Solving logarithmic equations
Solving Logarithmic Equations
Solving Logarithmic Equations is a topic in Exponential & Logarithmic Functions in the Texas Essential Knowledge and Skills (Algebra II, §111.40). It is aligned to Standard 2A.5(D), which requires students to solve single logarithmic equations, determining the reasonableness of the solution.
Logarithmic equations are solved by condensing to one log, rewriting in exponential form, and rejecting non-positive arguments.
Theory
To solve a logarithmic equation:
- Condense to a single logarithm using the laws.
- Rewrite in exponential form \(\log_b x=y\Rightarrow b^y=x\).
- Solve the resulting equation.
- Check — reject any negative or zero argument.
Convert to exponential form:
How to solve
- Combine logs into one with the laws.
- Rewrite in exponential form.
- Solve the equation.
- Reject solutions with a non-positive argument.
Rewrite in exponential form.
| \(x\) | \(=\) | \(2^5\) |
| \(=\) | \(32\) |
Condense, then use base 10.
| \(\log\big(x(x-3)\big)\) | \(=\) | \(1\) |
| \(x^2-3x\) | \(=\) | \(10\) |
| \((x-5)(x+2)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(5\) |
\(x=-2\) is rejected (negative argument).
Exponentiate with base \(e\).
| \(x\) | \(=\) | \(e^2\) |
| \(\approx\) | \(7.39\) |
A logarithm's argument must be positive, so any solution making it \(\le0\) is rejected.
Common pitfalls
Frequently asked questions
How do you solve \(\log_2 x=5\)?
Rewrite as \(x=2^5=32\).
Why check solutions of log equations?
The argument of a log must be positive; invalid ones are rejected.
How do you combine two logs?
Use the product law: \(\log x+\log y=\log(xy)\).
What makes a solution extraneous here?
It makes a logarithm's argument zero or negative.