The graph gives:

x ≈ 3.322

This agrees with the logarithmic solution:

x = ln(10)/ln(2)

Advantages of Graphical Methods

Graphs can help:

• Estimate solutions.

• Identify whether solutions exist.

• Determine the number of solutions.

• Check algebraic answers.

• Solve equations that are difficult to rearrange.

However, graphical answers may be approximate.

Numerical Methods

Some exponential equations cannot be solved using simple algebraic methods.

For example:

2ˣ = x + 3

The variable appears both in an exponent and outside it.

This equation has two real solutions.

Numerical methods can approximate them by finding the zeros of:

f(x) = 2ˣ − x − 3

Using graphing or numerical root-finding technology gives approximately:

x ≈ −2.863 and x ≈ 2.445

Numerical methods are useful when exact algebraic solutions are unavailable or impractical.

Practical Application: Compound Interest

Exponential equations are frequently used in finance.

For compound interest:

A = P(1 + r/n)^(nt)

where:

• A = final amount.

• P = initial principal.

• r = annual interest rate as a decimal.

• n = number of compounding periods per year.

• t = time in years.

Worked Example: Doubling an Investment

An investment earns 5% interest annually, compounded once per year.

How long will it take to double?

Step 1: Write the Equation

A = P(1.05)ᵗ

For doubling:

2P = P(1.05)ᵗ

Step 2: Simplify

2 = 1.05ᵗ

Step 3: Take Logarithms

ln(2) = t ln(1.05)

Step 4: Solve

t = ln(2)/ln(1.05)

t ≈ 14.21 years

Interpretation

The investment doubles after approximately 14.21 years according to the exponential model.

With interest credited only at the end of each year, the account first reaches at least twice its original value after 15 complete years.

 

 

 

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Practical Application: Population Growth

Suppose a population grows by 6% each year.

The population is modelled by:

P(t) = P₀(1.06)ᵗ

Worked Example

A town has an initial population of 12000.

How long will it take to reach 20000?

Step 1: Substitute

20000 = 12000(1.06)ᵗ

Step 2: Divide

20000/12000 = 1.06ᵗ

5/3 = 1.06ᵗ

Step 3: Take Logarithms

ln(5/3) = t ln(1.06)

Step 4: Solve

t = ln(5/3)/ln(1.06)

t ≈ 8.77 years

Interpretation

The population reaches 20000 after approximately 8.77 years under the model.

This assumes the annual proportional growth rate remains constant.

Practical Application: Radioactive Decay

Radioactive substances decay exponentially.

A common model is:

N(t) = N₀e^(−λt)

where:

• N(t) = amount remaining after time t.

• N₀ = initial amount.

• λ = decay constant.

• t = elapsed time.

Worked Example

A radioactive substance has a decay constant of 0.04 per year.

How long will it take for 25% of the original material to remain?

Step 1: Write the Equation

0.25N₀ = N₀e^(−0.04t)

Step 2: Divide

0.25 = e^(−0.04t)

Step 3: Take Natural Logarithms

ln(0.25) = −0.04t

Step 4: Solve

t = ln(0.25)/(−0.04)

t ≈ 34.66 years

Interpretation

Approximately 34.66 years are required for the substance to decrease to 25% of its original amount.

Practical Application: Bacterial Growth

Bacteria may reproduce through binary fission, causing populations to increase exponentially under suitable conditions.

Suppose a bacterial population doubles every 3 hours.

The model is:

N(t) = N₀ × 2^(t/3)

Worked Example

A bacterial population begins with 500 bacteria.

How long will it take to reach 8000 bacteria?

Step 1: Substitute

8000 = 500 × 2^(t/3)

Step 2: Divide

16 = 2^(t/3)

Step 3: Rewrite Using a Common Base

16 = 2⁴

Therefore:

2⁴ = 2^(t/3)

Step 4: Equate Exponents

4 = t/3

t = 12 hours

Interpretation

The population doubles four times:

500 → 1000 → 2000 → 4000 → 8000

Four doubling periods of 3 hours each give 12 hours.

Practical Application: Depreciation

Depreciation describes the decrease in the value of an asset over time.

A simplified depreciation model is:

V(t) = V₀(1 − r)ᵗ

where:

• V(t) = value after t years.

• V₀ = initial value.

• r = annual depreciation rate.

Worked Example

A vehicle costs $30000 and loses 15% of its value each year.

When will its value fall to $10000?

