Solving logarithmic equations: Exercises — Maths deep dive

Intuition How to use this page

Work top to bottom. Each problem sits at a difficulty level — from L1 Recognition (spot the move) up to L5 Mastery (chain several moves together). Try each problem before opening its solution. Every solution is hidden inside a collapsible box, so this page doubles as a self-test.

Three rules never change, no matter the level:

A logarithm asks a question: $lo g_{b} (N) = c$ means " $b$ to what power gives $N$ ?" — and its answer, rewritten, is $N = b^{c}$ .
Every log's argument (the thing inside) must be $> 0$ . Always check at the end.
Every log's base must satisfy $b > 0$ and $b \neq = 1$ (a base of $1$ can't grow — $1^{c} = 1$ always — and negative bases don't give a consistent curve). All problems below use standard bases like $2, 3, 5, 9, 10$ , so this holds automatically, but it is part of what "log" even means. See the parent note Solving logarithmic equations for the full theory.

Before we start, one picture to keep in mind: why some answers must be thrown away. The log graph only lives to the right of $x = 0$ .

Look at the figure carefully — we will refer back to it in the first few solutions:

The blue curve $y = lo g_{2} x$ never touches or crosses the red dashed line $x = 0$ . There is simply no point on the curve with a zero or negative input.
The green dot at $x = 4$ is a valid input: it lands on the curve, at height $lo g_{2} 4 = 2$ . Domain check passes.
The red cross at $x = - 0.6$ is a candidate the algebra might hand you — but it is nowhere on the curve. It is an extraneous root and gets rejected.

This single picture is the "domain check" in visual form: any candidate that falls in the shaded red zone ( $x \leq 0$ ) inside a log is off the curve and must go.

Level 1 — Recognition

You just have to spot the single move that undoes the log.

Recall Solution Q1

Step 1 — What/why. The log is already alone on one side. The move to undo it is exponentiate: rewrite the question " $4$ to what power is $x$ ?" as its answer. $x = 4^{3} = 64$ Step 2 — Check domain. Argument is $x = 64 > 0$ ✓ — well to the right of the red line in the figure. $x = 64$

Recall Solution Q2

Step 1. One log, already isolated → exponentiate. $lo g_{b} N = c ⟺ N = b^{c}$ . $2 x + 1 = 3^{2} = 9$ Step 2. $2 x = 8 \Rightarrow x = 4$ . Step 3 — Check. Argument $= 2 (4) + 1 = 9 > 0$ ✓ (a valid input, like the green dot in the figure). $x = 4$

Recall Solution Q3

Step 1 — What/why. Both sides are single logs of the same base $2$ . Because $lo g$ is one-to-one (it never gives the same output twice), equal outputs force equal inputs. $x - 5 = 11$ Step 2. $x = 16$ . Step 3 — Check. Argument $= 16 - 5 = 11 > 0$ ✓. $x = 16$

Level 2 — Application

Now you must combine logs (recall Laws of logarithms) before you can undo them.

Recall Solution Q4

Step 1 — Combine. Two separate logs can't be undone; the product law $lo g M + lo g N = lo g (M N)$ collapses them into one. $lo g_{2} (x (x - 2)) = 3$ Step 2 — Exponentiate. $lo g_{2} N = 3 ⟺ N = 2^{3} = 8$ . $x (x - 2) = 8 \Rightarrow x^{2} - 2 x - 8 = 0$ Step 3 — Solve. $(x - 4) (x + 2) = 0 \Rightarrow x = 4$ or $x = - 2$ . Step 4 — Check (domain). Need $x > 0$ and $x - 2 > 0$ , i.e. $x > 2$ .

