Logarithms, Exponents and Surds Previous Year Questions (PYQs) for CAT: 20+ Solved Questions with Step-by-Step Solutions

Solve 20+ Logarithms, Exponents and Surds previous year 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

Selected Topic

log

Reverse the power

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 MCQ

The sum of all possible values of x satisfying the equation $2^{4 x^{2}} - 2^{2 x^{2} + x + 16} + 2^{2 x + 30} = 0$ , is

A.
3
B.
$\frac{3}{2}$
C.
$\frac{5}{2}$
D.
$\frac{1}{2}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a quadratic-in-disguise exponential equation. Recognisable because the exponents are related algebraically and can form a perfect square.

Step 1: Analyze the exponents to find a common substitution.

The given equation is:

$2^{4 x^{2}} - 2^{2 x^{2} + x + 16} + 2^{2 x + 30} = 0$

Notice the relationships between the powers. Let $u = 2^{2 x^{2}}$ and $v = 2^{x + 15}$ .

Then:

$u^{2} = (2^{2 x^{2}})^{2} = 2^{4 x^{2}}$

$u \cdot v = 2^{2 x^{2}} \cdot 2^{x + 15} = 2^{2 x^{2} + x + 15}$

$v^{2} = (2^{x + 15})^{2} = 2^{2 x + 30}$

Step 2: Rewrite the equation using $u$ and $v$ .

The middle term is $2^{2 x^{2} + x + 16} = 2^{1} \cdot 2^{2 x^{2} + x + 15} = 2 uv$ .

Substitute these into the original equation:

$u^{2} - 2 uv + v^{2} = 0$

Step 3: Solve the algebraic equation.

This is a perfect square:

$(u - v)^{2} = 0 ⟹ u = v$

Step 4: Equate the original exponential expressions.

$2^{2 x^{2}} = 2^{x + 15}$

Since the bases are equal, the exponents must be equal:

$2 x^{2} = x + 15 ⟹ 2 x^{2} - x - 15 = 0$

Step 5: Find the sum of all possible values of $x$ .

By Vieta's formulas, the sum of the roots of $A x^{2} + B x + C = 0$ is $- B / A$ .

Sum of roots = $- \frac{- 1}{2} = \frac{1}{2}$ .

Answer: 1/2

Question 2 · Quantitative Ability NAT

If a, b and c are positive real numbers such that $a > 10 \geq b \geq c$ and $\frac{l o g _{8} ( a + b )}{l o g _{2} c} + \frac{l o g _{27} ( a - b )}{l o g _{3} c} = \frac{2}{3}$ , then the greatest possible integer value of a is

Correct Answer:

Step-by-Step Solution

This is a change-of-base logarithm question, recognisable because the bases of the logarithms in the numerators (8 and 27) are powers of the bases in the denominators (2 and 3).

Step 1: Apply the change-of-base formula to each term.

Recall that \ $\log_{b^k}(M) = \frac{\log_b(M)}{k}\$ . Therefore:

\ $\log_8(a+b) = \frac{\log_2(a+b)}{\log_2 8} = \frac{\log_2(a+b)}{3}\$

\ $\log_{27}(a-b) = \frac{\log_3(a-b)}{\log_3 27} = \frac{\log_3(a-b)}{3}\$

Step 2: Substitute into the given equation.

\ $\frac{\log_2(a+b)/3}{\log_2 c} + \frac{\log_3(a-b)/3}{\log_3 c} = \frac{2}{3}\$

Factor out \ $\frac{1}{3}\$ :

\ $\frac{1}{3}\left[\frac{\log_2(a+b)}{\log_2 c} + \frac{\log_3(a-b)}{\log_3 c}\right] = \frac{2}{3}\$

Multiply both sides by 3:

\ $\frac{\log_2(a+b)}{\log_2 c} + \frac{\log_3(a-b)}{\log_3 c} = 2\$

Step 3: Convert back to single-base logarithms.

Using \ $\frac{\log_b M}{\log_b N} = \log_N M\$ :

\ $\log_c(a+b) + \log_c(a-b) = 2\$

By log product rule:

\ $\log_c[(a+b)(a-b)] = 2 \implies \log_c(a^2 - b^2) = 2\$

Converting from log form to exponential form:

\ $a^2 - b^2 = c^2 \implies a^2 = b^2 + c^2\$

Step 4: Maximize \ $a\$ subject to the constraints.

