Exponents, Logarithms and Infinite Series Short Notes for GATE DA

Exponents, Logarithms and Infinite Series short notes for GATE DA: 2 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.

Quick Revision: Laws and Identities

Quick Revision

Laws and Identities Summary

Exponent Laws

$a^{m} a^{n} = a^{m + n}$
$(a^{m})^{n} = a^{mn}$

Logarithm Laws

$lo g (x y) = lo g x + lo g y$
$lo g (x^{n}) = n lo g x$
$lo g_{a} b = \frac{l o g b}{l o g a}$

Strategy

Unify bases for exponential equations.
Take logs to bring down variables.
Verify domain ( $a r g > 0, ba se > 0, \neq = 1$ ).

Revision Checklist: Series Summation

Summary

Revision Checklist

S_{\infty} = \frac{a}{1 - r}

for

|r| &lt; 1

Identify Type: Geometric? Arithmetic-Geometric? Mixed?

Check Convergence: Ensure

|r| &lt; 1

for geometric parts

Find Parameters: Explicitly determine

a

(first term) and

r

(ratio)

Split if Needed: Break complex sums into independent convergent series

Calculate: Apply formula and sum the results

! If

r \geq 1

, do not use the formula. The sum diverges.

Exponents, Logarithms and Infinite Series: Solved Questions with Step-by-Step Explanations (10 Problems)

Question 1 · Quantitative Aptitude MCQ

For $x > 0$ , let $f (x) = x^{0} + x^{1}$ . What is the maximum value of $f (x)$ on the interval $[1, 3]$ ?

A.
4
B.
3
C.
2
D.
5

Correct Answer:

Step-by-Step Solution

Key idea: This tests the zero exponent rule ( $x^{0} = 1$ for $x > 0$ ) and evaluating a function on a closed interval.

Step 1: Simplify $f (x)$ . Since $x > 0$ , $x^{0} = 1$ . So $f (x) = 1 + x^{1} = 1 + x$ .

Step 2: Find the maximum on $[1, 3]$ . Since $f (x) = 1 + x$ is a linear increasing function, the maximum occurs at the right endpoint $x = 3$ .

Step 3: Evaluate $f (3) = 1 + 3 = 4$ .

Answer: 4

Common trap: Forgetting that $x^{0} = 1$ and treating it as $x^{0} = 0$ or $x^{0} = x$ , leading to wrong simplification.

Question 2 · Quantitative Aptitude MCQ

For $a > 0, a \neq = 1$ and $x, y > 0$ , which of the following is always true?

A.
$lo g_{a} (x + y) = lo g_{a} x + lo g_{a} y$
B.
$lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$
C.
$lo g_{a} (x - y) = lo g_{a} x - lo g_{a} y$
D.
$lo g_{a} (x / y) = lo g_{a} x / lo g_{a} y$

Correct Answer:

Step-by-Step Solution

Key idea: This tests the fundamental logarithmic identities, specifically the product rule.

Step 1: Recall the product rule for logarithms: $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$ for $x, y > 0$ .

Step 2: Check each option:

A: $lo g_{a} (x + y) \neq = lo g_{a} x + lo g_{a} y$ (log of sum is not sum of logs)
B: $lo g_{a} (x y) = lo g_{a} x + lo g_{a} y$ ✓ (product rule)
C: $lo g_{a} (x - y) \neq = lo g_{a} x - lo g_{a} y$ (log of difference is not difference of logs)
D: $lo g_{a} (x / y) = lo g_{a} x - lo g_{a} y$ , not $lo g_{a} x / lo g_{a} y$ (quotient rule)

Answer: B

Common trap: Assuming log distributes over addition/subtraction like multiplication, or confusing the quotient rule.

Question 3 · Quantitative Aptitude MCQ

For $x > 0$ , let $f (x) = 2 x^{1/2} - x^{0}$ . What is the maximum value of $f (x)$ on the interval $[4, 16]$ ?

A.
7
B.
3
C.
8
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: This tests the fractional exponent rule (roots) and the zero exponent rule, combined with finding the maximum of an increasing function on a closed interval.

Step 1: Simplify $f (x)$ . Since $x > 0$ , $x^{0} = 1$ . Also, $x^{1/2} = x$ .

So, $f (x) = 2 x - 1$ .

