AP Precalculus: Solving Exponential & Logarithmic Equations

Master equation-solving techniques, applications, and word problems

🔢 Exponential 📊 Logarithmic 💰 Applications ✏️ Word Problems

📚 Solving Exponential & Logarithmic Equations

Exponential and logarithmic equations appear throughout mathematics and real-world applications. This guide covers multiple solution methods — from rewriting bases to using logarithms — along with essential formulas for growth, decay, and compound interest problems on the AP Precalculus exam.

1 Solving Exponential Equations by Rewriting the Base

When both sides of an exponential equation can be expressed with the same base, use the one-to-one property: if $a^{f(x)} = a^{g(x)}$, then $f(x) = g(x)$.

Same Base Property If $a^{f(x)} = a^{g(x)}$, then $f(x) = g(x)$

Steps to Solve

Identify a common base that both sides can be written with
Rewrite each side as a power of that base
Set the exponents equal to each other
Solve the resulting equation

📌 Example 1

Solve: $3^{2x} = 9^{x+1}$

Rewrite 9 as $3^2$: $3^{2x} = (3^2)^{x+1} = 3^{2(x+1)} = 3^{2x+2}$

Set exponents equal: $2x = 2x + 2$

Solve: $0 = 2$ — No solution!

📌 Example 2

Solve: $4^{x-1} = 8^{2x}$

Rewrite both as powers of 2: $(2^2)^{x-1} = (2^3)^{2x}$

$2^{2(x-1)} = 2^{6x}$ → $2^{2x-2} = 2^{6x}$

Set exponents equal: $2x - 2 = 6x$

Solve: $-2 = 4x$ → $x = -\frac{1}{2}$

💡 Common Bases to Know

4 = 2², 8 = 2³, 16 = 2⁴, 27 = 3³, 81 = 3⁴, 25 = 5², 125 = 5³, 1000 = 10³

2 Solving Exponential Equations Using Logarithms

When you cannot rewrite both sides with the same base, take the logarithm of both sides to bring the exponent down using the power rule.

Steps to Solve

Isolate the exponential expression on one side
Take the logarithm of both sides (usually $\log$ or $\ln$)
Use the power rule: $\log(a^x) = x \cdot \log a$
Solve for the variable

General Solution Pattern $ab^{cx} = d$ → $b^{cx} = \frac{d}{a}$ → $cx = \log_b\left(\frac{d}{a}\right)$ → $x = \frac{\log_b\left(\frac{d}{a}\right)}{c}$

📌 Example

Solve: $5^{2x-1} = 17$

Take ln of both sides: $\ln(5^{2x-1}) = \ln 17$

Apply power rule: $(2x-1) \ln 5 = \ln 17$

Solve for x:

$2x - 1 = \frac{\ln 17}{\ln 5} ≈ \frac{2.833}{1.609} ≈ 1.760$

$2x = 2.760$ → $x ≈ 1.380$

💡 Which Log to Use?

Use $\ln$ (natural log) or $\log$ (common log) — both give the same answer. Choose whichever your calculator handles best!

3 Solving Logarithmic Equations with One Log

When an equation has a single logarithm, convert to exponential form and solve. Remember to check that your solution doesn't make the argument negative or zero!

Convert to Exponential Form $\log_a(f(x)) = b$ → $f(x) = a^b$

Steps to Solve

Isolate the logarithm on one side
Convert to exponential form: $\log_a(f(x)) = b$ becomes $f(x) = a^b$
Solve the resulting equation for $x$
Check: Verify the argument of the log is positive

📌 Example

Solve: $\log_3(2x - 1) = 4$

Convert to exponential: $2x - 1 = 3^4 = 81$

Solve: $2x = 82$ → $x = 41$

Check: $2(41) - 1 = 81 > 0$ ✓

⚠️ Always Check Solutions!

A solution is extraneous if it makes the argument of any logarithm ≤ 0. Always substitute back to verify!

4 Solving Logarithmic Equations with Multiple Logs

When an equation has multiple logarithms, use logarithm properties to combine them into a single log, then solve.

📦 Combine with Product Rule

$\log_a f(x) + \log_a g(x) = \log_a[f(x) \cdot g(x)]$

➗ Combine with Quotient Rule

$\log_a f(x) - \log_a g(x) = \log_a\left[\frac{f(x)}{g(x)}\right]$

One-to-One Property If $\log_a f(x) = \log_a g(x)$, then $f(x) = g(x)$

📌 Example 1: Combine and Convert

Solve: $\log_2 x + \log_2(x - 2) = 3$

Combine: $\log_2[x(x-2)] = 3$

Convert: $x(x-2) = 2^3 = 8$

Expand: $x^2 - 2x - 8 = 0$

Factor: $(x-4)(x+2) = 0$ → $x = 4$ or $x = -2$

Check: $x = -2$ makes $\log_2(-2)$ undefined. Solution: $x = 4$

📌 Example 2: One-to-One

Solve: $\log(3x + 1) = \log(2x + 5)$

Apply one-to-one: $3x + 1 = 2x + 5$

Solve: $x = 4$

Check: $3(4)+1 = 13 > 0$ and $2(4)+5 = 13 > 0$ ✓

5 Exponential Growth & Decay Applications

Growth and decay problems model real-world phenomena where quantities increase or decrease by a constant percentage over equal time intervals.

