Graphs of logarithmic functions: Worked examples

Everything here rests on one fact from Graphs of logarithmic functions: reading $y = lo g_{a} x$ is the same as solving $x = a^{y}$ (see Inverse functions and reflection in y=x).

The scenario matrix

Every log-graph question falls into one of these cells. The examples below are labelled with the cell they cover, and together they hit all of them.

Cell	What varies	Danger you must handle
A Base $a > 1$	ordinary $ln$ , $l g$	rises slowly, $y$ any sign
B Base $0 < a < 1$	shrinking base	curve decreases (vertical flip)
C Negative / zero input	$x \leq 0$	undefined — not a small number, nothing
D Horizontal shift	$lo g_{a} (x - b)$	asymptote moves to $x = b$ ; domain moves
E Vertical shift via log law	$lo g_{a} (k x)$	inside-multiply becomes up-shift
F Reflection	$- lo g_{a} x$ , $lo g_{a} (- x)$	which axis flips? domain flips too
G Limiting behaviour	$x \to 0^{+}$ , $x \to \infty$	$- \infty$ vs slow $+ \infty$
H Real-world word problem	pH / Richter	plug into formula, keep units
I Exam twist	intersection / combined	two transforms at once

Example 1 — Cell A: ordinary base $a > 1$

Anchor at $x = 1$ . Set $y = lo g_{3} x$ , so $x = 3^{y}$ . Put $y = 0$ : $x = 3^{0} = 1$ . Point $(1, 0)$ . Why this step? $lo g_{a} 1 = 0$ for every base — it's the universal crossing point.
Anchor at $x = a$ . Put $y = 1$ : $x = 3^{1} = 3$ . Point $(3, 1)$ . Why? $lo g_{a} a = 1$ ; this fixes how "wide" the curve is.
A negative $y$ anchor. Put $y = - 1$ : $x = 3^{- 1} = \frac{1}{3}$ . Point $(\frac{1}{3}, - 1)$ . Why? Shows the curve diving toward the wall $x = 0$ .

Verify: $3^{0} = 1$ ✓, $3^{1} = 3$ ✓, $3^{- 1} = 0.333 \dots$ ✓ — all three points satisfy $x = 3^{y}$ .

Example 2 — Cell B: shrinking base $0 < a < 1$

Anchors first. $x = 1 \Rightarrow y = 0$ (point $(1, 0)$ ); $x = \frac{1}{2} \Rightarrow y = 1$ since $(\frac{1}{2})^{1} = \frac{1}{2}$ (point $(\frac{1}{2}, 1)$ ). Why? Same anchor logic — $x = a$ gives $y = 1$ , but here $a = \frac{1}{2} < 1$ .
Compare with $x = 2$ . $(\frac{1}{2})^{y} = 2 \Rightarrow y = - 1$ , so $(2, - 1)$ . Why? As $x$ grows past $1$ , $y$ goes negative — the tell-tale of a decreasing curve.
Use the flip identity. $lo g_{1/2} x = - lo g_{2} x$ . Why this tool and not another? One log law ( $lo g_{1/ a} x = - lo g_{a} x$ ) instantly proves it's the $a > 1$ curve reflected in the $x$ -axis, so it must decrease.

Verify: $- lo g_{2} 2 = - 1$ matches $(2, - 1)$ ✓, and $- lo g_{2} \frac{1}{2} = - (- 1) = 1$ matches $(\frac{1}{2}, 1)$ ✓.

Example 3 — Cell C: degenerate / illegal inputs

$lo g_{10} 1$ . $x = 1 0^{y}$ , want $x = 1 \Rightarrow y = 0$ . Answer $0$ . Why? Only power giving $1$ is $0$ .
$lo g_{10} 0$ . Need $1 0^{y} = 0$ . But $1 0^{y} > 0$ for all real $y$ ; it only approaches $0$ as $y \to - \infty$ . Why undefined, not $- \infty$ ? No finite $y$ works — $0$ is the asymptote value, never reached (Cell G idea).
$lo g_{10} (- 4)$ . Need $1 0^{y} = - 4$ . Impossible: $1 0^{y} > 0$ always. Why? The output of an exponential is strictly positive, so the input of a log must be strictly positive.

Verify: $1 0^{0} = 1$ ✓; there is no real $y$ with $1 0^{y} = 0$ or $1 0^{y} = - 4$ (checked numerically).

Example 4 — Cell D: horizontal shift

Find the new asymptote. The log blows up where its input is $0$ : $x - 3 = 0 \Rightarrow x = 3$ . Why? The wall is wherever the inside equals the old danger point $0$ .
$x$ -intercept. Set $y = 0$ : need inside $= 1$ , so $x - 3 = 1 \Rightarrow x = 4$ . Point $(4, 0)$ . Why? $lo g_{2} 1 = 0$ ; the crossing moves right by $3$ with everything else.
One more point. Inside $= 2$ : $x - 3 = 2 \Rightarrow x = 5$ , giving $y = 1$ . Point $(5, 1)$ .
Domain. Need inside $> 0$ : $x - 3 > 0 \Rightarrow x > 3$ . Why? Shifting right moves the allowed region right too.

Verify: at $x = 4$ , $lo g_{2} (4 - 3) = lo g_{2} 1 = 0$ ✓; at $x = 5$ , $lo g_{2} 2 = 1$ ✓; wall at $x = 3$ ✓.

Example 5 — Cell E: inside-multiply becomes vertical shift

Split with a log law. $lo g_{10} (1000 x) = lo g_{10} 1000 + lo g_{10} x$ . Why this tool? The Laws of logarithms product rule $lo g (M N) = lo g M + lo g N$ turns an inside change into an outside (vertical) one — the only clean way.
Evaluate the constant. $lo g_{10} 1000 = 3$ since $1 0^{3} = 1000$ .
Read the shift. $y = 3 + lo g_{10} x$ : every point moves up 3; asymptote $x = 0$ unchanged. Why unchanged? A vertical shift never touches the $x$ -position of the wall.

