A logarithmic equation such as log₃(x) = 4 has three parts: the base, the number inside the log, and the log value. Know any two and you can find the third. The key move is always the same: rewrite the equation in exponential form, where it becomes an ordinary power or root. This calculator does that for whichever part you choose, gives exact answers when they exist, and checks the result.
How to use the logarithm equation calculator
- Choose what to solve for: x (the number), b (the base) or y (the log value).
- Fill in the two remaining boxes. The base must be positive and not 1 (type
efor natural logs); x must be positive; y can be anything. - Read the answer on the tape and follow the steps, which always begin by converting to exponential form.
Formulas
Everything follows from the definition of a logarithm:
| Unknown | Formula | Example |
|---|---|---|
| x | x = bʸ | log₃(x) = 4 → x = 3⁴ = 81 |
| b | b = x^(1/y) | log_b(8) = 3 → b = ∛8 = 2 |
| y | y = ln x ÷ ln b | log₄(32) = y → y = 5/2 |
Worked examples
Solve for x: log₃(x) = 4. Exponential form: x = 3⁴ = 81. Check: 3 × 3 × 3 × 3 = 81, so log₃(81) = 4. ✓
Solve for b: log_b(1/16) = −4. Exponential form: b⁻⁴ = 1/16, so b⁴ = 16 and b = 2 (the negative fourth root is not a valid base).
Solve for y: log₄(32) = y. Since 4 = 2² and 32 = 2⁵, 4^(5/2) = 2⁵ = 32, so y = 5/2 = 2.5.
When the answer is irrational, the calculator gives a decimal. For log_b(10) = 2, b = √10 ≈ 3.1622776602; for ln(x) = 2, x = e² ≈ 7.3890560989.
Restrictions to remember
Logarithms have built-in limits, and most “no solution” results come from one of them:
- The argument must be positive. log_b(0) and log_b(−4) do not exist in the real numbers, because bʸ is always positive.
- The base must be positive and not 1. 1ʸ = 1 for every y, so base 1 cannot produce any other number, and negative bases do not have real powers for most exponents.
- log_b(1) = 0 for every base. If y = 0 and x = 1, every valid base is a solution; if y = 0 and x is anything else, no base works.
Solving harder log equations
Equations in textbooks often need a little algebra before they reach the form log_b(x) = y:
- Combine logs with the product and quotient laws: log₂(x) + log₂(x − 2) = 3 becomes log₂(x(x − 2)) = 3.
- Convert to exponential form: x(x − 2) = 2³ = 8, so x² − 2x − 8 = 0.
- Solve the resulting equation: (x − 4)(x + 2) = 0, so x = 4 or x = −2.
- Reject extraneous answers: x = −2 would put a negative number inside a log, so the only solution is x = 4.
Step 4 is where many solutions go wrong. Always substitute back into the original equation. For step 3, the quadratic formula calculator handles any quadratic, and the logarithm calculator evaluates the logs you need for checking.
Frequently asked questions
How do you solve log_b(x) = y for x?
Rewrite it in exponential form: x = b^y. For example, log₃(x) = 4 means x = 3⁴ = 81. This works for any base, including e: ln(x) = 2 gives x = e² ≈ 7.389.
How do you find an unknown base?
From log_b(x) = y, write b^y = x and take the y-th root of both sides: b = x^(1/y). For log_b(8) = 3, b = 8^(1/3) = 2. Because a base must be positive, only the positive root counts.
When does a logarithm equation have no solution?
When the input would have to be zero or negative (logs only accept positive numbers), when the base would have to be 1, or when log_b(x) = 0 but x is not 1. The calculator explains which rule rules out an answer.
Can the log value y be negative or a fraction?
Yes. log₂(x) = −3 gives x = 2⁻³ = 1/8, and log₄(x) = 1/2 gives x = √4 = 2. Only the base and the number x have restrictions; y can be any real number.