Logarithmic Functions

In 2.10 you built ${\log }_{b}x$ by swapping the columns of a table of $b^x$. Swapping columns swaps inputs and outputs, so the inverse trades domain and range.

The exponential function $b^x$ takes any real exponent, and every output it gives is positive. So for ${\log }_{b}x$:

• the domain is every real number greater than zero, because only positive numbers ever come out of $b^x$;
• the range is all real numbers, because every real number is an exponent of $b^x$.

Take ${\log }_{2}x$. It reaches 10 at $x=1024$ and $-10$ at $x=\frac{1}{1024}$. Pick any output you like, as big or as negative as you want, and some positive input produces it. What it never takes is 0 or a negative number as input.

The general form $f(x)=a{\log }_{b}x$ changes neither fact. The constant $a$ multiplies outputs, and a nonzero multiple of all real numbers is still all real numbers. The inputs are untouched, so they still have to be positive.

Reflecting a graph over $y=x$ cannot make it turn around. Since $b^x$ is always increasing or always decreasing, so is its inverse. The bend carries over too: a logarithmic graph is always concave up or always concave down.

Watch it in a table of ${\log }_{2}x$. The output rises from 0 to 1 over a run of 1 (from $x=1$ to 2), then from 1 to 2 over a run of 2, then over a run of 4. The graph keeps rising but needs longer and longer runs to do it. Rising and flattening is concave down.

A negative $a$ flips that: $-{\log }_{2}x$ is always decreasing and always concave up. So does a base between 0 and 1: ${\log }_{0.5}x$ gives 0, $-1$, $-2$ and $-3$ at 1, 2, 4 and 8, the same outputs as $-{\log }_{2}x$.

A graph that never turns around has no maximum or minimum on its whole domain, so a logarithmic function has extrema only on a closed interval, and they sit at the endpoints: on $\frac{1}{2}\le x\le 8$, ${\log }_{2}x$ runs from a minimum of $-1$ to a maximum of 3. And a bend that never switches means no points of inflection.

Walk toward 0 along the inputs of ${\log }_{2}x$: $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, and on to $\frac{1}{1024}$. The outputs are $-1$, $-2$, $-3$, and on to $-10$. Every halving of the input subtracts another 1, and you can halve forever without reaching 0, so the outputs have no floor:

The $0^{+}$ means you approach 0 from the right, the only side the domain allows. The graph hugs the line $x=0$ without touching it. That line is a vertical asymptote.

The other end is slow but has no ceiling either. Doubling the input adds 1, so ${\log }_{2}x$ is 10 at 1024 and 20 at $2^{20}$, which is about a million. Slow is not the same as leveling off:

That is the whole list for the general form $a{\log }_{b}x$. Each of the two limits is $\infty$ or $-\infty$. A negative $a$, or a base between 0 and 1, swaps both signs, so ${\lim }_{x\to 0^{+}}(-{\log }_{2}x)=\infty$. There is no horizontal asymptote at either end.

In a table of $f(x)={\log }_{2}x$, the outputs 0, 1, 2, 3 come at the inputs 1, 2, 4, 8. Over equal-length output intervals, the inputs are proportional: each is 2 times the one before. Every logarithmic function in general form has this pattern.

Now shift. The additive transformation $g(x)=f(x+k)$ with $k=3$ is $g(x)={\log }_2(x+3)$, and its outputs 0, 1, 2, 3 come at $-2$, $-1$, 1 and 5. Those inputs are not proportional; nothing multiplies $-2$ into $-1$ and also $-1$ into 1. For any $k\ne 0$, the pattern breaks.

It also works as a test. Suppose a function $f$ has outputs 0, 1, 2, 3 at the inputs $-2$, $-1$, 1, 5. Try $g(x)=f(x+k)$ with $k=-3$. Then $g(1)=f(-2)=0$, $g(2)=f(-1)=1$, $g(4)=f(1)=2$ and $g(8)=f(5)=3$. The inputs of $g$ are 1, 2, 4, 8, proportional over equal-length output intervals, so $f$ is logarithmic. In plain words: if moving every input by the same amount makes them proportional, you are looking at a logarithm.