Step 1: Write the Equation

10000 = 30000(0.85)ᵗ

Step 2: Divide

1/3 = 0.85ᵗ

Step 3: Take Logarithms

ln(1/3) = t ln(0.85)

Step 4: Solve

t = ln(1/3)/ln(0.85)

t ≈ 6.76 years

Interpretation

The vehicle's value falls to $10000 after approximately 6.76 years under the constant-rate depreciation model.

Practical Application: Medication Elimination

Some medications are eliminated from the body approximately exponentially over certain concentration ranges.

A simplified model is:

C(t) = C₀e^(−kt)

where:

• C(t) = concentration at time t.

• C₀ = initial concentration.

• k = elimination constant.

Worked Example

A medication concentration decreases according to:

C(t) = 80e^(−0.2t)

where t is measured in hours.

When will the concentration reach 20 units?

Step 1: Substitute

20 = 80e^(−0.2t)

Step 2: Divide

0.25 = e^(−0.2t)

Step 3: Take Natural Logarithms

ln(0.25) = −0.2t

Step 4: Solve

t = ln(0.25)/(−0.2)

t ≈ 6.93 hours

This is a mathematical model; actual medication elimination can depend on multiple biological processes.

Verifying Exponential Solutions

Verification is an important part of solving exponential equations.

A solution should satisfy the original equation, not merely a transformed version.

Verification Method

1. Substitute the calculated value into the original equation.

2. Evaluate the left-hand side.

3. Evaluate the right-hand side.

4. Compare the results.

5. Account for rounding if the solution is approximate.

Worked Example

Verify:

x = 2

for:

3^(x+1) = 27

Substitute:

3^(2+1) = 3³ = 27

The left side equals the right side.

Therefore:

x = 2 is valid.

Verifying Approximate Solutions

Suppose:

2ˣ = 13

and:

x ≈ 3.700

Substitution gives:

2³·⁷⁰⁰ ≈ 12.996

This is close to 13.

The slight difference is due to rounding.

For greater accuracy, use the unrounded logarithmic expression.

Checking for Extraneous or Invalid Solutions

When solving exponential equations by substitution, not every algebraic solution is necessarily valid.

For example:

2^(2x) + 2ˣ − 6 = 0

Let:

u = 2ˣ

Then:

u² + u − 6 = 0

Factorise:

(u + 3)(u − 2) = 0

Therefore:

u = −3 or u = 2

However:

2ˣ > 0

for every real x.

Thus:

u = −3 is impossible.

Only:

2ˣ = 2

is valid.

Therefore:

x = 1

The positivity of exponential expressions is an important restriction.

Choosing the Best Solution Method

The goal is not simply to memorise methods, but to recognise the structure of the equation and select an efficient strategy.

Common Mistakes

Confusing exponential equations with power equations

The variable's position determines the appropriate solution method.

Equating exponents when the bases are different

Exponents can be equated directly only when both sides have the same valid base.

Applying logarithms incorrectly

Logarithms must be applied to both sides of an equation, and their arguments must be positive.

Forgetting the logarithmic power rule

For example:

ln(3ˣ) = xln(3)

not ln(3x).

Taking logarithms of individual terms in a sum

In general:

ln(a + b) ≠ ln(a) + ln(b)

Forgetting to isolate the exponential expression

Simplify coefficients and constants before taking logarithms where possible.

Incorrectly rewriting exponential expressions

Remember:

a^(x+1) = a × aˣ

and:

a^(2x) = (aˣ)²

Ignoring the positivity of exponential expressions

If u = aˣ with a > 0, then u must be positive.

Keeping invalid solutions after substitution

Solutions to the transformed quadratic must satisfy the original exponential equation.

Rounding intermediate calculations too early

Keep exact logarithmic expressions until the final step.

Failing to verify solutions

Substitution into the original equation helps detect errors.

Ignoring practical limitations

Exponential models assume particular growth or decay patterns that may not continue indefinitely.

Key Terms

• Exponential Equation: An equation in which the unknown variable appears in an exponent.

• Base: The number raised to a power.

• Exponent: The power to which a base is raised.

• Exponential Function: A function in which the variable appears in an exponent.

• Common Base: A shared base used to rewrite both sides of an exponential equation.

• One-to-One Function: A function in which distinct inputs produce distinct outputs.

• Logarithm: The exponent required to produce a given positive number from a fixed base.

• Natural Logarithm: A logarithm with base e.

• Common Logarithm: A logarithm with base 10.

• Inverse Function: A function that reverses another function.

• Change-of-Base Formula: A formula used to rewrite logarithms using a different base.

• Power Rule of Logarithms: A rule allowing an exponent to become a multiplier outside a logarithm.

• Exponent Laws: Mathematical rules describing operations involving powers.

