Graphs of logarithmic functions: Exercises — Maths deep dive

Throughout, " $lo g_{a} x$ " means "the power you raise $a$ to, to get $x$ ". So $lo g_{a} x = y$ is exactly the same sentence as $x = a^{y}$ . That single equivalence solves almost everything below.

Level 1 — Recognition

These check you can read a log graph's basic features without computation.

Recall Solution L1.1

Answer: (C). Why: The log is the exponential reflected in $y = x$ (see Inverse functions and reflection in y=x). Reflecting swaps every $(p, q)$ to $(q, p)$ :

The exponential's point $(0, 1)$ becomes $(1, 0)$ — so the curve crosses the $x$ -axis at $x = 1$ .
The exponential's horizontal asymptote $y = 0$ becomes a vertical asymptote $x = 0$ .
$a > 1$ means the exponential rises, so its mirror image rises too — but slowly. (A) describes $0 < a < 1$ ; (B) describes an exponential, not a log.

Recall Solution L1.2

Domain: $== x > 0 ==$ . Range: all real numbers. Asymptote: == $x = 0$ == (vertical). Why the domain: $x = 5^{y}$ , and $5^{y}$ is always positive, so $x$ can never be $0$ or negative. Why the range: $y$ (the power) can be any real number, positive or negative, so the output covers everything.

Recall Solution L1.3

$lo g_{3} 1 = 0$ because $3^{0} = 1$ .
$lo g_{3} 3 = 1$ because $3^{1} = 3$ .
$lo g_{3} \frac{1}{3} = - 1$ because $3^{- 1} = \frac{1}{3}$ . These are the three anchor points $(1, 0)$ , $(3, 1)$ , $(\frac{1}{3}, - 1)$ — they pin the whole curve.

Level 2 — Application

Now compute intercepts and solve simple graph equations by rewriting logs as exponentials.

Recall Solution L2.1

Meeting means equal $y$ : set $lo g_{10} x = 3$ . Why rewrite: a log equation $lo g_{a} x = y$ is the same statement as $x = a^{y}$ (see Solving equations with logarithms). $x = 1 0^{3} = 1000.$ Intersection point: $(1000, 3)$ .

Recall Solution L2.2

Asymptote: the curve $lo g_{2} x$ has been shifted right 3 (change inside the bracket moves $x$ ). So the wall at $x = 0$ moves to $== x = 3 ==$ . Domain: $x > 3$ . $x$ -intercept ( $y = 0$ ): $lo g_{2} (x - 3) = 0 \Rightarrow x - 3 = 2^{0} = 1 \Rightarrow x = 4$ . Point $(4, 0)$ . A second easy point: $x - 3 = 2 \Rightarrow x = 5$ gives $y = 1$ , i.e. $(5, 1)$ .

Recall Solution L2.3

Rewrite as $x = 4^{- 1/2}$ . Why a negative fractional power: $4^{- 1/2} = \frac{1}{4 ^{1/2}} = \frac{1}{4} = \frac{1}{2}$ . So $x = \frac{1}{2}$ . (Notice $x$ is still positive, as every log input must be.)

Recall Solution L2.4

$(a, 1)$ on the curve means $lo g_{a} a = 1$ , always true. So if $(7, 1)$ is on it, the base is $a = 7$ . Check: $lo g_{7} 7 = 1$ because $7^{1} = 7$ . ✓

Level 3 — Analysis

Use log laws to rewrite transformed graphs, and reason about slow growth.

Recall Solution L3.1

Split the product using the log law $lo g_{a} (M N) = lo g_{a} M + lo g_{a} N$ (from Laws of logarithms): $lo g_{a} (a^{2} x) = lo g_{a} (a^{2}) + lo g_{a} x = 2 + lo g_{a} x .$ Why $lo g_{a} (a^{2}) = 2$ : it asks "what power of $a$ gives $a^{2}$ ?" → $2$ . The $+ 2$ sits outside the log of $x$ , so it's a vertical shift up 2. A horizontal-looking stretch (multiply inside by $a^{2}$ ) has become a pure vertical shift — that's the power of log laws.

Recall Solution L3.2

$lo g_{10} x = 6 \Rightarrow x = 1 0^{6} = 1000000$ . Comment: to raise $y$ from $1$ to $6$ (just $5$ more) the input exploded from $10$ to a million. Each extra $+ 1$ in $y$ costs a ×10 in $x$ . This is why logs "compress" huge ranges — the curve rises ever more lazily. See the figure.

Figure — Graphs of logarithmic functions

Recall Solution L3.3

The minus is outside, so it reflects the curve in the $x$ -axis (Transformations of graphs). Every $y$ flips sign:

$(1, 0) \to (1, 0)$ (unchanged — $y$ was $0$ ).
$(2, 1) \to (2, - 1)$ .
$(\frac{1}{2}, - 1) \to (\frac{1}{2}, 1)$ . So $y = - lo g_{2} x$ is decreasing: it starts at $+ \infty$ near $x = 0^{+}$ , falls through $(1, 0)$ , and heads to $- \infty$ slowly. Same vertical asymptote $x = 0$ , same domain $x > 0$ .

Level 4 — Synthesis

Combine several transformations, or work backwards from features to an equation.

Recall Solution L4.1

Asymptote $x = 2$ ⟹ inside change is $(x - 2)$ , so $b = 2$ .
$(3, 0)$ : $lo g_{a} (3 - 2) = lo g_{a} 1 = 0$ . ✓ for any base (this just confirms the shift is right).
$(5, 1)$ : $lo g_{a} (5 - 2) = lo g_{a} 3 = 1 \Rightarrow a = 3$ . Equation: $y = lo g_{3} (x - 2)$ . Check $(5, 1)$ : $lo g_{3} 3 = 1$ . ✓ Domain: $x > 2$ .

