Log equation solver
The log equation solver turns log_2(3x + 1) = 4 into 3x + 1 = 2^4 = 16, so x = 5.
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- Step 1: rewrite as an exponential log_b(u) = y says exactly that b^y = u: the log is the exponent.
- log_2(3x + 1) = 4 means 3x + 1 = 2^4
- Step 2: evaluate the power
- 2^4 = 16
- Step 3: solve the linear equation
- 3x + 1 = 16, so 3x = 16 - 1 = 15, and x = 15 / 3 = 5
- Check The argument of a log must be positive, and here it is 16. A solution of log_b(ax + c) = y always passes, because b^y is never zero or negative.
- log_2(3 × 5 + 1) = log_2(16) = 4
- Equations with a log on both sides, or a sum of logs, can produce a root where an argument is zero or negative. Those are extraneous: substitute each root back, as in the check, and discard any that fail.
How the Log Calculator works
A logarithmic equation solver starts from the definition: log_b(u) = y means b^y = u. The log on one side becomes a power on the other, and what is left is an ordinary equation in x.
Worked example: log_2(3x + 1) = 4.
1. Rewrite as an exponential: 3x + 1 = 2^4. 2. Evaluate the power: 2^4 = 16. 3. Solve the linear equation: 3x = 15, so x = 5. 4. Check: log_2(3 × 5 + 1) = log_2(16) = 4.
The check matters because a log accepts only positive arguments. A single log set equal to a number always passes, since b^y is positive, but equations that combine logs can produce roots that are not solutions. log_2(x) + log_2(x - 2) = 3 combines to log_2(x(x - 2)) = 3, so x² - 2x = 8, and x = 4 or x = -2. Only 4 works: log_2(-2) does not exist, so -2 is extraneous.
Exponential equations run the same rule the other way. To solve b^(ax + c) = k, take the log base b of both sides: ax + c = log_b(k). Most calculators have only ln and log10 keys, so log_b(k) is worked out as ln(k) / ln(b). For 3^(2x - 1) = 50, log_3(50) = 3.912023 / 1.0986123 = 3.5608768, so 2x - 1 = 3.5608768 and x = 2.2804384.
The same step answers doubling questions. Money growing 5% a year doubles when 1.05^x = 2, which gives x = ln 2 / ln 1.05 = 14.2 years. A natural log equation is the base-e case: ln(x) = 1 means x = e^1 = 2.718281828.
Log Calculator FAQ
What is the solution of log x = 2?
x = 100. With no base written, log means base 10, so log x = 2 says 10^2 = x.
How do you solve 2^x = 50?
Take a log of both sides: x = log_2(50) = ln 50 / ln 2 = 3.912023 / 0.6931472 = 5.6439. Check: 2^5.6439 is 50.
Can a logarithmic equation have no solution?
Yes. log_b(ax + c) = y always has one, but b^(ax + c) = k has none when k is zero or negative, because a positive base raised to any real power is positive. 2^x = -4 has no real solution.