IMAT Mathematics · Equations and Inequalities

Equations with Fractions, Roots, Powers and Logs

Clearing a denominator, squaring a root or substituting for a power turns these equations into ones you can solve, but every answer must then be checked against the original.

All Equations and Inequalities notes

Why this matters

Equations with x in a denominator appeared in 2018 and 2019, an inequality with a square root in 2025, and a count of the integers satisfying a condition with a root in 2011.

Concept 1 of 3: Equations with the unknown in a denominator

Multiplying through by the common denominator clears the fractions, but it also hides the fact that the denominator was never allowed to be zero. So write down the forbidden values first and throw out any answer that matches them.

Definition

State the excluded values: every $x$ that makes a denominator zero.
Multiply every term by the lowest common denominator.
Solve the resulting equation (often a quadratic).
Reject any solution that is an excluded value. If all are rejected, the equation has no solution.

Clearing the denominator

\frac{P(x)}{Q(x)} = R(x) \;\Rightarrow\; P(x) = R(x)\,Q(x), \quad Q(x) \ne 0

Worked example

Solve

\dfrac{2}{x} + \dfrac{3}{x + 1} = 2

Practice this conceptself-check · 4 quick reps

An excluded value can appear as an answer

After clearing denominators, the algebra may produce a value that makes an original denominator zero. It is not a solution. IMAT offers it as an option; the correct choice may be "no solution".

Concept 2 of 3: Equations and inequalities with square roots

Squaring both sides removes the root, but it also accepts solutions of a different equation: squaring

\sqrt{A} = B

gives the same result as squaring

\sqrt{A} = -B

. Since a square root is never negative, any answer that makes the right side negative is false and must be thrown out.

Definition

$\sqrt{A}$ is defined only for $A \ge 0$ , and $\sqrt{A} \ge 0$ always.
Equation $\sqrt{A} = B$ : square to get $A = B^2$ , solve, then check every answer in the original (or keep only those with $B \ge 0$ ).
Inequality $\sqrt{A} < B$ (in outline): you need $A \ge 0$ , $B > 0$ and $A < B^2$ . Squaring is safe only when both sides are non-negative.
Inequality $\sqrt{A} > B$ : if $B < 0$ it holds wherever $A \ge 0$ ; if $B \ge 0$ it needs $A > B^2$ .
Always state the domain ( $A \ge 0$ ) as part of the answer.

Squaring a root equation

\sqrt{A} = B \;\Rightarrow\; A = B^2 \quad\text{(keep only solutions with}\; B \ge 0)

Worked example

Solve

\sqrt{x + 7} = x - 5

Practice this conceptself-check · 4 quick reps

Squaring can add false solutions

Every solution of

\sqrt{A} = B

solves

A = B^2

, but not the other way round. Substitute each answer back into the original equation; any that makes the right-hand side negative is false.

A root inequality has a domain

\sqrt{x} < 3

0 \le x < 9

, not

x < 9

: negative

x

has no square root at all. Options that forget the lower limit are common.

Concept 3 of 3: Exponential and logarithmic equations

An exponential equation that mixes

4^x

and

2^x

is a quadratic in disguise, because

4^x = (2^x)^2

. A logarithmic equation is solved by combining the logs into one and undoing it, but a logarithm only accepts positive inputs, so every answer must be checked. The rules for powers and logarithms are in Numbers, Powers and Logarithms.

Definition

Common base: $a^{f(x)} = a^{g(x)}$ gives $f(x) = g(x)$ .
Substitution: in $a^{2x} + b\,a^x + c = 0$ , put $t = a^x$ and solve the quadratic in $t$ . Keep only $t > 0$ , since $a^x$ is always positive.
Logarithmic equations: first state the domain (every log argument $> 0$ ), then combine the logs with the log laws, undo the log, solve, and reject answers outside the domain.

Matching exponents or logs

a^{f(x)} = a^{g(x)} \;\Rightarrow\; f(x) = g(x) \qquad \log_a f(x) = \log_a g(x) \;\Rightarrow\; f(x) = g(x) > 0

Worked example

Solve

4^x - 6 \cdot 2^x + 8 = 0

Practice this conceptself-check · 4 quick reps

Check the domain of every logarithm

Combining logs can produce an answer that makes an original argument zero or negative, such as

x = -5

\log x

. Such an answer is not a solution. Likewise, in a substitution

t = a^x

, a negative

t

gives no

x

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Equations with the unknown in a denominator
Clearing the denominator
$\frac{P(x)}{Q(x)} = R(x) \;\Rightarrow\; P(x) = R(x)\,Q(x), \quad Q(x) \ne 0$
Equations and inequalities with square roots
Squaring a root equation
$\sqrt{A} = B \;\Rightarrow\; A = B^2 \quad\text{(keep only solutions with}\; B \ge 0)$
Exponential and logarithmic equations
Matching exponents or logs
$a^{f(x)} = a^{g(x)} \;\Rightarrow\; f(x) = g(x) \qquad \log_a f(x) = \log_a g(x) \;\Rightarrow\; f(x) = g(x) > 0$

Watch out for (4)

An excluded value can appear as an answer
→ Equations with the unknown in a denominator
Squaring can add false solutions
→ Equations and inequalities with square roots
A root inequality has a domain
→ Equations and inequalities with square roots
Check the domain of every logarithm
→ Exponential and logarithmic equations

Drill the past-year questions on this topic

Open the bank with this topic pre-filtered.

Open in the bank