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Exponential and Logarithmic Equations

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Exponential and Logarithmic Equations

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Exponential and Logarithmic Equations

🎯 Learning Objectives

By the end of this topic, you will be able to:

Solve exponential equations using logarithm properties
Solve logarithmic equations by converting to exponential form
Check for extraneous solutions in logarithmic equations
Apply exponential and logarithmic equations to real-world problems (growth, decay, Richter scale, pH)

📋 Prerequisites

Logarithm Properties — product, quotient, power, change of base laws
Exponential Functions — growth/decay modeling

📖 Core Content

9.1 Solving Exponential Equations

9.1.1 Intuition

An exponential equation has the variable in the exponent:

2^x = 32

3^{2x+1} = 27

5^{x} = 7

. To solve, we "bring the variable down" using logarithms.

9.1.2 Methods

Method 1: Same Base If both sides can be written with the same base, set exponents equal:

b^{f(x)} = b^{g(x)} \implies f(x) = g(x)

. Method 2: Logarithms (General) Take log of both sides (any base), use power rule to bring exponent down. Recipe:

Isolate the exponential term
Take log of both sides
Use $\log(a^n) = n \log a$ to bring variable down
Solve for variable

9.1.3 Worked Examples

Example 1.1 (Easy — Same Base): Solve

2^{x} = 32

32 = 2^5

, so

2^x = 2^5 \implies x = 5

\boxed{x = 5}

Example 1.2 (Medium): Solve

3^{2x-1} = 81

81 = 3^4

, so

3^{2x-1} = 3^4 \implies 2x - 1 = 4 \implies 2x = 5 \implies x = 2.5

\boxed{x = 2.5}

Example 1.3 (Medium — Logs needed): Solve

5^{x} = 7

. Step 1: Take

\log

of both sides:

\log(5^x) = \log 7

Step 2:

x \log 5 = \log 7

Step 3:

x = \frac{\log 7}{\log 5} \approx \frac{0.8451}{0.6990} \approx 1.209

\boxed{x \approx 1.209}

Example 1.4 (Hard): Solve

2^{x+1} = 3^{2x}

. Step 1:

\ln(2^{x+1}) = \ln(3^{2x})

Step 2:

(x+1)\ln 2 = 2x \ln 3

Step 3:

x\ln 2 + \ln 2 = 2x \ln 3

Step 4:

\ln 2 = 2x\ln 3 - x\ln 2 = x(2\ln 3 - \ln 2)

Step 5:

x = \frac{\ln 2}{2\ln 3 - \ln 2} \approx \frac{0.6931}{2(1.0986) - 0.6931} = \frac{0.6931}{1.5041} \approx 0.4605

\boxed{x \approx 0.4605}

9.2 Solving Logarithmic Equations

9.2.1 Intuition

A logarithmic equation has the variable inside a log:

\log_2(x+1) = 3

\ln x + \ln(x-1) = \ln 2

. To solve, "undo" the log by exponentiating.

9.2.2 Methods

Method 1: Convert to exponential form

\log_b f(x) = c \implies f(x) = b^c

Method 2: Combine logs, then convert Use log laws to combine into a single log, then exponentiate. ⚠️ Always check for extraneous solutions! Logs are only defined for positive arguments. Recipe:

Isolate the logarithmic term(s)
Use log laws to combine into a single log
Convert to exponential form
Solve for variable
Check! Plug back into original equation — reject any solution that makes argument $\leq 0$

9.2.3 Worked Examples

Example 2.1 (Easy): Solve

\log_2(x+3) = 4

. Step 1:

x + 3 = 2^4 = 16

Step 2:

x = 13

Check:

\log_2(13+3) = \log_2 16 = 4

✓

\boxed{x = 13}

Example 2.2 (Medium): Solve

\ln x + \ln(x-1) = \ln 2

. Step 1:

\ln[x(x-1)] = \ln 2

Step 2:

x(x-1) = 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x = 2

x = -1

Check:

x = 2

\ln 2 + \ln 1 = \ln 2 + 0 = \ln 2

✓

x = -1

\ln(-1)

is undefined ✗ (extraneous)

\boxed{x = 2}

Example 2.3 (Hard): Solve

\log_2(x+1) - \log_2(x-1) = 2

. Step 1:

