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Solving Exponential Equations

CSEC Additional Mathematics Essential Knowledge: Exponential equations feature variables in the exponent. They appear in diverse real-world contexts from population growth to radioactive decay. Mastering the techniques to solve these equations—both by expressing with common bases and using logarithms—is crucial for CSEC success and understanding advanced mathematical modeling.

Key Concept: An exponential equation is one in which the variable appears in the exponent. General form: \(a^{f(x)} = b\) or \(a^{f(x)} = a^{g(x)}\), where \(a > 0\), \(a \neq 1\). The two primary solution methods are: (1) Expressing both sides with the same base and equating exponents, or (2) Using logarithms to “bring down” the exponent.

Part 1: Understanding Exponential Equations

⚡

What Makes an Equation Exponential?

Exponential Equations

Variable in the exponent

\(3^x = 81\)
\(2^{x+1} = 32\)
\(5^{2x-3} = 125\)

Solution Methods: Same base or logarithms

Non-Exponential Equations

Variable in the base

\(x^3 = 27\)
\((x+1)^2 = 16\)
\(x^5 = 32\)

Solution Methods: Roots, factoring, algebra

🔍
Recognizing Exponential Forms

Exponential equations often appear in these patterns:

\(a^{f(x)} = b\)
\(a^{f(x)} = a^{g(x)}\)
\(a^{f(x)} = b^{g(x)}\)
\(e^{kx} = c\)

Where \(a, b > 0\), \(a, b \neq 1\), and \(f(x), g(x)\) are expressions containing the variable.

Graphical Insight: The equation \(a^x = b\) has:

  • Exactly one solution if \(b > 0\) (exponential functions are one-to-one)
  • No solution if \(b ≤ 0\) (exponential functions are always positive)

This explains why we require \(b > 0\) when taking logarithms of both sides.

Part 2: Method 1 – Same Base Technique

🔢

Expressing Both Sides with Common Base

The Fundamental Principle

If \(a^{f(x)} = a^{g(x)}\) and \(a > 0\), \(a \neq 1\), then \(f(x) = g(x)\)

Reason: Exponential functions are one-to-one. If the bases are equal and positive (not 1), then the exponents must be equal.

Step 1
\(2^{x+3} = 32\)
Original Equation
→
Step 2
\(2^{x+3} = 2^5\)
Express 32 as \(2^5\)
→
Step 3
\(x+3 = 5\)
Equate Exponents
→
Step 4
\(x = 2\)
Solve for x
📝 Example 1: Simple Same Base

Solve \(3^{2x-1} = 27\)

1
Express 27 as power of 3: \(27 = 3^3\)
2
Rewrite equation: \(3^{2x-1} = 3^3\)
3
Equate exponents: \(2x-1 = 3\)
4
Solve: \(2x = 4\) ⇒ \(x = 2\)
5
Check: \(3^{2(2)-1} = 3^{3} = 27\) ✓
📝 Example 2: More Complex Same Base

Solve \(8^{x+1} = 16^{x-1}\)

1
Express both with base 2: \(8 = 2^3\), \(16 = 2^4\)
2
Rewrite: \((2^3)^{x+1} = (2^4)^{x-1}\)
3
Apply power of a power: \(2^{3(x+1)} = 2^{4(x-1)}\)
4
Simplify exponents: \(2^{3x+3} = 2^{4x-4}\)
5
Equate exponents: \(3x+3 = 4x-4\)
6
Solve: \(3+4 = 4x-3x\) ⇒ \(7 = x\)
7
Check: \(8^{7+1} = 8^8\), \(16^{7-1} = 16^6 = (2^4)^6 = 2^{24} = (2^3)^8 = 8^8\) ✓

Prime Factorization Strategy: When bases don’t match, express both as powers of a common prime. Common conversions: \(4 = 2^2\), \(8 = 2^3\), \(9 = 3^2\), \(16 = 2^4\), \(25 = 5^2\), \(27 = 3^3\), \(32 = 2^5\), \(64 = 2^6 = 4^3 = 8^2\).

Part 3: Method 2 – Using Logarithms

📐

When Bases Cannot Be Made the Same

The Logarithmic Approach

If \(a^{f(x)} = b\), then \(f(x) = \log_a b\)
Or equivalently: \(f(x) = \frac{\log b}{\log a}\)

Why it works: Logarithms are the inverse operations of exponents. Taking log of both sides “brings down” the exponent.