$x = 4$ : $4 > 0$ and $2 > 0$ ✓
$x = - 2$ : $lo g_{2} (- 2)$ undefined ✗ — this is the red-cross case in the figure, reject. $x = 4$

Recall Solution Q5

Step 1 — Combine. Quotient law: $lo g M - lo g N = lo g \frac{M}{N}$ . $lo g_{5} (\frac{x + 3}{x - 1}) = 1$ Step 2 — Exponentiate. $lo g_{5} N = 1 ⟺ N = 5$ . $\frac{x + 3}{x - 1} = 5$ Step 3 — Solve. $x + 3 = 5 (x - 1) = 5 x - 5 \Rightarrow 8 = 4 x \Rightarrow x = 2$ . Step 4 — Check (domain). $x + 3 = 5 > 0$ and $x - 1 = 1 > 0$ ✓. $x = 2$

Recall Solution Q6

Step 1. Same base both sides → equate arguments (log is one-to-one). $x = 6 - x$ Step 2. $2 x = 6 \Rightarrow x = 3$ . Step 3 — Check (domain). $x = 3 > 0$ and $6 - x = 3 > 0$ ✓. $x = 3$

Level 3 — Analysis

Here the structure is disguised: a hidden quadratic, or a base you must change.

Recall Solution Q7

Step 1 — Substitute. The equation is quadratic in the log, not in $x$ . Let $u = lo g_{3} x$ (see Quadratic equations by substitution). $u^{2} - 4 u + 3 = 0$ Step 2 — Solve for $u$ . $(u - 1) (u - 3) = 0 \Rightarrow u = 1$ or $u = 3$ . Step 3 — Back-substitute.

$lo g_{3} x = 1 \Rightarrow x = 3^{1} = 3$
$lo g_{3} x = 3 \Rightarrow x = 3^{3} = 27$ Step 4 — Check (domain). Both $3 > 0$ and $27 > 0$ ✓. $x = 3 or x = 27$

Recall Solution Q8

Step 1 — Why we must act. The two logs have different bases ( $9$ and $3$ ), so we can't equate arguments yet. Convert one to the other with the Change of base formula, or notice $9 = 3^{2}$ . Step 2 — Rewrite $lo g_{9} x$ in base 3. Since $9 = 3^{2}$ , $lo g_{9} x = \frac{l o g _{3} x}{l o g _{3} 9} = \frac{l o g _{3} x}{2} .$ So the equation is $\frac{lo g _{3} x}{2} = lo g_{3} 4$ , giving $lo g_{3} x = 2 lo g_{3} 4$ . Step 3 — Power law. $2 lo g_{3} 4 = lo g_{3} (4^{2}) = lo g_{3} 16$ . $lo g_{3} x = lo g_{3} 16 \Rightarrow x = 16.$ Step 4 — Check (domain). $x = 16 > 0$ ✓. $x = 16$

Recall Solution Q9

Step 1 — Unify the base. We cannot add logs of different bases. Since $4 = 2^{2}$ , change $lo g_{4} x$ to base 2: $lo g_{4} x = \frac{l o g _{2} x}{l o g _{2} 4} = \frac{l o g _{2} x}{2} .$ Step 2 — Substitute. Let $u = lo g_{2} x$ : $u + \frac{u}{2} = 3 \Rightarrow \frac{3 u}{2} = 3 \Rightarrow u = 2.$ Step 3 — Back-substitute. $lo g_{2} x = 2 \Rightarrow x = 2^{2} = 4$ . Step 4 — Check (domain). $x = 4 > 0$ ✓. $x = 4$

Level 4 — Synthesis

Chain several ideas: combine, exponentiate to a quadratic, then filter by domain.

Recall Solution Q10

Step 1 — Combine. Product law: $lo g_{3} (x (x + 6)) = 3$ . Step 2 — Exponentiate. $lo g_{3} N = 3 ⟺ N = 3^{3} = 27$ . $x (x + 6) = 27 \Rightarrow x^{2} + 6 x - 27 = 0$ Step 3 — Solve. $(x + 9) (x - 3) = 0 \Rightarrow x = - 9$ or $x = 3$ . Step 4 — Check (domain). Need $x > 0$ and $x + 6 > 0$ , i.e. $x > 0$ .

