Why the argument cannot be zero or negative
The statement y = log_b(x) means b^y = x. For a base b that is positive and not equal to 1, the exponential b^y is always positive. It never hits 0, and it never becomes negative. There is therefore no real exponent y that satisfies b^y = 0 or b^y = −4. Those inputs are outside the domain of a real logarithm.
The graph has a vertical asymptote where the argument would be 0. As x approaches that line from the positive side, the logarithm drops without bound or rises without bound, depending on whether the base is greater than 1 or between 0 and 1. The function does not cross the asymptote.
Horizontal shift
For y = log_b(x − h) + k the argument is the expression x − h, not the letter x alone. The inequality is x − h > 0, which is x > h. In US interval notation the domain is (h, ∞). The endpoint h is open: substituting x = h makes the argument 0.
The constant k moves the graph up or down. It does not change which x-values are allowed. The range remains (−∞, ∞) for every valid base. That is why the calculator always reports the same range once the base has been accepted.
Illegal bases
Base 1 fails because 1 raised to any real power is still 1. The exponential is a horizontal line, not an invertible curve. A base of 0 or a negative base is not used for the real logarithm on this site. The calculator rejects those bases before it writes a domain. That is an input error, not a domain of a well-defined log function.
A number you can type
For y = log₂(x − 3) + 1 the domain is (3, ∞). The point x = 4 gives argument 1, and log₂(1) + 1 = 1. The endpoint x = 3 makes the argument 0, so that input is refused. The point x = 2 makes the argument −1, which is also refused. The Logarithm Calculator uses the same argument rule when you ask for a single value of log_b(x).
This guide does not treat complex logarithms, change-of-base proofs beyond the evaluator, or log inequalities with more than one factor. If a homework problem writes ln(5 − x), the same idea applies: 5 − x > 0, so x < 5. You would then translate that inequality into an interval by hand. The on-site calculator is built for the form log_b(x − h) + k, not for every rearrangement.