◆ QA · Algebra

Logarithms , formulas + CAT PYQs

Focused Algebra kit. The full chapter formula sheet (with explanations & basic examples) is tucked below; every CAT PYQ for Logarithms is here.

18CAT PYQs
Algebrachapter
Formulas PYQs (18)↩ Full Algebra chapter

Algebra, formula sheet

Show the full Algebra formula sheet (explanations + basic examples)
1Polynomials & zeroes
  • Plain English: a polynomial is just a sum of x-powers; its "zeroes" are the x-values that make it equal 0.
  • A polynomial of degree n: aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀
  • k is a zero of p(x) if p(k) = 0. Zeroes are the x-coordinates where y = p(x) cuts the x-axis.
  • Max zeroes = degree: linear → 1, quadratic → 2, cubic → 3, degree n → n.
  • e.g. p(x) = x² − 9 has degree 2, so at most 2 zeroes: x = 3 and x = −3.
2Sum & product of roots
  • Plain English: you can read the sum and product of the roots straight off the coefficients, no need to solve.
  • Quadratic ax²+bx+c: α + β = −b/a, αβ = c/a
  • Cubic ax³+bx²+cx+d: α+β+γ = −b/a, αβ+βγ+γα = c/a, αβγ = −d/a
  • Build the equation: x² − (sum)x + (product) = 0.
  • e.g. x² − 5x + 6 = 0: sum = 5, product = 6 ⇒ roots 2 and 3 (2+3=5, 2×3=6).
3Discriminant & nature of roots
  • Plain English: the discriminant D is a single number that tells you how many real roots a quadratic has, before solving.
  • For ax²+bx+c (a≠0): D = b² − 4ac
  • D > 0 → two distinct real roots; D = 0 → equal real roots; D < 0 → no real roots (complex).
  • D a perfect square (a,b,c rational) → roots are rational.
  • Roots: x = (−b ± √D)/2a
  • e.g. x² + x + 1: D = 1 − 4 = −3 < 0 ⇒ no real roots.
4Sum of squares of roots (trick)
  • Plain English: you can get α²+β² from the sum and product alone, never solve for the roots first.
  • α² + β² = (α+β)² − 2αβ
  • To minimise a sum-of-squares-of-roots expression in a parameter, complete the square, minimum is at the vertex.
  • e.g. if α+β = 3 and αβ = 2, then α²+β² = 9 − 4 = 5.
5Algebraic identities
  • Plain English: memorised expand/factor templates that turn ugly expressions into products (or vice-versa) instantly.
  • (a±b)² = a² ± 2ab + b²
  • a² − b² = (a+b)(a−b)
  • (a+b+c)² = a²+b²+c² + 2(ab+bc+ca)
  • a³ ± b³ = (a±b)(a² ∓ ab + b²)
  • a³+b³+c³ − 3abc = (a+b+c)(a²+b²+c² − ab−bc−ca)
  • e.g. 97×103 = (100−3)(100+3) = 100² − 3² = 9991.
6Linear equations in two variables
  • Plain English: comparing the coefficient ratios tells you whether two lines cross once, never, or lie on top of each other.
  • a₁x+b₁y+c₁=0 and a₂x+b₂y+c₂=0.
  • Unique solution (intersecting): a₁/a₂ ≠ b₁/b₂
  • No solution (parallel): a₁/a₂ = b₁/b₂ ≠ c₁/c₂
  • Infinite solutions (coincident): a₁/a₂ = b₁/b₂ = c₁/c₂
  • e.g. x+y=2 and 2x+2y=5: ratios 1/2 = 1/2 ≠ 2/5 ⇒ parallel, no solution.
7Inequalities, basic rules
  • Plain English: inequalities behave like equations, except multiplying or dividing by a negative reverses the arrow.
