Logarithms, Exponents and Surds Practice Questions for CAT: 257+ Solved Questions with Step-by-Step Solutions

Solve 257+ Logarithms, Exponents and Surds practice questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Logarithms, Exponents and Surds

∑

Chapter roadmap

Logarithms, Exponents and Surds

🔐 Logarithmic Equations and Identities

Convert logs, handle domains, compare bases, solve equations, and use log identities in CAT-style traps.

13 direct CAT PYQs | highest weight in this chapter

⚡ Exponential Equations and Inequalities

Bring powers to common bases, substitute expressions, and solve exponential inequalities.

8 direct CAT PYQs

√ Surds, Radicals and Reciprocal Powers

Simplify radicals, nested surds, reciprocal powers, and expressions like $x^{n} + \frac{1}{x ^{n}}$ .

7 direct CAT PYQs

By the end, you should be able to turn complicated powers, logs, and roots into clean algebraic equations.

Topic Hero: Logarithmic Equations and Identities

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Algebra → Logarithms, Exponents and Surds → Topic 1

Logarithmic Equations and Identities

The fastest way to handle hidden powers, changing bases, and tricky domains.

✅ Meaning of logarithm as reverse exponent

✅ Product, quotient, power and change-of-base identities

✅ Log domain and base restrictions

✅ CAT patterns: nested logs, inequalities, AP, and optimization

Logarithms, Exponents and Surds: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability NAT

If $x, y, z$ are positive real numbers satisfying $lo g_{2} x + lo g_{2} y + lo g_{2} z = 6$ , then the minimum possible value of $x + 2 y + 4 z$ is:

Correct Answer:

Step-by-Step Solution

Key idea: This combines log sum-to-product conversion with weighted AM-GM optimization. The log constraint becomes a product constraint, enabling AM-GM with carefully chosen weights to match the linear objective.

Step 1: Convert log sum to product. $lo g_{2} x + lo g_{2} y + lo g_{2} z = lo g_{2} (x y z) = 6 \Rightarrow x y z = 2^{6} = 64$ .

Step 2: Set up weighted AM-GM. We want to minimize $S = x + 2 y + 4 z$ subject to $x y z = 64$ . To apply AM-GM effectively, split terms so that equality condition aligns with the constraint. Try splitting into terms whose product involves $x y z$ :

Consider $x + 2 y + 4 z$ as sum of 3 terms. By AM-GM:

$x + 2 y + 4 z \geq 3 3 x \cdot 2 y \cdot 4 z = 3 3 8 x y z = 3 3 8 \cdot 64 = 33512 = 3 \cdot 8 = 24$

Step 3: Check equality condition. Equality holds when $x = 2 y = 4 z$ . Let $x = k$ , then $y = k /2$ , $z = k /4$ . Product: $k \cdot (k /2) \cdot (k /4) = k^{3} /8 = 64 \Rightarrow k^{3} = 512 \Rightarrow k = 8$ . So $x = 8, y = 4, z = 2$ . Check: $8 + 2 (4) + 4 (2) = 8 + 8 + 8 = 24$ . And $x y z = 8 \cdot 4 \cdot 2 = 64$ . Valid.

Step 4: Confirm minimum. Since AM-GM gives a lower bound and equality is achievable, 24 is indeed the minimum.

Answer: 24

Question 2 · Quantitative Ability NAT

The number of integers $n$ with $1 \leq n \leq 100$ satisfying

$lo g_{n} (2 n) > lo g_{2 n} (n)$

is:

Correct Answer:

Step-by-Step Solution

Key idea: This is a variable base logarithmic inequality requiring case analysis based on whether the base is greater than or less than 1. The bases $n$ and $2 n$ depend on $n$ , so their relationship to 1 changes across the domain.

Step 1: Determine valid domain. For $lo g_{n} (2 n)$ and $lo g_{2 n} (n)$ to be defined:

$n > 0$ , $n \neq = 1$ (base conditions)
$2 n > 0$ , $2 n \neq = 1 \Rightarrow n \neq = 1/2$

Given $1 \leq n \leq 100$ and $n$ integer, exclude $n = 1$ . So $n \in {2, 3, ..., 100}$ . Total candidates: 99.

Step 2: Simplify the inequality. Use change of base to express both sides in terms of $ln n$ :

$lo g_{n} (2 n) = \frac{ln ( 2 n )}{ln n} = 1 + \frac{ln 2}{ln n}$

$lo g_{2 n} (n) = \frac{ln n}{ln ( 2 n )} = \frac{ln n}{ln 2 + ln n}$

Let $u = ln n$ and $c = ln 2 > 0$ . Inequality becomes:

$1 + \frac{c}{u} > \frac{u}{c + u}$

Step 3: Analyze for $n \geq 2$ . Here $u = ln n > 0$ . Since $u > 0$ and $c + u > 0$ , we can multiply by $u (c + u)$ without reversing inequality:

$(u + c) (c + u) > u^{2} \Rightarrow (u + c)^{2} > u^{2} \Rightarrow u^{2} + 2 u c + c^{2} > u^{2} \Rightarrow 2 u c + c^{2} > 0$

Since $u > 0$ and $c > 0$ , this is ALWAYS TRUE.

