Log Equation- Solving Logarithmic Equations

What Log Equations Actually Are

A log equation is any equation where the variable sits inside a logarithm. That's it. If you see log, ln, or any variable trapped in a logarithm, you're dealing with a log equation.

The goal is always the same: isolate the variable and solve. But the process trips up most students because they forget the core relationship.

Remember this forever:

logb(x) = y is the same as by = x

This one conversion is the backbone of every log equation you'll ever solve. Everything else is just manipulation around this relationship.

Essential Logarithm Rules You Need

Before touching any equation, these rules must be automatic. If you're hunting for them mid-problem, you're already lost.

Most log equation errors come from forgetting these or applying them backwards. Keep them straight.

How to Solve Log Equations

Here's the process. No fluff.

Step 1: Identify the Domain

Logarithms only accept positive arguments. Before you do anything else, find what x cannot be.

If your equation has log(x - 3), then x - 3 > 0, so x > 3. That's your domain restriction. Solutions outside this range are invalid, no matter how clean the algebra looks.

Step 2: Combine Logs If Needed

Most equations have logs on both sides. Use the rules above to combine them into a single log expression per side.

Example:

log(x) + log(3) = log(6)

Combine left side: log(3x) = log(6)

Step 3: Drop the Logs

Once you have log(something) = log(something else), drop both logs. They're equal only if the somethings are equal.

3x = 6

Step 4: Solve the Resulting Equation

Now it's basic algebra. Solve for x.

3x = 6 โ†’ x = 2

Step 5: Check Your Answer

Plug x = 2 back into your original equation. Does it work? If not, discard it. Domain restrictions apply.

log(2) + log(3) = log(6) โœ“

Common Log Equation Patterns

Pattern 1: Variable Inside a Single Log

log2(x + 5) = 4

Convert to exponential form: 24 = x + 5

16 = x + 5

x = 11

Check: log2(16) = 4 โœ“

Pattern 2: Logs on Both Sides

log3(x + 1) = log3(2x - 4)

Drop the logs: x + 1 = 2x - 4

1 + 4 = 2x - x

x = 5

Check domain: x + 1 > 0 and 2x - 4 > 0. Both true when x = 5. โœ“

Pattern 3: Variable in the Base

logx(8) = 3

Convert: x3 = 8

x = 2

Domain note: base x must be positive and โ‰  1. x = 2 works.

Pattern 4: Natural Log Equations

ln(x) + ln(x - 2) = ln(8)

Combine left: ln(x(x - 2)) = ln(8)

Drop ln: x(x - 2) = 8

x2 - 2x - 8 = 0

(x - 4)(x + 2) = 0

x = 4 or x = -2

Check domain: ln requires positive arguments. x = -2 fails. Only x = 4 works.

Natural Log vs. Log Base 10

ln means loge โ€” base e (approximately 2.718).

log without a subscript usually means log10.

Solving works the same way for both. The rules don't change. Only the base does.

Mistakes That Will Cost You

Quick Reference: Log Rules at a Glance

Rule Name Log Form Expanded Form
Product Rule logb(MN) logbM + logbN
Quotient Rule logb(M/N) logbM - logbN
Power Rule logb(Mp) p ยท logbM
Change of Base logbM logkM / logkb
Zero Property logb1 0
Identity logbb 1

Practice: Solve These

Try these before checking answers below.

1. log5(x + 4) = 2

2. log2(x) + log2(x - 2) = 3

3. ln(2x) = 4

Answers

1. 52 = x + 4 โ†’ 25 = x + 4 โ†’ x = 21

2. Combine: log2(x(x - 2)) = 3 โ†’ x2 - 2x = 8 โ†’ x2 - 2x - 8 = 0 โ†’ (x - 4)(x + 2) = 0 โ†’ x = 4 (x = -2 fails domain)

3. e4 = 2x โ†’ x = e4/2 โ‰ˆ 27.3

The Bottom Line

Log equations aren't hard. They're mechanical. Combine logs, drop them, solve the leftover equation, check your work. That's the entire process.

The mistakes aren't conceptual โ€” they're procedural. You know the rules. Apply them correctly. Check your answers. That's it.