  • Adding/subtracting keeps direction; multiplying by a negative flips the sign.
  • If X > Y > 0 then 1/X < 1/Y.
  • For x > 0: x + 1/x ≥ 2 (equality at x = 1).
  • e.g. −2x > 6 ⇒ divide by −2 and flip ⇒ x < −3.
8Quadratic inequalities
  • Plain English: factor it, then "< 0" means between the roots and "> 0" means outside the roots.
  • (x−m)(x−n) < 0, n > m ⇒ m < x < n (between the roots).
  • (x−m)(x−n) > 0 ⇒ x < m or x > n (outside the roots).
  • Sign-of-product / wavy-curve method handles higher degree.
  • e.g. x² − 5x + 6 < 0 ⇒ (x−2)(x−3) < 0 ⇒ 2 < x < 3.
9Modulus (absolute value)
  • Plain English: |x| is the distance of x from 0, so it strips the sign and is never negative.
  • |x| = max(x, −x); −|x| ≤ x ≤ |x|.
  • |a+b| ≤ |a|+|b| and |a|−|b| ≤ |a−b|; |ab| = |a||b|.
  • |x| ≤ k ⇒ −k ≤ x ≤ k. |x| ≥ k ⇒ x ≥ k or x ≤ −k.
  • |f| + |g| = |f+g| only when f, g have the same sign.
  • e.g. |x| ≤ 3 ⇒ −3 ≤ x ≤ 3; |x − 4| = 2 ⇒ x = 6 or x = 2.
10AM-GM-HM inequality
  • Plain English: for positive numbers the plain average is always ≥ the geometric average, the go-to tool for "find the minimum".
  • For positive reals: AM ≥ GM ≥ HM, equality when all equal.
  • Two numbers: AM = (a+b)/2, GM = √(ab), HM = 2ab/(a+b).
  • AM × HM = GM²
  • If a₁a₂…aₙ = 1 then a₁+a₂+…+aₙ ≥ n.
  • e.g. for a = 2, b = 8: AM = 5 ≥ GM = √16 = 4. ✓
11Maxima & minima of a quadratic
  • Plain English: a parabola's turning point is at x = −b/2a; that's where the min (opens up) or max (opens down) lives.
  • ax²+bx+c: vertex at x = −b/2a; extreme value = (4ac − b²)/4a = −D/4a.
  • a > 0 → opens up → minimum; a < 0 → opens down → maximum.
  • min/max of max-of-two / min-of-two lines occurs where the two graphs intersect.
  • e.g. x² − 6x + 5: vertex at x = 3, minimum value = 9 − 18 + 5 = −4.
12Functions, domain, range, even/odd
  • Plain English: domain is what you may feed in, range is what comes out; even/odd describe the graph's symmetry.
  • Domain = allowed inputs; range = resulting outputs.
  • Even: f(−x) = f(x) (graph symmetric about y-axis), e.g. x², |x|.
  • Odd: f(−x) = −f(x) (symmetric about origin), e.g. x³, 1/x.
  • Inverse exists only if f is one-to-one.
  • e.g. f(x) = x³ is odd: f(−2) = −8 = −f(2). ✓
13Functional equations
  • Plain English: the form of a functional rule reveals the function, "turns + into ×" means exponential, etc.
  • f(x+y) = f(x)·f(y) ⇒ exponential type, f(x) = aˣ.
  • f(xy) = f(x)·f(y) ⇒ power/multiplicative; f(1) = 1.
  • If f(a+x) = f(a−x), the graph is symmetric about x = a; roots pair around a (sum of 4 roots = 4a).
  • e.g. f(x+y) = f(x)f(y) with f(1) = 3 ⇒ f(2) = f(1)² = 9.
14Graph shifting
  • Plain English: changes outside f() move the graph vertically; changes inside f() move it horizontally (and oppositely).
  • f(x)+c → shift up c; f(x)−c → shift down c.
  • f(x+c) → shift left c; f(x−c) → shift right c.
