Log Equation Calculator: Solve Logarithmic Equations with Steps
How to use the log equation calculator
Enter a logarithmic equation such as log2(x)+log2(x-2)=3. The calculator combines the logarithms, converts to exponential form, solves, and checks every candidate in the original equation, rejecting any value that makes a logarithm argument zero or negative.
Worked example
log2(x)+log2(x-2)=3 becomes log2(x(x-2))=3, so x(x-2)=2^3=8, giving x^2-2x-8=0, so x=4 or x=-2. Only x=4 is valid: at x=-2 both log arguments are negative, so x=-2 is extraneous and rejected.
For a fully worked example of this check in action, see extraneous roots in log equations.
Frequently asked questions
A logarithm log_b(A) asks for the power that turns the base b into A. Since a positive base raised to any power stays positive, A must be positive for a real logarithm to exist.
An extraneous solution appears during solving (for example after combining logs or squaring) but does not satisfy the original equation. It must be checked and discarded.
The exponential form behind log equations
Most log equations are solved by turning the logarithm back into a power. Once an equation has a single log on one side and a plain number on the other, you can rewrite it in exponential form and finish with ordinary algebra. Write down the domain limits first, so you know which answers are even allowed.
If two logs are added, the product rule merges them into one log first. If a log is multiplied by a number, the power rule moves that number up as an exponent inside the log.
Example: Solve log_5(2x + 1) = 2
Step 1, Note the domain: The argument must be positive, so 2x + 1 > 0, which means x > -0.5.
Step 2, Convert to exponential form: log_5(2x + 1) = 2 means 2x + 1 = 5^2 = 25.
Step 3, Solve the linear equation: Subtract 1 to get 2x = 24, then divide by 2, so x = 12.
Step 4, Check in the original equation: 2(12) + 1 = 25 and log_5(25) = 2, and 12 is inside the domain, so x = 12 is the solution.
Common mistakes
- Skipping the final check: Candidates from the algebra can fall outside the domain. Always test each one in the original equation before accepting it.
- Splitting a sum inside a log: log_b(A + B) is not log_b(A) + log_b(B). The product rule only applies to a product inside the log.
- Using the wrong base as the power: In log_5(2x + 1) = 2 the power is built on base 5, so the exponential form uses 5^2, not 2^5.
When to use this calculator
Use it when the unknown sits inside a logarithm, as in growth, sound level, and acidity problems, or when a homework equation mixes logs with different arguments. The tool is free to use and its domain check saves you from handing in an answer that cannot actually work.
More questions
log_b(MN) = log_b(M) + log_b(N), log_b(M/N) = log_b(M) – log_b(N), and log_b(M^p) = p log_b(M). These rules let you merge separate logs into one, or split one log apart, before converting to exponential form.
Yes. If every candidate from the algebra makes an argument zero or negative, all of them are rejected and the equation has no real solution. That outcome is valid, and stating it with the failed check shown is a complete answer.
Convert them to one base first with the change of base rule, log_b(A) = log(A) / log(b), using any common base such as 10. Once the bases match, the usual product and quotient rules apply.