📈 Growth Model

$y = a(1 + r)^t$

Use when quantity increases by rate $r$ per time period

📉 Decay Model

$y = a(1 - r)^t$

Use when quantity decreases by rate $r$ per time period

🔄 Continuous Model

$y = ae^{kt}$

$k > 0$: growth; $k < 0$: decay

Variable	Meaning
$a$	Initial value (when $t = 0$)
$r$	Rate of growth or decay (as decimal)
$t$	Time (in consistent units)
$k$	Continuous rate constant
$y$	Final value after time $t$

📌 Example: Population Growth

Problem: A population of 5000 grows at 3% per year. When will it reach 10000?

Model: $10000 = 5000(1.03)^t$

Solve: $2 = 1.03^t$ → $\log 2 = t \log 1.03$

$t = \frac{\log 2}{\log 1.03} ≈ \frac{0.301}{0.0128} ≈ 23.4$ years

6 Compound Interest Formulas

Compound interest problems are a common application of exponential equations. Interest compounds at specific intervals (monthly, quarterly, etc.) or continuously.

💵 Periodic Compounding

$A = P\left(1 + \frac{r}{n}\right)^{nt}$

Interest compounds $n$ times per year

♾️ Continuous Compounding

$A = Pe^{rt}$

Interest compounds every instant

Variable	Meaning
$P$	Principal (initial investment)
$r$	Annual interest rate (as decimal)
$n$	Number of compounding periods per year
$t$	Time in years
$A$	Final amount (principal + interest)

Common Values of $n$

Annually

$n = 1$

Semi-annually

$n = 2$

Quarterly

$n = 4$

Monthly

$n = 12$

Daily

$n = 365$

Continuous

Use $A = Pe^{rt}$

📌 Example

Problem: $5000 invested at 6% compounded monthly. Find the amount after 10 years.

Given: $P = 5000$, $r = 0.06$, $n = 12$, $t = 10$

Calculate: $A = 5000\left(1 + \frac{0.06}{12}\right)^{12 \times 10}$

$A = 5000(1.005)^{120} ≈ 5000(1.8194) ≈ \$9,097$

7 Half-Life Problems

The half-life is the time required for a quantity to reduce to half its initial value. This is commonly used for radioactive decay and drug metabolism.

Half-Life Formula \[N(t) = N_0 \left(\frac{1}{2}\right)^{t/h}\] where $h$ = half-life, $t$ = time, $N_0$ = initial amount

📌 Example

Problem: A radioactive substance has a half-life of 5 years. If you start with 200g, how much remains after 12 years?

Model: $N(12) = 200 \cdot \left(\frac{1}{2}\right)^{12/5}$

$N(12) = 200 \cdot (0.5)^{2.4} ≈ 200 \cdot 0.189 ≈ 37.9$ grams

💡 Finding Half-Life

To find half-life when given two data points, set up the equation $\frac{1}{2} = b^h$ and solve for $h$.

8 Common Mistakes to Avoid

Exponential and logarithmic equations have specific rules that students often misapply. Review these common errors!

Forgetting to check for extraneous solutions — Solutions that make log arguments ≤ 0 are invalid
Incorrectly distributing logs: $\log(a + b) \neq \log a + \log b$ — Only products work!
Confusing $\log a^b$ with $(\log a)^b$ — The power rule gives $b \log a$, not $(\log a)^b$
Forgetting to isolate before taking logs — Isolate the exponential term first!
Using wrong compound interest formula — Know when to use periodic vs. continuous
Rate confusion — Convert percentages to decimals (5% = 0.05)

⚠️ Common Log Error

$\log(x + y) \neq \log x + \log y$
$\log(x - y) \neq \log x - \log y$
Only $\log(xy) = \log x + \log y$ is correct!

📋 Quick Reference: Key Formulas

Same Base Property

$a^{f(x)} = a^{g(x)}$ → $f(x) = g(x)$

Log Equation

$\log_a(f(x)) = b$ → $f(x) = a^b$

Growth Model

$y = a(1 + r)^t$

Decay Model

$y = a(1 - r)^t$

Compound Interest

$A = P(1 + \frac{r}{n})^{nt}$

Continuous

$A = Pe^{rt}$

Variable	Meaning
\(a\)	Initial value (when \(t = 0\))
\(r\)	Rate of growth or decay (as decimal)
\(t\)	Time (in consistent units)
\(k\)	Continuous rate constant
\(y\)	Final value after time \(t\)

AP Precalculus: Solving Exponential & Logarithmic Equations

📚 Solving Exponential & Logarithmic Equations

1 Solving Exponential Equations by Rewriting the Base

Steps to Solve

2 Solving Exponential Equations Using Logarithms

Steps to Solve

3 Solving Logarithmic Equations with One Log

Steps to Solve

4 Solving Logarithmic Equations with Multiple Logs

5 Exponential Growth & Decay Applications

6 Compound Interest Formulas

Common Values of \(n\)

Annually

Semi-annually

Quarterly

Monthly

Daily

Continuous

7 Half-Life Problems

8 Common Mistakes to Avoid

📋 Quick Reference: Key Formulas

Same Base Property

Log Equation

Growth Model

Decay Model

Compound Interest

Continuous

Need Help with Exponential & Logarithmic Equations?

Variable	Meaning
\(P\)	Principal (initial investment)
\(r\)	Annual interest rate (as decimal)
\(n\)	Number of compounding periods per year
\(t\)	Time in years
\(A\)	Final amount (principal + interest)