Solving logarithmic equations: Worked examples

The scenario matrix

Every log-equation problem falls into one of these cells. We will hit all of them below.

#	Cell (scenario)	Trap it hides	Example
A	Single isolated log $=$ number	none — warm-up	Ex 1
B	Two logs added, $=$ number → quadratic, both roots survive	forget domain?	Ex 2
C	Two logs, one root is extraneous (makes argument $\leq 0$ )	reject the bad root	Ex 3
D	Logs on both sides, same base → equate arguments	quadratic, sign check	Ex 4
E	Different bases → change of base needed	can't combine directly	Ex 5
F	Hidden quadratic in the log (substitution)	it's quadratic in $u = lo g x$ , not $x$	Ex 6
G	Degenerate / no-solution case (empty domain)	answer set is empty	Ex 7
H	Real-world word problem (units, meaning of the number)	interpret + round sensibly	Ex 8
I	Exam twist: log of a log / unknown appears twice awkwardly	peel outer log first	Ex 9

Prerequisites feeding this page: Laws of logarithms, Exponential equations, Change of base formula, Quadratic equations by substitution, Domain and range of log functions, Inverse functions.

Before we start, one picture reminds us what "log" even is and why we must always check the domain.

Figure s01 — Log and exp are mirror inverses. The yellow curve $y = 2^{x}$ never dips below the horizontal axis: a positive base raised to any power stays positive. The blue curve $y = lo g_{2} x$ is its mirror image across the pink dashed line $y = x$ , and it only exists to the right of the vertical axis. The shaded pink strip ( $x \leq 0$ ) is the forbidden zone — you can never feed a log a zero or negative argument. This is the visual root of every domain check below.

Cell A — single isolated log

Cell B — two logs added, $=$ number, both roots survive

Worked example Example 2 (Cell B)

Solve $lo g_{3} (x) + lo g_{3} (2 x + 3) = lo g_{3} (2)$ .

Forecast: two logs added, equal to a single number ( $lo g_{3} 2$ is a constant). Combining gives a quadratic — bet now whether both its roots survive the domain, or only one.

Step 1. Combine the left with the product law: $lo g_{3} (x (2 x + 3)) = lo g_{3} 2$ . Why this step? Two separate logs can't be undone; one log can. See Laws of logarithms.

Step 2. Both sides are now single base-3 logs → equate arguments: $x (2 x + 3) = 2$ . Why this step? $lo g$ is one-to-one, so equal logs force equal arguments.

Step 3. Expand and solve: $2 x^{2} + 3 x - 2 = 0 \Rightarrow (2 x - 1) (x + 2) = 0$ , so $x = \frac{1}{2}$ or $x = - 2$ . Why this step? Standard quadratic factorisation.

Step 4 (full domain check). We need both original arguments positive: $x > 0$ and $2 x + 3 > 0$ (i.e. $x > - \frac{3}{2}$ ).

$x = \frac{1}{2}$ : $x = \frac{1}{2} > 0$ ✓ and $2 (\frac{1}{2}) + 3 = 4 > 0$ ✓ — keep.
$x = - 2$ : $x = - 2 < 0$ ✗ → $lo g_{3} (- 2)$ undefined — reject. Why this step? Every candidate must make all original log arguments $> 0$ ; checking one is not enough.

Verify: $lo g_{3} (\frac{1}{2}) + lo g_{3} (4) = lo g_{3} (\frac{1}{2} \cdot 4) = lo g_{3} 2$ ✓. $x = \frac{1}{2}$

Cell C — extraneous root must be rejected

This is the most-tested trap. We picture why a root dies below.

Figure s02 — Why the extraneous root $- 3$ dies. Each straight line is one of the two log arguments from Example 3: blue is $x - 1$ , yellow is $x + 2$ . A log only exists where its argument sits above the horizontal axis. Both lines are positive only inside the pink band $x > 1$ . The candidate $x = 2$ (dashed white) lands safely inside the band, so it is kept; the candidate $x = - 3$ (pink cross) lies far to the left where $x - 1 = - 4 < 0$ , so $lo g_{2} (- 4)$ is undefined and the root is rejected.

Worked example Example 3 (Cell C)

Solve $lo g_{2} (x - 1) + lo g_{2} (x + 2) = 2$ .

Forecast: you'll get a quadratic with two roots. Bet now: do BOTH survive, or does one die? Look at the domain first — we need $x - 1 > 0$ and $x + 2 > 0$ , so $x > 1$ .

Step 1. Combine with the product law: $lo g_{2} ((x - 1) (x + 2)) = 2$ . Why this step? Two separate logs can't be undone; one log can. See Laws of logarithms.

Step 2. Exponentiate: $(x - 1) (x + 2) = 2^{2} = 4$ . Why this step? $lo g_{2} N = 2 ⟺ N = 4$ .

Step 3. Expand and solve: $x^{2} + x - 2 = 4 \Rightarrow x^{2} + x - 6 = 0 \Rightarrow (x + 3) (x - 2) = 0$ , so $x = - 3$ or $x = 2$ . Why this step? Standard quadratic factorisation.

