Logarithms, Exponents and Surds Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Logarithms, Exponents and Surds short notes for CAT: 38 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Domain Rules: The Silent Marks Killer

Log domain checklist

Summary

For

lo g_{a} M

to be defined:

M > 0, a > 0, a \neq = 1

Argument

$M$ must be positive.

Base positive

$a$ must be greater than zero.

Base not one

$a = 1$ is not allowed.

Always check these before accepting final values.

Core Log Identities

Must-know identities

Identity	Use
$lo g_{a} (M N) = lo g_{a} M + lo g_{a} N$	Break products
$lo g_{a} (\frac{M}{N}) = lo g_{a} M - lo g_{a} N$	Break fractions
$lo g_{a} (M^{k}) = k lo g_{a} M$	Pull powers down
$lo g_{a} M = \frac{l o g _{b} M}{l o g _{b} a}$	Change the base

These identities work only when all logs involved are defined.

Base Less Than 1: Inequality Reverses

Small base, reversed behavior

a > 1

$lo g_{a} x$ increases as $x$ increases.

0 < a < 1

$lo g_{a} x$ decreases as $x$ increases.

For

0 < a < 1

lo g_{a} M > 0 ⟺ 0 < M < 1

This is one of the most common CAT log inequality traps.

Change of Base: Make Logs Speak One Language

One base = simpler equation

lo g_{a} b = \frac{lo g _{c} b}{lo g _{c} a}

\frac{lo g _{8} ( a + b )}{lo g _{2} c} = \frac{\frac{l o g _{2} ( a + b )}{3}}{lo g _{2} c} = \frac{1}{3} lo g_{c} (a + b)

\frac{lo g _{27} ( a - b )}{lo g _{3} c} = \frac{1}{3} lo g_{c} (a - b)

This identity is a major weapon for CAT logarithm equations.

Logarithms, Exponents and Surds: Solved Questions with Step-by-Step Explanations (2 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

More short notes in this unit

chapter

Logarithms, Exponents and Surds Short Notes for CAT: Concepts, Formulas, Worked Examples & Practice

Logarithms, Exponents and Surds short notes for CAT: 38 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

A question from this chapter

Question 1

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:

Question 2

The number of integers n with 1 \leq n \leq 100 satisfying lo g_{n} (2 n) > lo g_{2 n} (n) is:

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