  • −f(x) → reflect in x-axis; f(−x) → reflect in y-axis.
  • e.g. y = (x−2)² is y = x² shifted 2 units right.
15Logarithm, definition
  • Plain English: log_b x just asks "what power of b gives x?", it's the inverse of raising to a power.
  • y = log_b x ⇔ x = bʸ (b > 0, b ≠ 1, x > 0).
  • log_a a = 1; log_a 1 = 0; a^(log_a m) = m.
  • e.g. log₂8 = 3 because 2³ = 8.
16Logarithm laws
  • Plain English: logs turn multiplication into addition, division into subtraction, and powers into multipliers.
  • log_a(xy) = log_a x + log_a y
  • log_a(x/y) = log_a x − log_a y
  • log_a(xᵐ) = m·log_a x
  • log_(aⁿ)(xᵐ) = (m/n)·log_a x
  • Change of base: log_a x = (log x)/(log a); log_a x = 1/log_x a
  • e.g. log₂40 = log₂(8×5) = log₂8 + log₂5 = 3 + log₂5.
17Indices (laws of exponents)
  • Plain English: same base, add exponents when multiplying, subtract when dividing, multiply when raising a power to a power.
  • pᵐ·pⁿ = pᵐ⁺ⁿ; pᵐ/pⁿ = pᵐ⁻ⁿ; (pᵐ)ⁿ = pᵐⁿ
  • pⁿ·qⁿ = (pq)ⁿ; (p/q)ⁿ = pⁿ/qⁿ
  • p⁻ⁿ = 1/pⁿ; p⁰ = 1; p^(1/n) = ⁿ√p
  • e.g. 2³·2⁴ = 2⁷ = 128; 8^(2/3) = (∛8)² = 2² = 4.
18Surds & rationalisation
  • Plain English: a surd is an unresolved root like √2; "rationalising" clears it from a denominator using the conjugate.
  • √(ab) = √a·√b; √(a/b) = √a/√b.
  • Rationalise a/(b+√c) by multiplying top & bottom by the conjugate (b−√c).
  • If a+√b is a root of a rational quadratic, so is its conjugate a−√b.
  • e.g. 1/(√3 − 1) × (√3 + 1)/(√3 + 1) = (√3 + 1)/2.
19Arithmetic Progression (AP)
  • Plain English: an AP adds the same step d each time; its sum is just "how many terms × the average of first and last".
  • Constant difference d. nth term: Tₙ = a + (n−1)d
  • Sum: Sₙ = n/2 · [2a + (n−1)d] = n/2 · (first + last)
  • Arithmetic mean of a, b: A = (a+b)/2. Middle term = average of an odd count of AP terms.
  • e.g. 2, 5, 8, …: T₄ = 2 + 3×3 = 11; sum of first 4 = 4/2·(2+11) = 26.
20Geometric Progression (GP)
  • Plain English: a GP multiplies by the same ratio r each time; if |r| < 1 the infinite sum settles to a finite value.
  • Constant ratio r. nth term: Tₙ = a·rⁿ⁻¹
  • Sum: Sₙ = a(rⁿ − 1)/(r − 1), r ≠ 1.
  • Infinite sum (|r| < 1): S∞ = a/(1 − r)
  • Geometric mean: G = √(ab).
  • e.g. 1 + ½ + ¼ + … = 1/(1 − ½) = 2.
21Harmonic Progression (HP)
  • Plain English: an HP is just an AP flipped, take reciprocals and you're back to a normal AP.
  • a, b, c… in HP ⇔ 1/a, 1/b, 1/c… in AP.
  • Harmonic mean of a, b: H = 2ab/(a+b)
  • nth term of HP = 1/(nth term of the corresponding AP).
  • e.g. 1, ½, ⅓, ¼ is an HP (reciprocals 1, 2, 3, 4 form an AP).
22Standard summation formulas
  • Plain English: ready-made closed forms for adding up the first n numbers, their squares, and their cubes.
  • Σn = n(n+1)/2
  • Σn² = n(n+1)(2n+1)/6
  • Σn³ = [n(n+1)/2]²
  • Telescoping: 1/(k·(k+1)) = 1/k − 1/(k+1).
  • e.g. 1 + 2 + … + 10 = 10×11/2 = 55.
23Common terms of two APs
