Solving Logarithmic Equations with Squared Terms
What Are Logarithmic Equations with Squared Terms?
These are equations where you find something squared inside or outside the logarithm. The most common forms look like:
- (log x)² + 3 log x + 2 = 0
- log(x² + 1) = 2
- log₂(x) + log₂(x+1) = 1
The first type is trickier because you have a quadratic expression with log terms. The second type just requires you to handle the squared argument inside the log. Don't confuse them — they need different approaches.
Prerequisites You Need to Know
Before touching these problems, you need solid ground on:
- Logarithm properties (product, quotient, power rules)
- Domain restrictions — logs only accept positive arguments
- Basic quadratic factoring or quadratic formula
- Exponential form conversion
If any of these are shaky, fix that first. Building on weak foundations guarantees failure.
The Substitution Method — Your Main Tool
When you see something like (log x)², the move is substitution. Let t = log x. Now you have a regular quadratic equation: t² + bt + c = 0. Solve for t, then convert back.
Example Walkthrough
Solve: (log x)² - 5 log x + 6 = 0
Step 1: Substitute. Let t = log x
The equation becomes: t² - 5t + 6 = 0
Step 2: Factor. (t - 2)(t - 3) = 0
Step 3: Solve. t = 2 or t = 3
Step 4: Convert back. If t = log x, then x = 10² = 100 or x = 10³ = 1000
Step 5: Check domain. Both 100 and 1000 are positive, so both work.
That's it. The whole thing hinges on recognizing the pattern and making the substitution.
Common Mistakes That Sink Students
Ignoring domain restrictions. Logs require positive arguments. Every solution you get must be plugged back into the original log to verify it doesn't produce log(negative) or log(0).
Forgetting to check for extraneous solutions. Squaring both sides of an equation, or manipulating logs algebraically, can introduce solutions that don't actually satisfy the original equation. Always verify.
Mixing up log types. log(x) usually means base 10. ln(x) means natural log (base e). Mixing these up in a single problem is a fast track to wrong answers.
Not isolating the log before exponentiating. If you have log(x) = 3, you convert to x = 10³. But if it's 2 + log(x) = 5, you must isolate first: log(x) = 3, then x = 10³.
Comparing Solution Approaches
| Equation Type | Best Method | Key Step |
|---|---|---|
| (log x)² + b(log x) + c = 0 | Substitution | Let t = log x, solve quadratic |
| log(x²) = k | Exponentiate directly | x² = 10^k, then square root |
| log(f(x)) = log(g(x)) | Set arguments equal | f(x) = g(x), check domain |
| Nested logs: log(log(x)) = k | Work inside out | Solve inner log first |
Getting Started: A Step-by-Step Process
Here's the system to use every time:
- Identify the structure. Is it a quadratic in log form, or a log of something squared?
- Apply the right method. Substitution for quadratic forms, direct exponentiation for simpler cases.
- Solve the resulting equation. Algebra, factoring, or quadratic formula as needed.
- Convert back to x. If you substituted, undo it now.
- Check every solution. Plug back into original equation. Discard anything that breaks the log domain.
Quick Reference: Logarithm Rules
- log(ab) = log(a) + log(b)
- log(a/b) = log(a) - log(b)
- log(a^n) = n · log(a)
- log_b(x) = y means b^y = x
These are not optional. Know them cold.
Practice Problem
Solve: 2(log x)² + log(x³) - 5 = 0
Hint: Use log(x³) = 3 log x to combine terms into a quadratic in log x, then substitute.
Solution approach:
- Rewrite: 2(log x)² + 3 log x - 5 = 0
- Let t = log x: 2t² + 3t - 5 = 0
- Use quadratic formula: t = (-3 ± √49) / 4
- t = 1 or t = -3.5
- x = 10¹ = 10 or x = 10^(-3.5) ≈ 0.000316
- Both positive, both valid
The practice problem contains the full pattern you'll see on tests. Master this and you're set.