Common Logarithm (base 10)

\(\log(a^{f(x)}) = \log b\)
\(f(x) \cdot \log a = \log b\)
\(f(x) = \frac{\log b}{\log a}\)

Use when: No obvious common base, calculator needed

Natural Logarithm (base e)

\(\ln(a^{f(x)}) = \ln b\)
\(f(x) \cdot \ln a = \ln b\)
\(f(x) = \frac{\ln b}{\ln a}\)

Use when: Equations involve \(e\), calculus contexts

📝 Example 3: Simple Logarithmic Solution

Solve \(5^x = 20\) (give answer to 3 decimal places)

1
Take log of both sides: \(\log(5^x) = \log 20\)
2
Apply power law: \(x \log 5 = \log 20\)
3
Isolate x: \(x = \frac{\log 20}{\log 5}\)
4
Calculate: \(\log 20 ≈ 1.30103\), \(\log 5 ≈ 0.69897\)
5
Divide: \(x ≈ \frac{1.30103}{0.69897} ≈ 1.86135\)
6
Round: \(x ≈ 1.861\) (to 3 decimal places)
7
Check: \(5^{1.861} ≈ 20.0\) ✓
📝 Example 4: More Complex Logarithmic Equation

Solve \(3^{2x+1} = 7^{x-2}\)

1
Take log of both sides: \(\log(3^{2x+1}) = \log(7^{x-2})\)
2
Apply power law: \((2x+1)\log 3 = (x-2)\log 7\)
3
Expand: \(2x\log 3 + \log 3 = x\log 7 – 2\log 7\)
4
Collect x terms: \(2x\log 3 – x\log 7 = -2\log 7 – \log 3\)
5
Factor x: \(x(2\log 3 – \log 7) = -(2\log 7 + \log 3)\)
6
Solve: \(x = \frac{-(2\log 7 + \log 3)}{2\log 3 – \log 7}\)
7
Calculate: \(\log 3 ≈ 0.4771\), \(\log 7 ≈ 0.8451\)
8
Numerator: \(-(2×0.8451 + 0.4771) = -(1.6902 + 0.4771) = -2.1673\)
9
Denominator: \(2×0.4771 – 0.8451 = 0.9542 – 0.8451 = 0.1091\)
10
Divide: \(x ≈ \frac{-2.1673}{0.1091} ≈ -19.87\)

Critical Check: Always verify that the solution makes sense in the original equation. For \(a^{f(x)} = b\), we must have \(b > 0\). If you get a solution that makes \(b ≤ 0\), you’ve made an error.

Part 4: Equations with Base e (Natural Exponential)

𝑒

Working with Natural Exponentials

🎯
The Special Base e

The number \(e ≈ 2.71828\) is a mathematical constant that appears naturally in growth/decay problems. Equations with base \(e\) are solved using natural logarithms (\(\ln\)), since \(\ln e = 1\).

If \(e^{f(x)} = b\), then \(f(x) = \ln b\)

Why e is special: The function \(y = e^x\) has the unique property that its derivative is itself: \(\frac{d}{dx}e^x = e^x\). This makes it ideal for modeling continuous growth and decay.

📝 Example 5: Solving with Base e

Solve \(e^{3x-2} = 10\)

1
Take natural log of both sides: \(\ln(e^{3x-2}) = \ln 10\)
2
Apply \(\ln(e^a) = a\): \(3x-2 = \ln 10\)
3
Solve: \(3x = \ln 10 + 2\)
4
Calculate: \(\ln 10 ≈ 2.30259\)
5
Thus: \(3x ≈ 2.30259 + 2 = 4.30259\)
6
Final: \(x ≈ 1.4342\) (to 4 decimal places)
📝 Example 6: Population Growth Model

A bacteria culture grows according to \(P(t) = 500e^{0.03t}\), where \(t\) is in hours. When will the population reach 2000?