• Substitution: Replacing a complicated expression with a simpler variable.

• Quadratic Equation: An equation containing a squared variable as its highest power.

• Factorisation: Rewriting an expression as a product of factors.

• Quadratic Formula: A formula used to solve quadratic equations.

• Graphical Solution: A solution found by examining intersections or zeros on a graph.

• Numerical Method: A procedure for approximating solutions using calculations.

• Verification: Checking a solution by substituting it into the original equation.

• Extraneous Solution: A candidate solution produced during manipulation that does not satisfy the original equation.

• Exponential Growth: Growth by a constant multiplicative factor over equal time intervals.

• Exponential Decay: Decrease by a constant multiplicative factor over equal time intervals.

• Growth Rate: The rate at which a quantity increases.

• Decay Constant: A parameter describing continuous exponential decay.

• Doubling Time: The time required for a growing quantity to become twice its initial value.

• Half-Life: The time required for a decaying quantity to decrease to half its original value.

• Compound Interest: Interest calculated on an initial principal and previously accumulated interest.

• Depreciation: A decrease in an asset's value over time.

• Mathematical Model: A mathematical representation of a real-world system.

• Approximate Solution: A numerical value close to an exact solution.

• Exact Solution: A solution expressed without numerical rounding.

Key Takeaways

• Exponential equations contain variables in exponents.

• Exponential equations can often be solved by rewriting both sides using a common base.

• When the bases are equal and valid, their exponents can be equated.

• Exponent laws help simplify exponential expressions before solving.

• Fractional and negative exponents can be rewritten using familiar bases.

• Logarithms allow exponential equations to be solved when common bases are unavailable.

• The natural logarithm is especially useful for equations involving e.

• The logarithmic power rule transforms an exponential equation into an algebraic equation.

• Exponential equations with multiple terms may be simplified by factoring.

• Substitution can transform some exponential equations into quadratic equations.

• When substituting u = aˣ, only positive values of u are valid.

• Some exponential equations require numerical or graphical methods.

• Graphs help identify the number and approximate locations of solutions.

• Verification is essential, particularly after substitution or algebraic manipulation.

• Exact logarithmic expressions should be retained until the final calculation.

• Exponential equations are widely used in finance, population biology, radioactive decay, and other scientific models.

• Growth models involve increasing exponential functions, while decay models involve decreasing exponential functions.

• Practical solutions must be interpreted using appropriate units and assumptions.

• Selecting the correct method depends on recognising the mathematical structure of the equation.

• Mastering exponential equations provides an essential foundation for logarithmic equations, exponential modelling, and advanced mathematical problem solving.

Check Your Understanding

Recall and Understanding

1. Define an exponential equation and explain how it differs from a power equation.

2. Explain when exponents can be equated directly.

3. Describe why logarithms are useful for solving exponential equations.

4. Explain how substitution can transform an exponential equation into a quadratic equation.

5. Explain why the substitution u = 3ˣ requires u > 0.

Application

6. Solve 2^(3x−1) = 64.

7. Solve 9ˣ = 27^(x−1).

8. Solve 4ˣ = 30, giving your answer to three decimal places.

9. Solve e^(0.5x) = 18.

10. Solve 2^(2x) − 6(2ˣ) + 8 = 0.

Analysis and Challenge

11. Solve 3^(2x) − 12(3ˣ) + 27 = 0, checking all solutions.

12. Solve 5^(x+1) − 5ˣ = 100 using factorisation.

13. Solve 2^(x+2) = 7^(x−1), giving your answer to three decimal places.

14. A bacterial population starts at 800 and doubles every 4 hours. Determine when the population will reach 20000.

15. An investment of $10000 earns 4.5% annual interest, compounded monthly. Determine how long it takes to reach $15000, and identify the first complete month when the balance reaches or exceeds that amount.

Final Summary

Exponential equations are used to determine unknown exponents and model processes involving repeated multiplication.

When both sides of an equation can be expressed using a common base, the exponents can be equated and solved algebraically.

When common bases are unavailable, logarithms provide a powerful alternative.

More complicated equations involving multiple exponential terms can often be simplified using factoring or substitution.

Graphical and numerical methods provide additional techniques when exact algebraic solutions are difficult to obtain.

Exponential equations are particularly important in real-world situations involving population growth, compound interest, radioactive decay, depreciation, and other changing quantities.

A complete solution should use an appropriate method, preserve mathematical accuracy, verify the result, and interpret the answer in context.

The central principle is that exponential equations become manageable when we recognise their structure and use common bases, logarithms, factoring, or substitution to isolate the unknown exponent.