Given \ $a > 10 \geq b \geq c > 0\$ , to maximize \ $a\$ we maximize \ $b^2 + c^2\$ .

The maximum occurs when \ $b = 10\$ and \ $c = 10\$ (since \ $c \leq b \leq 10\$ ):

\ $a^2 \leq 10^2 + 10^2 = 200\$

\ $a \leq \sqrt{200} = 10\sqrt{2} \approx 14.142\$

Step 5: Find the greatest integer value.

The greatest integer \ $\leq 14.142\$ is 14. Verify that \ $a=14\$ is achievable: if \ $b=10\$ , then \ $c^2 = 196 - 100 = 96\$ , so \ $c = 4\sqrt{6} \approx 9.8\$ . This satisfies \ $10 \geq 9.8 > 0\$ and \ $c \neq 1\$ , so the logs are defined.

Answer: 14

Question 3 · Quantitative Ability MCQ

If $lo g_{64} x^{2} + lo g_{8} y + 3 lo g_{512} (y z) = 4$ , where x,y and z are positive real numbers, then the minimum possible value of $(x + y + z)$ is

A.
48
B.
36
C.
24
D.
96

Correct Answer:

Step-by-Step Solution

This is a log-to-single-constraint optimization problem: recognize it because every log term can be converted to the same base, collecting everything into one equation of the form $($ power $) lo g_{2} x + ($ power $) lo g_{2} y + ($ power $) lo g_{2} z =$ constant — after which AM-GM finds the minimum sum. (Note: inside the third log, $y$ multiplies $z$ , i.e. $y \cdot z$ .)

Step 1 — Convert every log to base 2. Using $64 = 2^{6}, 8 = 2^{3}, 512 = 2^{9}$ :

$lo g_{64} x^{2} = \frac{2 lo g _{2} x}{6} = \frac{lo g _{2} x}{3}$

$lo g_{8} y = \frac{\frac{1}{2} lo g _{2} y}{3} = \frac{lo g _{2} y}{6}$

$3 lo g_{512} (y \cdot z) = 3 \cdot \frac{\frac{1}{2} lo g _{2} y + lo g _{2} z}{9} = \frac{lo g _{2} y}{6} + \frac{lo g _{2} z}{3}$

Step 2 — Add all three terms.

$\frac{lo g _{2} x}{3} + \frac{lo g _{2} y}{6} + \frac{lo g _{2} y}{6} + \frac{lo g _{2} z}{3} = \frac{lo g _{2} x}{3} + \frac{lo g _{2} y}{3} + \frac{lo g _{2} z}{3} = 4$

Step 3 — Multiply through by 3.

$lo g_{2} x + lo g_{2} y + lo g_{2} z = 12 \Rightarrow lo g_{2} (x y z) = 12 \Rightarrow x y z = 2^{12} = 4096$

Step 4 — Minimize $x + y + z$ given $x y z = 4096$ . By AM-GM for positive reals:

$\frac{x + y + z}{3} \geq 3 x y z = 34096 = 16 \Rightarrow x + y + z \geq 48$

with equality when $x = y = z = 16$ .

Step 5 — Verify. $16 \times 16 \times 16 = 4096$ ✓. So the minimum value of $x + y + z$ is $48$ .

Trap to avoid: miscombining the two separate $lo g_{2} y$ contributions (one from the second log term, one from the third) — forgetting either piece changes the final constraint on $x y z$ and shifts the answer away from 48.

Question 4 · Quantitative Ability NAT

If $3^{a} = 4$ , $4^{b} = 5$ , $5^{c} = 6$ , $6^{d} = 7$ , $7^{e} = 8$ and $8^{f} = 9$ , then the value of the product abcdef is

Correct Answer:

Step-by-Step Solution

This is a telescoping log product question, recognisable because each equation defines a variable as a logarithm, and the variables appear in a chain from 3 to 9.

Step 1: Convert each exponential equation to logarithmic form.

\ $3^a = 4 \implies a = \log_3 4\$

\ $4^b = 5 \implies b = \log_4 5\$

\ $5^c = 6 \implies c = \log_5 6\$

\ $6^d = 7 \implies d = \log_6 7\$

\ $7^e = 8 \implies e = \log_7 8\$

\ $8^f = 9 \implies f = \log_8 9\$

Step 2: Write out the product explicitly.