1
Set up equation: \(500e^{0.03t} = 2000\)
2
Divide by 500: \(e^{0.03t} = 4\)
3
Take natural log: \(\ln(e^{0.03t}) = \ln 4\)
4
Simplify: \(0.03t = \ln 4\)
5
Calculate: \(\ln 4 ≈ 1.38629\)
6
Solve: \(t = \frac{1.38629}{0.03} ≈ 46.21\) hours
7
Answer: Approximately 46.2 hours

Part 5: Real-World Applications

🌍

Exponential Equations in Context

Common Applications of Exponential Equations:

Population growth: \(P(t) = P_0 e^{rt}\)
Radioactive decay: \(A(t) = A_0 e^{-kt}\)
Compound interest: \(A = P(1 + r/n)^{nt}\)
Newton’s Law of Cooling: \(T(t) = T_s + (T_0 – T_s)e^{-kt}\)
Sound intensity (decibels): \(L = 10\log(I/I_0)\)
pH scale: \(\text{pH} = -\log[H^+]\)
💰
Compound Interest Example

Problem: How long will it take for $1000 to double at 5% annual interest compounded annually?

1
Formula: \(A = P(1 + r)^t\) where \(A = 2000\), \(P = 1000\), \(r = 0.05\)
2
Equation: \(1000(1.05)^t = 2000\)
3
Divide by 1000: \((1.05)^t = 2\)
4
Take logs: \(\log(1.05^t) = \log 2\)
5
Apply power law: \(t\log 1.05 = \log 2\)
6
Solve: \(t = \frac{\log 2}{\log 1.05}\)
7
Calculate: \(t ≈ \frac{0.30103}{0.02119} ≈ 14.21\) years
8
Answer: Approximately 14.2 years
📈
Half-Life Problem

Problem: A radioactive substance decays according to \(A(t) = 100e^{-0.0231t}\). Find its half-life (time for half to decay).

1
Half-life means \(A(t) = 50\) (half of 100)
2
Equation: \(100e^{-0.0231t} = 50\)
3
Divide by 100: \(e^{-0.0231t} = 0.5\)
4
Take natural log: \(\ln(e^{-0.0231t}) = \ln 0.5\)
5
Simplify: \(-0.0231t = \ln 0.5\)
6
Calculate: \(\ln 0.5 ≈ -0.69315\)
7
Solve: \(t = \frac{-0.69315}{-0.0231} ≈ 30.0\) years
8
Answer: Half-life is 30.0 years

Part 6: CSEC Past Paper Questions

📚

Exam-Style Questions

CSEC May/June 2019 Paper 2

Question: Solve the equation \(2^{x+1} = 5^{x-1}\), giving your answer to 3 decimal places.

1
Take logs: \(\log(2^{x+1}) = \log(5^{x-1})\)
2
Apply power law: \((x+1)\log 2 = (x-1)\log 5\)
3
Expand: \(x\log 2 + \log 2 = x\log 5 – \log 5\)
4
Collect x terms: \(x\log 2 – x\log 5 = -\log 5 – \log 2\)
5
Factor: \(x(\log 2 – \log 5) = -(\log 5 + \log 2)\)
6
Solve: \(x = \frac{-(\log 5 + \log 2)}{\log 2 – \log 5}\)
7
Calculate: \(\log 2 ≈ 0.3010\), \(\log 5 ≈ 0.6990\)
8
Numerator: \(-(0.6990 + 0.3010) = -1.0000\)
9
Denominator: \(0.3010 – 0.6990 = -0.3980\)
10
Divide: \(x = \frac{-1.0000}{-0.3980} ≈ 2.5126\)
11
Answer: \(x ≈ 2.513\) (to 3 decimal places)
CSEC January 2018 Paper 2

Question: Solve the equation \(3^{2x} – 4(3^x) + 3 = 0\)

1
Recognize quadratic form: Let \(y = 3^x\)
2
Substitute: \(y^2 – 4y + 3 = 0\)
3
Factor: \((y-1)(y-3) = 0\)
4
So \(y = 1\) or \(y = 3\)
5
Substitute back: \(3^x = 1\) or \(3^x = 3\)
6
Solve \(3^x = 1\): \(x = 0\) (since \(3^0 = 1\))
7
Solve \(3^x = 3\): \(x = 1\) (since \(3^1 = 3\))
8
Answer: \(x = 0\) or \(x = 1\)

Hidden Quadratic Technique: Equations of form \(a^{2x} + b(a^x) + c = 0\) can be solved by substituting \(y = a^x\). This creates a quadratic in \(y\): \(y^2 + by + c = 0\). Solve for \(y\), then solve \(a^x = y\).