$x = 3$ : $3 > 0$ and $9 > 0$ ✓
$x = - 9$ : $lo g_{3} (- 9)$ undefined ✗ — reject. $x = 3$

Recall Solution Q11

Step 1 — Power law first. $2 lo g_{2} x = lo g_{2} (x^{2})$ . Now the left side is $lo g_{2} (x^{2}) - lo g_{2} (x + 3)$ . Step 2 — Quotient law. $lo g_{2} (\frac{x ^{2}}{x + 3}) = 2$ . Step 3 — Exponentiate. $= 2^{2} = 4$ . $\frac{x ^{2}}{x + 3} = 4 \Rightarrow x^{2} = 4 (x + 3) = 4 x + 12$ Step 4 — Solve. $x^{2} - 4 x - 12 = 0 \Rightarrow (x - 6) (x + 2) = 0 \Rightarrow x = 6 or x = - 2.$ Step 5 — Check (domain). Need $x > 0$ (from $lo g_{2} x$ ) and $x + 3 > 0$ . So $x > 0$ .

$x = 6$ : $6 > 0$ , $9 > 0$ ✓
$x = - 2$ : $lo g_{2} (- 2)$ undefined ✗ — reject. $x = 6$

Recall Solution Q12

Step 1 — Combine. $lo g_{10} (x (x - 3)) = 1$ . Step 2 — Exponentiate. $= 1 0^{1} = 10$ . $x (x - 3) = 10$ Step 3 — Solve. $x^{2} - 3 x - 10 = 0 \Rightarrow (x - 5) (x + 2) = 0 \Rightarrow x = 5 or x = - 2.$ Step 4 — Check (domain). Need $x > 0$ and $x - 3 > 0$ , i.e. $x > 3$ .

$x = 5$ : $5 > 0$ , $2 > 0$ ✓
$x = - 2$ : undefined ✗ — reject. $x = 5$

Level 5 — Mastery

Full-strength: exponential–log mixtures, and cases where all candidates may survive or die.

Recall Solution Q13

Step 1 — Combine & exponentiate. $lo g_{2} (x (x - 6)) = 4 \Rightarrow x (x - 6) = 2^{4} = 16$ . Step 2 — Solve. $x^{2} - 6 x - 16 = 0 \Rightarrow (x - 8) (x + 2) = 0 \Rightarrow x = 8 or x = - 2.$ Step 3 — Check (domain). Domain needs $x > 0$ and $x - 6 > 0$ , i.e. $x > 6$ .

$x = 8$ : $8 > 6$ ✓
$x = - 2$ : fails $x > 0$ ✗ — reject. Why one survives: the tighter of the two conditions is $x > 6$ ; only $x = 8$ clears that bar, so the quadratic's second (negative) root is always extraneous here. $x = 8$

Recall Solution Q14

Step 1 — Simplify the log equation. Quotient law: $lo g_{2} \frac{x}{y} = 1 \Rightarrow \frac{x}{y} = 2^{1} = 2 \Rightarrow x = 2 y$ . Step 2 — Substitute into the linear equation. $2 y + y = 12 \Rightarrow 3 y = 12 \Rightarrow y = 4$ , hence $x = 8$ . Step 3 — Check (domain). Both $x = 8 > 0$ and $y = 4 > 0$ ✓ (needed since both appear inside logs). And $x + y = 12$ ✓ (algebra check). $x = 8, y = 4$

Recall Solution Q15

Step 1 — Substitute. This is quadratic in $2^{x}$ (see Exponential equations). Let $u = 2^{x}$ ; note $2^{2 x} = (2^{x})^{2} = u^{2}$ . $u^{2} - 5 u + 4 = 0 \Rightarrow (u - 1) (u - 4) = 0 \Rightarrow u = 1 or u = 4.$ Step 2 — Undo the exponential with a log. We now need $2^{x} = 1$ and $2^{x} = 4$ . To free $x$ from the exponent we take $lo g_{2}$ (the inverse of $2^{(\cdot)}$ ).

$2^{x} = 1 \Rightarrow x = lo g_{2} 1 = 0$
$2^{x} = 4 \Rightarrow x = lo g_{2} 4 = 2$ Step 3 — Check (algebra, not domain). Here $x$ is not inside a log, so there's no positivity restriction — instead do the algebra check by substituting back: at $x = 0$ , $2^{0} - 5 \cdot 2^{0} + 4 = 1 - 5 + 4 = 0$ ✓; at $x = 2$ , $2^{4} - 5 \cdot 2^{2} + 4 = 16 - 20 + 4 = 0$ ✓. Both valid (any real $x$ is a legal exponent). $x = 0 or x = 2$

Connections

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