: Calculate the result according to the expression, verify the rationality in combination with the actual situation (e.g., the quantity cannot be negative), and compare with the linear model for analysis if necessary.