Quiz: Test Your Understanding

Solving Exponential Equations Quiz
Question 1: Solve \(4^{x-1} = 64\)
Answer:
Express 64 as power of 4: \(64 = 4^3\)
Equation: \(4^{x-1} = 4^3\)
Equate exponents: \(x-1 = 3\)
\(x = 4\)
Final answer: \(x = 4\)
Question 2: Solve \(5^{2x} = 20\), giving answer to 2 decimal places
Answer:
Take logs: \(\log(5^{2x}) = \log 20\)
\(2x\log 5 = \log 20\)
\(2x = \frac{\log 20}{\log 5}\)
\(\log 20 ≈ 1.30103\), \(\log 5 ≈ 0.69897\)
\(2x ≈ \frac{1.30103}{0.69897} ≈ 1.86135\)
\(x ≈ 0.9307\)
Final answer: \(x ≈ 0.93\) (to 2 decimal places)
Question 3: Solve \(e^{2x+1} = 15\)
Answer:
Take natural log: \(\ln(e^{2x+1}) = \ln 15\)
\(2x+1 = \ln 15\)
\(\ln 15 ≈ 2.70805\)
\(2x = 2.70805 – 1 = 1.70805\)
\(x ≈ 0.8540\)
Final answer: \(x ≈ 0.854\)
Question 4: Solve \(9^x – 10(3^x) + 9 = 0\)
Answer:
Let \(y = 3^x\), then \(9^x = (3^2)^x = 3^{2x} = (3^x)^2 = y^2\)
Equation: \(y^2 – 10y + 9 = 0\)
Factor: \((y-1)(y-9) = 0\)
\(y = 1\) or \(y = 9\)
\(3^x = 1\) ⇒ \(x = 0\)
\(3^x = 9 = 3^2\) ⇒ \(x = 2\)
Final answer: \(x = 0\) or \(x = 2\)
Question 5: A bacteria culture doubles every 3 hours. If initially there are 100 bacteria, when will there be 10,000 bacteria?
Answer:
Growth model: \(P(t) = 100(2)^{t/3}\) where t is in hours
Set \(100(2)^{t/3} = 10000\)
Divide by 100: \(2^{t/3} = 100\)
Take logs: \(\log(2^{t/3}) = \log 100\)
\((t/3)\log 2 = \log 100 = 2\)
\(t/3 = \frac{2}{\log 2} ≈ \frac{2}{0.3010} ≈ 6.644\)
\(t ≈ 19.93\) hours
Final answer: Approximately 19.9 hours

🎯 Key Concepts Summary

  • Exponential Equation: Variable in the exponent: \(a^{f(x)} = b\)
  • Method 1: Same Base:
    • Express both sides as powers of the same base
    • Equate exponents: \(a^{f(x)} = a^{g(x)}\) ⇒ \(f(x) = g(x)\)
    • Use prime factorization: \(4=2^2\), \(8=2^3\), \(9=3^2\), etc.
  • Method 2: Logarithms:
    • Take log of both sides: \(\log(a^{f(x)}) = \log b\)
    • Apply power law: \(f(x)\log a = \log b\)
    • Solve: \(f(x) = \frac{\log b}{\log a}\)
    • For base e: Use natural logs: \(f(x) = \ln b\)
  • Hidden Quadratics: For \(a^{2x} + b(a^x) + c = 0\), substitute \(y = a^x\)
  • Real-World Models:
    • Growth/decay: \(A = A_0e^{kt}\) or \(A = A_0(1+r)^t\)
    • Half-life/doubling time problems
    • Compound interest calculations
  • Common CSEC Question Types:
    • Solve exponential equations using same base
    • Solve using logarithms (often to specified decimal places)
    • Solve equations with base e
    • Hidden quadratic exponential equations
    • Applied problems (growth, decay, interest)
  • Exam Strategy:
    • Always try same base method first if possible
    • When using logs, state which base you’re using (log or ln)
    • For decimal answers, round to specified places
    • Check solutions in original equation when possible
    • For applied problems, include units and interpret answers

CSEC Exam Strategy: When solving exponential equations: (1) First check if bases can be made the same—this is quicker and avoids calculator use. (2) If bases differ, use logarithms—show clear steps: take logs, apply power law, isolate variable. (3) For equations like \(a^{2x} + ba^x + c = 0\), recognize the quadratic form and substitute. (4) Always check if your answer makes sense in context (e.g., time can’t be negative, population must be positive). Show all working for method marks!

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