Recall Solution L4.2

$lo g_{5} (25 x) = lo g_{5} 25 + lo g_{5} x = 2 + lo g_{5} x,$ since $lo g_{5} 25 = 2$ (as $5^{2} = 25$ ). It's $y = lo g_{5} x$ shifted up 2. At $x = 1$ : $lo g_{5} 1 + 2 = 0 + 2 = 2$ . So the point $(1, 2)$ .

Recall Solution L4.3

Yes — identical. Using log laws: $lo g_{10} (10 x) = lo g_{10} 10 + lo g_{10} x = 1 + lo g_{10} x = lo g_{10} x + 1$ . Why this matters: a horizontal-looking rescale (multiply input by $10$ ) and a vertical shift (add $1$ ) are the same transformation for log graphs. Both pass through $(1, 1)$ and $(10, 2)$ , asymptote $x = 0$ .

Recall Solution L4.4

Set equal: $lo g_{2} x = lo g_{2} (x - 3) + 2$ . Move the $2$ across and write it as a log: $2 = lo g_{2} 4$ , so $lo g_{2} x - lo g_{2} (x - 3) = 2 \Rightarrow lo g_{2} (\frac{x}{x - 3}) = 2.$ Why divide inside: subtracting logs = log of a quotient (Laws of logarithms). Rewrite as exponential: $\frac{x}{x - 3} = 2^{2} = 4$ . $x = 4 (x - 3) = 4 x - 12 \Rightarrow 3 x = 12 \Rightarrow x = 4$ . Then $y = lo g_{2} 4 = 2$ . Check domain: both need $x > 3$ ; $x = 4$ is fine. Intersection $(4, 2)$ .

Level 5 — Mastery

Full modelling and multi-step reasoning that ties graph, laws, and inverse together.

Recall Solution L5.1

(a) $lo g_{a} 8 = 3$ means $a^{3} = 8 \Rightarrow a = 2$ . (b) $lo g_{2} x = - 2 \Rightarrow x = 2^{- 2} = \frac{1}{4}$ . Point $(\frac{1}{4}, - 2)$ . (c) Reflecting a log in $y = x$ undoes the inverse — it gives back the exponential: $y = 2^{x}$ . (Reflecting swaps $x$ and $y$ : $y = lo g_{2} x ⟺ x = lo g_{2} y ⟺ y = 2^{x}$ .)

Recall Solution L5.2

(a) $I = I_{0} \Rightarrow I / I_{0} = 1 \Rightarrow lo g_{10} 1 = 0 \Rightarrow L = 0$ . The reference is the "zero" of the scale. (b) We want $L$ up by $10$ : $10 lo g_{10} (I / I_{0})$ must gain $10$ , so $lo g_{10} (I / I_{0})$ gains $1$ , meaning $I / I_{0}$ is multiplied by $10$ . Answer: $\times 10$ . Why: adding $1$ to a base- $10$ log means multiplying the input by $10$ — the slow-growth fact turned into a design rule. (c) Let $u = lo g_{10} (I / I_{0})$ . Then $L = 10 u$ : a straight line through the origin with slope $10$ . Plotting against the log flattens the exponential range into a line — the whole point of a log scale.

Recall Solution L5.3

Combine with the product law: $lo g_{3} (x (x - 6)) = 3$ . Rewrite as exponential: $x (x - 6) = 3^{3} = 27$ . $x^{2} - 6 x - 27 = 0 \Rightarrow (x - 9) (x + 3) = 0 \Rightarrow x = 9 or x = - 3.$ Domain check: need $x > 0$ and $x - 6 > 0$ , i.e. $x > 6$ . So reject $x = - 3$ (and it would fail anyway). Only $x = 9$ survives. Verify: $lo g_{3} 9 + lo g_{3} 3 = 2 + 1 = 3$ . ✓

Recall Solution L5.4

The exponential $y = a^{x}$ ( $a > 1$ ) is strictly increasing with no turning point — it just keeps climbing. A reflection in $y = x$ preserves "strictly monotonic" (it only swaps the axes' roles), so $y = lo g_{a} x$ is strictly increasing everywhere on $x > 0$ . A strictly increasing function has no highest value: for any $x$ you pick, $lo g_{a} (a x) = 1 + lo g_{a} x$ is larger. Hence no maximum — it rises without bound, just slowly.

Active recall

Rewrite

lo g_{a} x = y

as an exponential statement.

x = a^{y}

Inside change

lo g_{a} (x - b)

moves the graph and asymptote how?

Right by

b

; asymptote goes to

x = b

Outside change

lo g_{a} x + c

does what?

Shifts the whole curve up by

c

(asymptote unchanged).

lo g_{a} (a^{2} x)

equals which vertical shift?

lo g_{a} x + 2

Why must you domain-check solutions of log equations?

Every log argument must be

> 0

; algebra can produce roots that break this.

Reflecting

y = lo g_{a} x

y = x

gives what?

y = a^{x}

(the exponential — the inverse undone).

Adding 1 to a base-10 log input requires multiplying

x

by what?

10.

Connections

Graphs of logarithmic functions — the parent topic these exercises drill.
Solving equations with logarithms — the rewrite $lo g_{a} x = y ⟺ x = a^{y}$ used throughout.
Laws of logarithms — product/quotient laws in L3–L5.
Transformations of graphs — shifts and reflections in L2–L4.
Inverse functions and reflection in y=x — reflection reasoning in L5.
Exponential functions and their graphs — the reflection partner.
Natural logarithm ln and e — the base- $e$ special case.