Step 4: Count valid integers. All integers from 2 to 100 inclusive satisfy the inequality. Count = $100 - 2 + 1 = 99$ .

Answer: 99

Question 3 · Quantitative Ability MCQ

Which expression correctly rewrites $lo g_{a} x$ using a new base $c$ ?

A.
$\frac{l o g _{c} a}{l o g _{c} x}$
B.
$lo g_{c} a - lo g_{c} x$
C.
$(lo g_{c} x) (lo g_{c} a)$
D.
$\frac{l o g _{c} x}{l o g _{c} a}$

Correct Answer:

Step-by-Step Solution

Key idea: this is a change-of-base question, recognisable because one logarithm is being rewritten in terms of a different base.

Step 1: The change-of-base formula says that to rewrite $lo g_{a} x$ in base $c$ , divide the log of the argument by the log of the old base.

Step 2: Apply the formula:

$lo g_{a} x = \frac{lo g _{c} x}{lo g _{c} a} .$

Answer: $\frac{l o g _{c} x}{l o g _{c} a}$ .

Question 4 · Quantitative Ability NAT

The number of real solutions to the equation

$4^{x^{2} - 4 x + 5} - 5 \cdot 2^{x^{2} - 4 x + 5} + 4 = 0$

is:

Correct Answer:

Step-by-Step Solution

Key idea: This is an exponential quadratic in disguise combined with a surds domain/range constraint. The exponent $x^{2} - 4 x + 5$ is not arbitrary — its range restricts which exponential solutions are valid.

Step 1: Substitute to reveal quadratic structure. Let $t = x^{2} - 4 x + 5$ . Note $t \geq 0$ by definition of principal square root. The equation becomes:

$4^{t} - 5 \cdot 2^{t} + 4 = 0 \Rightarrow (2^{t})^{2} - 5 (2^{t}) + 4 = 0$

Step 2: Solve the quadratic in $2^{t}$ . Factor: $(2^{t} - 1) (2^{t} - 4) = 0$ . So $2^{t} = 1$ or $2^{t} = 4$ , giving $t = 0$ or $t = 2$ .

Step 3: Check feasibility against the range of $t$ . Complete the square inside the radical:

$x^{2} - 4 x + 5 = (x - 2)^{2} + 1 \geq 1$

So $t = (x - 2)^{2} + 1 \geq 1 = 1$ . Therefore $t \geq 1$ .

Step 4: Filter solutions. $t = 0$ is rejected because $t \geq 1$ . Only $t = 2$ is valid.

Step 5: Solve for $x$ when $t = 2$ :

$(x - 2)^{2} + 1 = 2 \Rightarrow (x - 2)^{2} + 1 = 4 \Rightarrow (x - 2)^{2} = 3 \Rightarrow x = 2 \pm 3$

Both are real and distinct.

Step 6: Verify no extraneous roots. Both values satisfy the original domain (radicand always positive) and produce $t = 2$ .

Answer: 2

Question 5 · Quantitative Ability MCQ

Assertion (A): If $p = lo g_{2} 5$ and $q = lo g_{2} 3$ , then $lo g_{3} 5 = \frac{p}{q}$ .

Reason (R): The change-of-base rule gives $lo g_{3} 5 = \frac{l o g _{2} 5}{l o g _{2} 3}$ .

A.
A is false, R is true
B.
Both A and R are true, and R is the correct explanation of A
C.
Both A and R are true, but R is not the correct explanation of A
D.
A is true, R is false

Correct Answer:

Step-by-Step Solution

Key idea: this is a change-of-base question, recognisable because a logarithm with one base is being rewritten using logs of another common base.

Step 1: Recall the change-of-base identity:

$lo g_{b} x = \frac{lo g _{c} x}{lo g _{c} b}$

for any valid new base $c$ .

Step 2: Apply it with target base $3$ , argument $5$ , and common base $2$ :

$lo g_{3} 5 = \frac{lo g _{2} 5}{lo g _{2} 3} .$

Step 3: Substitute the given symbols $p = lo g_{2} 5$ and $q = lo g_{2} 3$ :

$lo g_{3} 5 = \frac{p}{q} .$

Step 4: Therefore assertion A is true, reason R is true, and R directly explains A.

Answer: B.

Common trap: reversing the ratio and writing $\frac{q}{p}$ . The argument's log goes in the numerator, and the original base's log goes in the denominator.

Logarithms, Exponents and Surds Practice Questions for CAT: 257+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Logarithms, Exponents and Surds

Logarithms, Exponents and Surds

Topic Hero: Logarithmic Equations and Identities

Logarithmic Equations and Identities

Logarithms, Exponents and Surds: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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