\log_2\frac{x+1}{x-1} = 2

Step 2:

\frac{x+1}{x-1} = 2^2 = 4 \implies x+1 = 4(x-1) \implies x+1 = 4x - 4 \implies 1+4 = 4x - x \implies 5 = 3x \implies x = \frac{5}{3}

Check:

x = 5/3

\log_2(8/3) - \log_2(2/3) = \log_2\left(\frac{8/3}{2/3}\right) = \log_2 4 = 2

✓

\boxed{x = \frac{5}{3}}

9.3 Applications

9.3.1 Exponential Growth/Decay

N(t) = N_0 e^{kt}

N(t) = N_0 a^t

Doubling time: Solve

2N_0 = N_0 e^{kt} \implies 2 = e^{kt} \implies \ln 2 = kt \implies t = \frac{\ln 2}{k}

Half-life: Solve

\frac{1}{2}N_0 = N_0 e^{kt} \implies \frac{1}{2} = e^{kt} \implies \ln(1/2) = kt \implies t = -\frac{\ln 2}{k}

9.3.2 Richter Scale (Earthquakes)

M = \log_{10}\left(\frac{I}{I_0}\right)

where

I

is intensity and

I_0

is reference intensity. A magnitude 7 earthquake has

10^{7-5} = 100

times the intensity of a magnitude 5 earthquake.

9.3.3 pH Scale

\text{pH} = -\log_{10}[H^+]

where

[H^+]

is hydrogen ion concentration.

9.3.4 Worked Example

Example 3.1: The population of a city grows exponentially:

P(t) = 100000 e^{0.02t}

. How long until the population reaches 200,000? Step 1:

200000 = 100000 e^{0.02t}

Step 2:

2 = e^{0.02t}

Step 3:

\ln 2 = 0.02t

Step 4:

t = \frac{\ln 2}{0.02} \approx \frac{0.6931}{0.02} = 34.66

years

\boxed{\approx 34.7 \text{ years}}

📐 Key Formulas — Summary Table

Equation Type	Method	Check
$b^{f(x)} = b^{g(x)}$	Set $f(x) = g(x)$	Same base required
$b^{f(x)} = c$	Take $\log$ : $f(x) \log b = \log c$
$\log_b f(x) = c$	$f(x) = b^c$	$f(x) > 0$
$\log_b f(x) = \log_b g(x)$	$f(x) = g(x)$	Both $f(x), g(x) > 0$
Doubling time	$t = \frac{\ln 2}{k}$	Exponential growth
Half-life	$t_{1/2} = \frac{\ln 2}{k}$	Exponential decay ( $k > 0$ is decay rate)

⚠️ Common Pitfalls

Pitfall 1: Forgetting to Check for Extraneous Solutions

Mistake: Solving

\log(x-1) + \log(x+1) = \log 3

and accepting all algebraic solutions. Correct:

\log(x-1) + \log(x+1) = \log 3 \implies \log(x^2-1) = \log 3 \implies x^2-1=3 \implies x^2=4 \implies x = \pm 2

. But

x = -2

makes

\log(-3)

undefined → extraneous. Only

x = 2

is valid.

Pitfall 2: Wrong Log Base Assumption

Mistake: Using

\log

(base 10) when the problem uses

\ln

(base

e

). Correct: Pay attention to the base.

\log x

is base 10,

\ln x

is base

e

. They differ by a constant factor.

Pitfall 3: Applying Log to Negative Arguments

Mistake: Taking

\log(-5)

which is undefined in real numbers. Correct:

\log_b x

requires

x > 0

. If solving gives

x \leq 0

, it's extraneous.

Pitfall 4: Solving $2^{x^2} = 2^{4x}$ by Setting $x^2 = 4x$ Then Dividing by $x$

Mistake:

x^2 = 4x \implies x = 4

(losing

x = 0

). Correct:

x^2 - 4x = 0 \implies x(x-4) = 0 \implies x = 0

x = 4

📝 Practice Questions

Q1: Solve $4^{x} = 64$ .

Strategy Hint: $64 = 4^3$ , so $x = 3$ .

$\boxed{3}$ Q2: Solve $2^{x} = 10$ to 3 decimal places.