Step 2: Analyze the function on $[4, 16]$ . Since $x$ is strictly increasing for $x > 0$ , $f (x)$ is also strictly increasing. The maximum occurs at the right endpoint, $x = 16$ .

Step 3: Evaluate $f (16)$ .

$f (16) = 216 - 1 = 2 (4) - 1 = 8 - 1 = 7$ .

Answer: 7

Common trap: Forgetting the $- 1$ from $x^{0}$ and just calculating $216 = 8$ .

Question 4 · Quantitative Aptitude MCQ

For $x > 0$ , which of the following expressions is equivalent to $lo g_{3} (x^{2})$ ?

A.
$2 lo g_{3} x$
B.
$lo g_{3} (2 x)$
C.
$2 + lo g_{3} x$
D.
$(lo g_{3} x)^{2}$

Correct Answer:

Step-by-Step Solution

Key idea: This tests the power rule for logarithms.

Step 1: Recall the power rule: $lo g_{a} (x^{n}) = n lo g_{a} x$ .

Step 2: Apply the rule to the given expression.

$lo g_{3} (x^{2}) = 2 lo g_{3} x$ .

Step 3: Compare with the options. Option A matches exactly.

Answer: $2 lo g_{3} x$

Common trap: Confusing the power rule with the product rule, bringing the exponent down as an addition ( $2 + lo g_{3} x$ ) or squaring the entire log expression.

Question 5 · Quantitative Aptitude MCQ

For $x \in [1, 16]$ , let $f (x) = 10 - x^{1/2}$ . What is the maximum value of $f (x)$ ?

A.
6
B.
9
C.
10
D.
-6

Correct Answer:

Step-by-Step Solution

Key idea: This tests the fractional exponent rule (roots) and finding the maximum of a decreasing function on a closed interval.

Step 1: Simplify $f (x)$ . The term $x^{1/2}$ is equivalent to $x$ .

So, $f (x) = 10 - x$ .

Step 2: Analyze the function on $[1, 16]$ . Since $x$ is strictly increasing for $x > 0$ , $- x$ is strictly decreasing. Therefore, $f (x)$ is strictly decreasing. The maximum occurs at the left endpoint, $x = 1$ .

Step 3: Evaluate $f (1)$ .

$f (1) = 10 - 1 = 10 - 1 = 9$ .

Answer: 9

Common trap: Evaluating at the right endpoint $x = 16$ instead of recognizing the function is decreasing, leading to $10 - 4 = 6$ .

Question 6 · Quantitative Aptitude MCQ

For $x > 0$ and $x \neq = 1$ , consider the statements:

P: $lo g_{x} (x^{4}) = 4$

Q: $lo g_{x} (x + x^{3}) = 4$

R: $lo g_{x} (x^{2}) = 2 x$

Which of the following options is correct?

A.
Only Q is true
B.
Only P is true
C.
Both P and R are true
D.
All are true

Correct Answer:

Step-by-Step Solution

Key idea: This tests the fundamental logarithmic identities, specifically the power rule, and checks for common misconceptions about log distributing over addition.

Step 1: Evaluate Statement P.

Using the power rule $lo g_{a} (x^{n}) = n lo g_{a} x$ :

$lo g_{x} (x^{4}) = 4 lo g_{x} x = 4 (1) = 4$ .

So, P is true.

Step 2: Evaluate Statement Q.

The product rule applies to multiplication, not addition. $lo g_{x} (x + x^{3}) \neq = lo g_{x} x + lo g_{x} (x^{3})$ .

In fact, $lo g_{x} (x + x^{3}) = lo g_{x} (x (1 + x^{2})) = 1 + lo g_{x} (1 + x^{2})$ , which is not generally equal to 4.

So, Q is false.

Step 3: Evaluate Statement R.

Using the power rule: $lo g_{x} (x^{2}) = 2 lo g_{x} x = 2 (1) = 2$ .

The statement claims it is $2 x$ , which is false (unless $x = 1$ , but $x \neq = 1$ ).

So, R is false.

Answer: Only P is true.

Common trap: Assuming log distributes over addition, making Q true.

Question 7 · Quantitative Aptitude MCQ

For $x \in [1, 4]$ , let $f (x) = x^{- 1/2}$ . What is the maximum value of $f (x)$ ?

A.
1
B.
0.5
C.
2
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: This tests the fractional and negative exponent rules, combined with finding the maximum of a decreasing function on a closed interval.