  • Plain English: numbers shared by two APs themselves form an AP whose step is the LCM of the two steps.
  • Common terms of two APs form a new AP with common difference = LCM of the two differences.
  • Find the first common term, then count multiples of the LCM up to the smaller upper limit.
  • e.g. 2,5,8,… and 3,7,11,…: first common term 11, new step = LCM(3,4) = 12 ⇒ 11, 23, 35, …
24Recurrence & tₙ from Sₙ
  • Plain English: if you know the running total Sₙ, each term is just this total minus the previous total.
  • If Sₙ given: aₙ = Sₙ − Sₙ₋₁ (and a₁ = S₁).
  • Alternating-sum sequences: subtract consecutive defining equations to isolate a term.
  • e.g. Sₙ = n² ⇒ a₅ = S₅ − S₄ = 25 − 16 = 9.
25Integer / Diophantine solutions
  • Plain English: once you spot one whole-number solution, all the rest come by stepping x and y in fixed jumps.
  • ax + by = c with one integer solution (x₀, y₀): all others are x₀ + (b/g)t, y₀ − (a/g)t, where g = gcd(a,b).
  • Bound the count using the given ranges on x and y.
  • e.g. 2x + 3y = 12: (x,y) = (3,2) works; next is (0,4), then (6,0), x jumps by 3, y by 2.
26When does Aᴮ = 1?
  • Plain English: a power equals 1 in exactly three situations, check all three or you'll miss cases.
  • Base = 1 (any exponent), or
  • Exponent = 0 (base ≠ 0), or
  • Base = −1 with an even exponent.
  • e.g. (−1)⁴ = 1 (base −1, even power); 7⁰ = 1 (zero power); 1⁹⁹ = 1 (base 1).
27Three terms in AP / GP
  • Plain English: centering three terms on a middle value makes their sum (AP) or product (GP) collapse to one symbol.
  • Three in AP: take a−d, a, a+d (their sum = 3a).
  • Three in GP: take a/r, a, ar (product = a³).
  • Three consecutive integers as roots: n−1, n, n+1.
  • e.g. three numbers in AP summing to 18 ⇒ middle = 6, so 6−d, 6, 6+d.
28|x − a| as distance (modulus sums)
  • Plain English: read |x−a| as "distance from a", and sums of such distances are smallest when x sits among the points.
  • |x−a| = distance of x from a on the number line.
  • |x−p|+|x−q| is minimised for any x between p and q; minimum value = |p−q|.
  • |x−p| = |x−q| at the midpoint x = (p+q)/2.
  • e.g. |x−2| + |x−7| ≥ 5, achieved for any x in [2, 7].
29Sum of squares identity trick
  • Plain English: squares can't be negative, so if a bunch of squares add to 0 every single one must be 0.
  • If a sum of squares equals 0, each square = 0: e.g. (x−2y)² + (y−z)² = 0 ⇒ x = 2y and y = z.
  • Group given expressions into perfect squares to pin exact values.
  • e.g. (a−3)² + (b+1)² = 0 forces a = 3 and b = −1.
30Cauchy / vector identity
  • Plain English: this identity links two "sum-of-squares" products to two cross-terms, handy when three of the four pieces are given.
  • (a²+b²)(x²+y²) = (ax+by)² + (ay−bx)².
  • Useful when given a²+b², x²+y² and ax+by to find ay−bx.
  • e.g. (1²+2²)(3²+4²) = 5·25 = 125 = 11² + 2² = (1·3+2·4)² + (1·4−2·3)².
18 CAT questions

