Solving logarithmic equations
Solving Logarithmic Equations
Solving Logarithmic Equations is a topic in Exponential & Logarithmic Functions in the Common Core State Standards. It is aligned to Standard F-LE.4, which requires students to solve logarithmic equations and interpret the solution using logarithms.
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.