Step 1: Simplify $f (x)$ using exponent rules.

$x^{- 1/2} = \frac{1}{x ^{1/2}} = \frac{1}{x}$ .

Step 2: Analyze the function on $[1, 4]$ . Since $x$ is strictly increasing for $x > 0$ , its reciprocal $\frac{1}{x}$ is strictly decreasing.

Step 3: For a strictly decreasing function on a closed interval, the maximum value occurs at the left endpoint, $x = 1$ .

Step 4: Evaluate $f (1)$ .

$f (1) = \frac{1}{1} = 1$ .

Answer: 1

Common trap: Evaluating at the right endpoint $x = 4$ instead of recognizing the function is decreasing, leading to $1/ 4 = 0.5$ .

Question 8 · Quantitative Aptitude MCQ

For $x > 0$ , which of the following expressions is equivalent to $lo g_{5} (\frac{1}{x})$ ?

A.
$- lo g_{5} x$
B.
$\frac{1}{l o g _{5} x}$
C.
$lo g_{5} x - 1$
D.
$5 - lo g_{5} x$

Correct Answer:

Step-by-Step Solution

Key idea: This tests the reciprocal identity for logarithms, which is a direct consequence of the power rule.

Step 1: Rewrite the argument using a negative exponent.

$\frac{1}{x} = x^{- 1}$ .

Step 2: Apply the power rule for logarithms: $lo g_{a} (x^{n}) = n lo g_{a} x$ .

$lo g_{5} (x^{- 1}) = - 1 \cdot lo g_{5} x = - lo g_{5} x$ .

Step 3: Alternatively, use the quotient rule: $lo g_{a} (\frac{u}{v}) = lo g_{a} u - lo g_{a} v$ .

$lo g_{5} (\frac{1}{x}) = lo g_{5} 1 - lo g_{5} x = 0 - lo g_{5} x = - lo g_{5} x$ .

Answer: $- lo g_{5} x$

Common trap: Confusing the reciprocal of the argument with the reciprocal of the entire logarithm, leading to $\frac{1}{l o g _{5} x}$ .

Question 9 · Quantitative Aptitude MCQ

For how many of the following infinite geometric series is it IMPOSSIBLE to compute a finite sum using the formula $S = \frac{a}{1 - r}$ ?

(P) $1 + \frac{1}{3} + \frac{1}{9} + \dots$

(Q) $2 + 4 + 8 + \dots$

(R) $5 - \frac{5}{2} + \frac{5}{4} - \dots$

(S) $1 + 1 + 1 + \dots$

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: The geometric series formula $S = \frac{a}{1 - r}$ is only valid when $∣ r ∣ < 1$ . If $∣ r ∣ \geq 1$ , the series diverges and the formula gives nonsense.

Step 1: Check (P). Common ratio $r = \frac{1}{3}$ . Since $∣1/3∣ < 1$ , the series converges. Formula applies.

Step 2: Check (Q). Common ratio $r = 2$ . Since $∣2∣ > 1$ , the series diverges. Formula does NOT apply.

Step 3: Check (R). Common ratio $r = - \frac{1}{2}$ . Since $∣ - 1/2∣ < 1$ , the series converges. Formula applies.

Step 4: Check (S). Common ratio $r = 1$ . Since $∣1∣ = 1$ , the series diverges. Formula does NOT apply.

Series (Q) and (S) cannot have finite sums computed. Count = 2.

Answer: B

Question 10 · Quantitative Aptitude MCQ

The sum of the infinite geometric series $1 + \frac{1}{4} + \frac{1}{16} + \dots$ is:

A.
3/4
B.
4/3
C.
5/4
D.
4/5

Correct Answer:

Step-by-Step Solution

Key idea: This is a direct application of the infinite geometric series sum formula.

Step 1: Identify the first term $a$ and the common ratio $r$ .

The series is $1 + 1/4 + 1/16 + \dots$

Here, the first term is $a = 1$ .

The common ratio is $r = \frac{1/4}{1} = \frac{1}{4}$ .

Step 2: Check for convergence.

Since $∣ r ∣ = 1/4 < 1$ , the series converges to a finite sum.

Step 3: Apply the geometric series sum formula.

The formula is $S = \frac{a}{1 - r}$ .

Substitute the values: $S = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$ .

Answer: B