Logarithms, CAT PYQs

Logarithms

EasyCAT 1994

log₆ 216√6 is:

  • (1) 3
  • (2) 3/2
  • (3) 7/2
  • (4) None of these
Show solution
(3) 7/2. If log₆ 216√6 = x, then 6ˣ = 216√6 = 6³(6^(1/2)) = 6^(7/2). ∴ x = 7/2.
EasyCAT 1994

If log₇ log₅ (√(x + 5) + √x) = 0, find the value of x.

  • (1) 1
  • (2) 0
  • (3) 2
  • (4) None of these
Show solution
(2) 0. log₇ log₅ (√(x + 5) + √x) = 0, ∴ log₅ (√(x + 5) + √x) = 7⁰ = 1, or (√(x + 5) + √x) = 5¹ = 5 ∴ 2√x = 0 or x = 0.
ModerateCAT 1997

If log₂ [log₇ (x² − x + 37)] = 1, then what could be the value of 'x'?

  • (1) 3
  • (2) 5
  • (3) 4
  • (4) None of these
Show solution
(3) 4. log₇ (x² − x + 37) = 2¹ = 2, and furthermore (x² − x + 37) = 7² = 49. Thus x² − x − 12 = 0, whose solutions are x = 4 or x = −3. The value that matches the given answer-choices is x = 4.
HardCAT 2003

If log₃ 2, log₃ (2ˣ − 5), log₃ (2ˣ − 7/2) are in arithmetic progression, Then the value of x is equal to ___.

  • (1) 5
  • (2) 4
  • (3) 2
  • (4) 3
Show solution
(4) 3. log₃ 2, log₃ (2ˣ − 5), log₃ (2ˣ − 7/2) are in A.P. ∴ 2 × log₃ (2ˣ − 5) = log₃ 2 + log₃ (2ˣ − 7/2) ⇒ log₃ (2ˣ − 5)² = log₃ [2 × (2ˣ − 7/2)]. Let 2ˣ = a, then (a − 5)² = 2(a − 7/2) ⇒ a² − 10a + 25 = 2a − 7 ⇒ a² − 12a + 32 = 0 ⇒ (a − 8)(a − 4) = 0 ⇒ a = 8 or 4 ⇒ 2ˣ = 8 or 2ˣ = 4 ⇒ x = 3 and x = 2. x = 2 cannot be the answer as (2ˣ − 5) would become negative and logarithms of negative numbers are not defined. ∴ x = 3.
HardCAT 2003

If log₁₀ x − log₁₀ √x = 2 log_x 10, then a possible value of x is given by

  • (1) 10
  • (2) 1/100
  • (3) 1/1000
  • (4) None of these
Show solution
(2) 1/100. log₁₀ x − log₁₀ √x = 2/log₁₀ x ⇒ log₁₀ (x/√x) = 2/log₁₀ x ⇒ log₁₀ √x = 2/log₁₀ x ⇒ (1/2) log₁₀ x = 2/log₁₀ x ⇒ (1/2)(log₁₀ x)² = 2 ⇒ (log₁₀ x)² = 4 ⇒ log₁₀ x = ±2 ⇒ x = 10^±2 ⇒ x = 100 or 1/100.
HardCAT 2004

Let u = (log₂ x)² − 6 log₂ x + 12 where x is a real number. Then the equation xᵘ = 256, has

  • (1) no solution for x
  • (2) exactly one solution for x
  • (3) exactly two distinct solutions for x
  • (4) exactly three distinct solutions for x
Show solution
(2) exactly one solution for x. We have u = (log₂ x)² − 6 log₂ x + 12. Put log₂ x = y. Then xᵘ = 256 ⇒ xᵘ = 2⁸ ⇒ 2^(uy) = 2⁸ ⇒ uy = 8 ⇒ u = 8/y. So 8/y = y² − 6y + 12 ⇒ 8 = y³ − 6y² + 12y ⇒ y³ − 6y² + 12y − 8 = 0 ⇒ (y − 2)(y² − 4y + 4) = 0, i.e. either y − 2 = 0 or (y − 2)² = 0 ⇒ y = 2. Hence the equation has exactly one solution for x.
ModerateCAT 2005

If x ≥ y and y > 1, then the value of the expression log_x(x/y) + log_y(y/x) can never be

  • (1) −1
  • (2) −0.5
  • (3) 0
  • (4) 1
Show solution
(4) 1. log_x(x/y) + log_y(y/x) = (log x − log y)/log x + (log y − log x)/log y = 1 − log_x y + 1 − log_y x = 2 − (log_x y + log_y x). As x ≥ y and y > 1, log_x y ≤ 1 and log_y x ≥ 1, so log_x y + log_y x > 1. ∴ 2 − (log_x y + log_y x) < 1, i.e. the expression can never equal 1.
ModerateCAT 2017