\ $abcdef = (\log_3 4)(\log_4 5)(\log_5 6)(\log_6 7)(\log_7 8)(\log_8 9)\$

Step 3: Apply the chain rule for logarithms.

The key identity is: \ $\log_a b \cdot \log_b c = \log_a c\$ .

This works because:

\ $\log_a b \cdot \log_b c = \frac{\ln b}{\ln a} \cdot \frac{\ln c}{\ln b} = \frac{\ln c}{\ln a} = \log_a c\$

Apply this repeatedly (the intermediate bases cancel like a telescope):

\ $(\log_3 4)(\log_4 5) = \log_3 5\$

\ $(\log_3 5)(\log_5 6) = \log_3 6\$

\ $(\log_3 6)(\log_6 7) = \log_3 7\$

\ $(\log_3 7)(\log_7 8) = \log_3 8\$

\ $(\log_3 8)(\log_8 9) = \log_3 9\$

Step 4: Evaluate the final log.

\ $\log_3 9 = \log_3 (3^2) = 2\$

Answer: 2

Question 5 · Quantitative Ability MCQ

For a real number x, if $\frac{1}{2}$ , $\frac{l o g _{3} ( 2 ^{x} - 9 )}{l o g _{3} 4}$ , and $\frac{l o g _{5} ( 2 ^{x} + \frac{17}{2} )}{l o g _{5} 4}$ are in an arithmetic progression, then the common difference is

A.
$lo g_{4} (\frac{3}{2})$
B.
$lo g_{4} 7$
C.
$lo g_{4} (\frac{23}{2})$
D.
$lo g_{4} (\frac{7}{2})$

Correct Answer:

Step-by-Step Solution

This is a logs in arithmetic progression with change of base question, recognisable because the second and third terms have the form $\frac{l o g _{b} ( something )}{l o g _{b} 4}$ , which simplifies via change of base.

Step 1: Apply change of base.

$\frac{l o g _{3} ( 2 ^{x} - 9 )}{l o g _{3} 4} = lo g_{4} (2^{x} - 9)$

$\frac{l o g _{5} ( 2 ^{x} + 17/2 )}{l o g _{5} 4} = lo g_{4} (2^{x} + 17/2)$

So the three AP terms are: $\frac{1}{2}$ , $lo g_{4} (2^{x} - 9)$ , $lo g_{4} (2^{x} + 17/2)$ .

Step 2: Apply the AP condition.

For three terms in AP: $2 \times middle = first + third$ .

$2 lo g_{4} (2^{x} - 9) = \frac{1}{2} + lo g_{4} (2^{x} + \frac{17}{2})$

Step 3: Combine logs.

$lo g_{4} (2^{x} - 9)^{2} = lo g_{4} (4^{1/2}) + lo g_{4} (2^{x} + \frac{17}{2})$

$lo g_{4} (2^{x} - 9)^{2} = lo g_{4} (2 (2^{x} + \frac{17}{2}))$

$(2^{x} - 9)^{2} = 2 \cdot 2^{x} + 17$

Step 4: Solve the quadratic.

Let $t = 2^{x}$ :

$t^{2} - 18 t + 81 = 2 t + 17$

$t^{2} - 20 t + 64 = 0$

$(t - 4) (t - 16) = 0$

So $t = 4$ or $t = 16$ .

Step 5: Check domain.

We need $2^{x} - 9 > 0$ , so $2^{x} > 9$ . Since $4 < 9$ , reject $t = 4$ . Accept $t = 16$ , giving $x = 4$ .

Step 6: Find the common difference.

Second term: $lo g_{4} (16 - 9) = lo g_{4} 7$

Common difference: $lo g_{4} 7 - \frac{1}{2} = lo g_{4} 7 - lo g_{4} 2 = lo g_{4} (\frac{7}{2})$

Answer: $lo g_{4} (7/2)$ (Option D)

Logarithms, Exponents and Surds Previous Year Questions (PYQs) for CAT: 20+ 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

More previous year questions (pyqs) in this unit

A question from this chapter

Login to view the complete previous-year questions and solutions

Personalised first. High quality throughout.

Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course