Step 4. Test against $x > 1$ :

$x = 2$ : $x - 1 = 1 > 0$ , $x + 2 = 4 > 0$ ✓ — keep.
$x = - 3$ : $x - 1 = - 4 < 0$ → $lo g_{2} (- 4)$ undefined ✗ — reject (extraneous root, born when we multiplied the two arguments together). Why this step? Every candidate must make all original arguments positive.

Verify: $lo g_{2} 1 + lo g_{2} 4 = 0 + 2 = 2$ ✓. $x = 2$

Cell D — logs on both sides, same base

Cell E — different bases (change of base)

Worked example Example 5 (Cell E)

Solve $lo g_{2} (x) = lo g_{4} (x) + 3$ .

Forecast: two different bases — you cannot combine them yet. What one tool converts them to a common base?

Step 1. Convert $lo g_{4} x$ to base 2 using the Change of base formula $lo g_{4} x = \frac{lo g _{2} x}{lo g _{2} 4} = \frac{lo g _{2} x}{2}$ . Why this step? $lo g_{2} 4 = 2$ , so this rewrites the base-4 log in base 2. Now everything speaks the same language.

Step 2. Let $u = lo g_{2} x$ : the equation becomes $u = \frac{u}{2} + 3$ . Why this step? A substitution turns a log equation into plain algebra (same trick as Quadratic equations by substitution, here linear).

Step 3. Solve: $u - \frac{u}{2} = 3 \Rightarrow \frac{u}{2} = 3 \Rightarrow u = 6$ . Why this step? Isolate $u$ .

Step 4. Back-substitute: $lo g_{2} x = 6 \Rightarrow x = 2^{6} = 64$ . Why this step? Undo the substitution and exponentiate.

Verify: $lo g_{2} 64 = 6$ ; $lo g_{4} 64 = 3$ (since $4^{3} = 64$ ); $3 + 3 = 6$ ✓. Domain $x = 64 > 0$ ✓. $x = 64$

Cell F — hidden quadratic in the log

Cell G — degenerate case, no solution

Worked example Example 7 (Cell G)

Solve $lo g_{2} (x) + lo g_{2} (x - 2) = lo g_{2} (- 3)$ .

Forecast: look at the right-hand side before doing any algebra. Notice anything fatal?

Step 1. First state the left-hand domain: for $lo g_{2} (x)$ and $lo g_{2} (x - 2)$ to exist we need $x > 0$ and $x - 2 > 0$ , i.e. $x > 2$ . So any solution must live in $x > 2$ . Why this step? We must know where the LHS is even defined before comparing sides.

Step 2. Now inspect the RHS: $lo g_{2} (- 3)$ . Why this step? A log of a negative number does not exist (a positive base can never give a negative output). The equation is broken on the right, independent of $x$ .

Step 3. Since the RHS is undefined for every $x$ — including the allowed region $x > 2$ — there is no value making both sides defined and equal. Why this step? "Solving" means both sides defined and equal — impossible here.

Verify: $2^{y} = - 3$ has no real solution for any $y$ , so $lo g_{2} (- 3)$ genuinely has no value ✓. $No solution$

Cell H — real-world word problem

Worked example Example 8 (Cell H)

A radioactive sample decays so its mass (in grams) after $t$ days is $M = 200 \times 2^{- t /5}$ . After how many days is the mass $25 g$ ? Give $t$ in days.

Forecast: the unknown $t$ is trapped in an exponent. What undoes an exponent?

Step 1. Set up: $200 \times 2^{- t /5} = 25$ . Why this step? Translate "mass is $25$ g" into the model.

Step 2. Isolate the power: $2^{- t /5} = \frac{25}{200} = \frac{1}{8}$ . Why this step? We must get the base-and-power alone before logging (this is an exponential equation).

Step 3. Recognise $\frac{1}{8} = 2^{- 3}$ , so $2^{- t /5} = 2^{- 3}$ . Equal bases → equal exponents: $- \frac{t}{5} = - 3$ . Why this step? $2^{a} = 2^{b} \Rightarrow a = b$ (the exponential is one-to-one — the mirror image of the log rule).

Step 4. Solve: $t = 15$ days. Why this step? Simple rearrangement. Units: $t$ measures days, matching the model.

Verify: $200 \times 2^{- 15/5} = 200 \times 2^{- 3} = 200 \times \frac{1}{8} = 25$ g ✓. $t = 15 days$

Cell I — exam twist: a log of a log

Recall Which cell is this? (quick self-test)

$lo g_{5} x + lo g_{5} (x - 4) = 1$ — which cell and what survives? ::: Cell C — combine to a quadratic, roots $5$ and $- 1$ ; reject $- 1$ , keep $x = 5$ . $lo g_{2} (x) = lo g_{8} (x) + 4$ — which move? ::: Cell E — change of base ( $lo g_{8} x = \frac{l o g _{2} x}{3}$ ), then solve. $lo g_{3} (lo g_{2} x) = 0$ — first move? ::: Cell I — outer first: $lo g_{2} x = 3^{0} = 1$ , so $x = 2$ . $lo g_{7} (x) = lo g_{7} (- x^{2} - 1)$ — answer? ::: Cell G — RHS argument $- x^{2} - 1 < 0$ always, so no solution.

Connections

Laws of logarithms
Exponential equations
Change of base formula
Inverse functions
Quadratic equations by substitution
Domain and range of log functions