Suppose log₃ x = log₁₂ y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log₆ G is equal to

  • (1) √a
  • (2) 2a
  • (3) a/2
  • (4) a
Show solution
(4) a. log₃ x = a and log₁₂ y = a. Hence, x = 3ᵃ and y = 12ᵃ. Therefore, the geometric mean of x and y equals √(xy) = √(3ᵃ × 12ᵃ) = 6ᵃ. Hence, G = 6ᵃ or, log₆ G = a.
ModerateCAT 2017

If x is a real number such that log₃ 5 = log₅ (2 + x), then which of the following is true?

  • (1) 0 < x < 3
  • (2) 23 < x < 30
  • (3) x > 30
  • (4) 3 < x < 23
Show solution
(4) 3 < x < 23. 1 < log₃ 5 < 2 ⇒ 1 < log₅ (2 + x) < 2 ⇒ 5 < 2 + x < 25 ⇒ 3 < x < 23.
ModerateCAT 2017TITA

If log (2ᵃ × 3ᵇ × 5ᶜ) is the arithmetic mean of log (2² × 3³ × 5), log (2⁶ × 3 × 5⁷), and log (2 × 3² × 5⁴), then a equals:

Show solution
3. log (2ᵃ × 3ᵇ × 5ᶜ) = [log (2² × 3³ × 5) + log (2⁶ × 3 × 5⁷) + log (2 × 3² × 5⁴)]/3 ⇒ 3 log (2ᵃ × 3ᵇ × 5ᶜ) = log (2^(2+6+1) × 3^(3+1+2) × 5^(1+7+4)) ⇒ log (2^(3a) × 3^(3b) × 5^(3c)) = log (2⁹ × 3⁶ × 5¹²). Hence, 3a = 9 or a = 3.
ModerateCAT 2018

If log₂ (5 + log₃ a) = 3 and log₅ (4a + 12 + log₂ b) = 3, then a + b is equal to:

  • (1) 59
  • (2) 40
  • (3) 32
  • (4) 67
Show solution
(1) 59. log₂ (5 + log₃ a) = 3 ⇒ 5 + log₃ a = 2³ = 8 ⇒ log₃ a = 3 ⇒ a = 3³ = 27. Now log₅ (4a + 12 + log₂ b) = 3 ⇒ 4a + 12 + log₂ b = 125 ⇒ log₂ b = 125 − 12 − 4 × 27 = 5 ⇒ b = 2⁵ = 32. ∴ a + b = 59.
HardCAT 2019

If x and y be positive real numbers such that log₅ (x + y) + log₅ (x − y) = 3, and log₂ y − log₂ x = 1 − log₂ 3. Then xy equals

  • (1) 150
  • (2) 25
  • (3) 100
  • (4) 250
Show solution
(1) 150. log₅ (x + y) + log₅ (x − y) = 3 ⇒ log₅ [(x + y)(x − y)] = 3 ⇒ (x + y)(x − y) = 5³ = 125 ⇒ x² − y² = 125. And log₂ y − log₂ x = 1 − log₂ 3 ⇒ log₂ (y/x) = log₂ 2 − log₂ 3 ⇒ log₂ (y/x) = log₂ (2/3) ⇒ y/x = 2/3. Let x = 3k, y = 2k: (3k)² − (2k)² = 125 ⇒ 5k² = 125 ⇒ k = 5. Hence, x × y = 3k × 2k = 6 × 25 = 150.
ModerateCAT 2019

Let A be a real number. Then the roots of the equation x² − 4x − log₂ A = 0 are real and distinct if and only if

  • (1) A > 1/16
  • (2) A < 1/16
  • (3) A < 1/8
  • (4) A > 1/8
Show solution
(1) A > 1/16. For ax² + bx + c = 0, the roots are real and distinct if b² − 4ac > 0. For x² − 4x − log₂ A = 0: (−4)² − 4 × 1 × (−log₂ A) > 0 ⇒ 16 + 4 log₂ A > 0 ⇒ log₂ A > −4 ⇒ A > 2⁻⁴ = 1/16.
ModerateCAT 2021 · Slot 1TITA