$x = \log_2 10 = \frac{\ln 10}{\ln 2} \approx \frac{2.3026}{0.6931} \approx 3.322$

$\boxed{3.322}$ Q3: Solve $\log_3(x+5) = 2$ .

$x+5 = 3^2 = 9 \implies x = 4$

$\boxed{4}$ Q4: Solve $e^{2x} = 5$ .

$2x = \ln 5 \implies x = \frac{\ln 5}{2} \approx 0.805$

$\boxed{\frac{\ln 5}{2}}$ Q5: Solve $\log_2(x-1) + \log_2(x+3) = \log_2 5$ .

$\log_2[(x-1)(x+3)] = \log_2 5 \implies (x-1)(x+3) = 5$ $x^2 + 2x - 3 = 5 \implies x^2 + 2x - 8 = 0 \implies (x+4)(x-2) = 0 \implies x = -4$ or $x = 2$

Check: $x = -4 \to \log_2(-5)$ undefined ✗ $x = 2 \to \log_2(1) + \log_2(5) = 0 + \log_2 5$ ✓

$\boxed{2}$ Q6: Solve $5^{2x} = 125$ .

$125 = 5^3$ , so $5^{2x} = 5^3 \implies 2x = 3 \implies x = 1.5$

$\boxed{1.5}$ Q7: Solve $\ln(x) + \ln(x-3) = \ln 4$ .

$\ln[x(x-3)] = \ln 4 \implies x(x-3) = 4 \implies x^2 - 3x - 4 = 0 \implies (x-4)(x+1) = 0$ $x = 4$ or $x = -1$ (extraneous)

$\boxed{4}$ Q8: A bacteria culture triples every 2 hours. If initial population is 1000, how long until it reaches 27,000?

$N(t) = 1000 \cdot 3^{t/2}$ $27000 = 1000 \cdot 3^{t/2} \implies 27 = 3^{t/2} \implies 3^3 = 3^{t/2} \implies 3 = t/2 \implies t = 6$

$\boxed{6 \text{ hours}}$ Q9: Solve $2 \cdot 3^{x} = 54$ .

$3^{x} = 27 \implies 3^{x} = 3^3 \implies x = 3$

$\boxed{3}$ Q10: Solve $\log(x) + \log(2x) = 3$ (base 10).

$\log(2x^2) = 3 \implies 2x^2 = 10^3 = 1000 \implies x^2 = 500 \implies x = \sqrt{500} = 10\sqrt{5} \approx 22.36$

(Only positive solution, since $\log x$ requires $x > 0$ .)

$\boxed{10\sqrt{5}}$ Q11: Solve $e^{x} + e^{-x} = 4$ .

Strategy Hint: Multiply by $e^x$ : $e^{2x} + 1 = 4e^x \implies e^{2x} - 4e^x + 1 = 0$ . Let $u = e^x$ : $u^2 - 4u + 1 = 0$ . $u = \frac{4 \pm \sqrt{16-4}}{2} = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$ $e^x = 2 + \sqrt{3} \implies x = \ln(2+\sqrt{3})$ $e^x = 2 - \sqrt{3} \approx 0.268 \implies x = \ln(2-\sqrt{3})$ (also valid, since $2-\sqrt{3} > 0$ )

$\boxed{x = \ln(2 \pm \sqrt{3})}$ Q12: The half-life of Carbon-14 is 5730 years. If a fossil has 25% of its original Carbon-14, how old is it?

$N(t) = N_0 (1/2)^{t/5730}$ $0.25 N_0 = N_0 (1/2)^{t/5730} \implies 0.25 = (1/2)^{t/5730} \implies \ln(0.25) = \frac{t}{5730} \ln(0.5)$ $t = 5730 \cdot \frac{\ln(0.25)}{\ln(0.5)} = 5730 \cdot \frac{-1.3863}{-0.6931} = 5730 \cdot 2 = 11460$

$\boxed{11,460 \text{ years}}$

🔗 Cross-References

Previous topic: Logarithm Properties
Next topic: Sequences and Limits
Across courses: BSMA1002 Stats 1 (logistic regression, log-likelihood); BSMA1003 Maths 2 (logarithmic differentiation) Join Discord Previous6.1 Logarithmic Functions & PropertiesNext7.1 Sequences & Convergence