The graph of $y = \log_b x$ is the graph of $y = b^x$ reflected across the line $y = x$, because they are inverse functions. Everything about it follows from that reflection and from what a log is.
Key idea
$y = \log_b x$ (with $b > 1$) has domain $(0, \infty)$, range all real numbers, $x$-intercept $(1, 0)$, the point $(b, 1)$, and a vertical asymptote $x = 0$. For $y = \log_b(x - h) + k$: the asymptote is $x = h$ and the domain is $(h, \infty)$.
Why it works
Reflecting across $y = x$ swaps $x$ and $y$. The exponential's horizontal asymptote $y = 0$ becomes a vertical asymptote $x = 0$; its range $(0, \infty)$ becomes the log's domain; its $y$-intercept $(0, 1)$ becomes the log's $x$-intercept $(1, 0)$. And logs of numbers near 0 are hugely negative: $\log_2 \frac{1}{1024} = -10$.
Worked examples
Worked example: Asymptote and domain
Find the asymptote, domain and $x$-intercept of $y = \log_2(x + 3) - 1$.
- The input must be positive: $x + 3 > 0$, so the domain is $(-3, \infty)$ and the asymptote is $x = -3$.
- $x$-intercept: $\log_2(x + 3) = 1$, so $x + 3 = 2$ and $x = -1$.
Answer: Asymptote $x = -3$; domain $(-3, \infty)$; $x$-intercept $(-1, 0)$
Worked example: A few points to plot
Plot $y = \log_3 x$.
- Use powers of 3: $(1, 0)$, $(3, 1)$, $(9, 2)$, $(\frac{1}{3}, -1)$.
Answer: A slow rise to the right, plunging down near $x = 0$.
Common mistake
Giving the log graph a horizontal asymptote. Logs keep growing, slowly but without limit ($\log_2 1024 = 10$, $\log_2 2^{100} = 100$), so there is no horizontal asymptote. The asymptote is vertical, where the input approaches 0.