If 5 − log₁₀ √(1 + x) + 4 log₁₀ √(1 − x) = log₁₀ (1/√(1 − x²)), then 100x equals

Show solution
99. 5 log₁₀ 10 − log₁₀ √(1 + x) + 4 log₁₀ √(1 − x) = log₁₀ (1/√(1 − x²)) ⇒ log₁₀ [10⁵ × (1 − x)²/√(1 + x)] = log₁₀ [1/(√(1 + x)·√(1 − x))]. Removing log from both sides: 10⁵ × (1 − x)²/√(1 + x) = 1/(√(1 + x)·√(1 − x)) ⇒ 10⁵ × (1 − x)² = 1/(1 − x)^(1/2) ⇒ 10⁵ = 1/(1 − x)^(5/2). Taking the 5th root: 10 = 1/(1 − x)^(1/2). Squaring: 100 = 1/(1 − x) ⇒ 100 − 100x = 1 ⇒ 100x = 99.
ModerateCAT 2021 · Slot 2TITA

If log₂ [3 + log₃ {4 + log₄ (x − 1)}] − 2 = 0, then 4x equals

Show solution
5. log₂ [3 + log₃ {4 + log₄ (x − 1)}] = 2 × 1 = 2 × log₂ 2 ⇒ log₂ [3 + log₃ {4 + log₄ (x − 1)}] = log₂ 2² ⇒ 3 + log₃ {4 + log₄ (x − 1)} = 4 ⇒ log₃ {4 + log₄ (x − 1)} = 1 = log₃ 3 ⇒ 4 + log₄ (x − 1) = 3 ⇒ log₄ (x − 1) = −1 ⇒ log₄ (x − 1) = log₄ (4)⁻¹ ⇒ x − 1 = 4⁻¹ = 1/4 ⇒ 4x − 4 = 1 ⇒ 4x = 5.
HardCAT 2022 · Slot 2TITA

The number of distinct integer values of n satisfying (4 − log₂ n)/(3 − log₄ n) < 0 is:

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47. (4 − log₂ n)/(3 − log₄ n) < 0 = (4 − log₂ n)/(3 − ½ log₂ n) < 0. Let log₂ n = x ⇒ (4 − x)/(6 − x) < 0 ⇒ (x − 4)/(x − 6) < 0 ⇒ 4 < x < 6 ⇒ 4 < log₂ n < 6 ⇒ 2⁴ < n < 2⁶ ⇒ 16 < n < 64. ∴ The number of integer values of n is 47.
ModerateCAT 2023 · Slot 1

If x and y are positive real numbers such that log_x (x² + 12) = 4 and 3 log_y x = 1, then x + y equals

  • (1) 10
  • (2) 68
  • (3) 20
  • (4) 11
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(1) 10. log_x (x² + 12) = 4 ⇒ x⁴ = x² + 12 ⇒ x⁴ − x² − 12 = 0 ⇒ (x² − 4)(x² + 3) = 0. Since x² + 3 ≠ 0, x² − 4 = 0 ⇒ x = +2 (x ≠ −2). Now 3 log_y x = 1 ⇒ log_y x³ = 1 ⇒ x³ = y ⇒ y = 2³ = 8. Now x + y = 2 + 8 = 10.
HardCAT 2023 · Slot 2TITA

For some positive real number x, if log_√3 (x) + (log_x 25)/(log_x 0.008) = 16/3, then the value of log₃ (3x²) is

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7. log_√3 (x) + (log_x 25)/(log_x 0.008) = 16/3 ⇒ log_√3 (x) + log_(8/1000) 25 = 16/3 ⇒ log_√3 (x) + log_(5⁻³) 5² = 16/3 ⇒ log_√3 (x) − (2/3) log_5 5 = 16/3 ⇒ log_√3 (x) = 16/3 + 2/3 = 6 ⇒ x = (√3)⁶ ⇒ x² = 3⁶ ⇒ 3x² = 3⁷ ⇒ log₃ (3x²) = 7.