Verify: at $x = 1$ , LHS $= lo g_{10} 1000 = 3$ ; RHS $= 3 + lo g_{10} 1 = 3$ ✓.

Example 6 — Cell F: reflections (two different ones)

$y = - lo g_{2} x$ . The outside minus negates the output — reflection in the $x$ -axis. Domain still $x > 0$ . Why? Changing sign of $y$ -values only, so the picture flips vertically.
$y = lo g_{2} (- x)$ . The inside minus negates the input — reflection in the $y$ -axis. Now we need $- x > 0 \Rightarrow x < 0$ . Why? The allowed inputs flip to the negative side of the $x$ -axis.
Check a point. For $y = - lo g_{2} x$ at $x = 2$ : $y = - 1$ . For $y = lo g_{2} (- x)$ at $x = - 2$ : $y = lo g_{2} 2 = 1$ .

Verify: $- lo g_{2} 2 = - 1$ ✓; $lo g_{2} (- (- 2)) = lo g_{2} 2 = 1$ ✓. See Transformations of graphs.

Example 7 — Cell G: limiting behaviour both ends

Left end $x \to 0^{+}$ . $x = e^{y}$ ; to make $x$ tiny we need $y \to - \infty$ . So $y \to - \infty$ : vertical asymptote $x = 0$ . Why? Only hugely negative powers of $e$ give near-zero values.
Right end $x \to \infty$ . Need $e^{y}$ huge, so $y \to + \infty$ — but slowly. Why slow? Each extra $+ 1$ in $y$ multiplies $x$ by $e \approx 2.718$ ; input grows geometrically, output only linearly (see Natural logarithm ln and e).
$y = 10$ . $x = e^{10} \approx 22026$ . Why? You needed a five-figure input just to reach $y = 10$ — proof of the crawl.

Verify: $e^{10} = 22026.4658 \dots$ ✓, so $ln (22026.4658) = 10$ ✓.

Example 8 — Cell H: real-world word problem (pH)

Plug in. $pH = - lo g_{10} (1 0^{- 2}) = - (- 2) = 2$ . Why the minus sign? Concentrations are tiny ( $< 1$ ), so $lo g$ is negative; the $-$ makes pH a friendly positive number — the whole point of using a log scale (compression, Cell H).
Effect of pH $- 1$ . pH $1$ means $[H^{+}] = 1 0^{- 1} = 0.1$ . Ratio to before: $0.1/0.01 = 10$ . Why? Each unit of pH is one power of $10$ — the graph's "multiply $x$ by $a$ per step" made physical.

Verify: $- lo g_{10} (1 0^{- 2}) = 2$ ✓, and $1 0^{- 1} /1 0^{- 2} = 10$ ✓ (units cancel — a pure ratio).

Example 9 — Cell I: exam twist (intersection + combined transform)

Isolate the log. $lo g_{2} (x - 1) + 3 = 5 \Rightarrow lo g_{2} (x - 1) = 2$ . Why? Undo the outside $+ 3$ first — vertical shift is the outer layer.
Undo the log. $x - 1 = 2^{2} = 4$ . Why this tool? Rewriting $lo g_{2} u = 2$ as $u = 2^{2}$ is the defining move (log ↔ exponential), see Solving equations with logarithms.
Solve. $x = 5$ . Intersection at $(5, 5)$ .
Asymptote & domain. Inside $= 0$ at $x = 1$ , so asymptote $x = 1$ ; domain $x > 1$ . Why? The $+ 3$ shift doesn't move the wall; only the $x - 1$ does.

Verify: $lo g_{2} (5 - 1) + 3 = lo g_{2} 4 + 3 = 2 + 3 = 5$ ✓ — lands exactly on $y = 5$ .

Example 10 — Cell I companion: base you must switch

Rewrite as exponential. $x = 9^{1/2}$ . Why? $lo g_{9} x = \frac{1}{2}$ asks "what input gives output $\frac{1}{2}$ ?" — read straight off $x = a^{y}$ .
Evaluate. $9^{1/2} = 9 = 3$ . So $(3, \frac{1}{2})$ . Why smaller than 9? Output $\frac{1}{2} < 1$ , and outputs below $1$ correspond to inputs below the base $a$ .

Verify: $9^{1/2} = 3$ ✓, and $lo g_{9} 3 = 0.5$ ✓.

Recall Quick self-test (cover the answers)

Which cell handles $lo g (- 3)$ ? ::: Cell C — undefined, input must be $> 0$ . $lo g_{2} (x - 3)$ asymptote? ::: $x = 3$ (Cell D). $lo g_{10} (100 x)$ as a shift? ::: up $2$ (Cell E, since $lo g_{10} 100 = 2$ ). $- lo g_{2} x$ reflects in which axis? ::: the $x$ -axis (Cell F). pH of $[H^{+}] = 1 0^{- 2}$ ? ::: $2$ (Cell H). Solve $lo g_{9} x = \frac{1}{2}$ . ::: $x = 3$ (Cell I).

Connections

Graphs of logarithmic functions — the parent this deep-dive expands.
Laws of logarithms — split $lo g (k x)$ into a vertical shift (Cell E).
Transformations of graphs — shifts and reflections (Cells D, F, I).
Solving equations with logarithms — the intersection twists (Cells I).
Natural logarithm ln and e — the $ln$ limiting behaviour (Cell G).
Inverse functions and reflection in y=x — why $x = a^{y}$ reading works everywhere.
Exponential functions and their graphs — the reflection partner behind every anchor.