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Logarithm for CAT Exam – Concept Builder Questions with Detailed Solutions and Tips

Saturday, November 7th, 2020

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This article was contributed by Sandip Gupta.

Logarithm is simple. The simplest way to master it is to do more number and varieties of problems.

I think you have already understood the basic concepts of logarithm from the previous blog.

Before we start, let me suggest two things,

1. Write down all the formulae and try to remember them,
2. When you see a problem, first try to solve it yourself and then see the solution.

Let’s start

At first let me give you two useful formulae.

1. log(base n⁡)m = log(base a)⁡m/log(base a)⁡n, (where a is any positive number other than 1)
2. log(base n⁡)m = 1/log(base m)⁡n

Note: In the following section,

1. All the logarithms are taken to the base 10, unless mentioned otherwise.
2. log(base n⁡)m is written as log (base n) m.

Problem 1.  Find the least integer n, such that 7ⁿ > 10⁵, given log₁₀343 = 2.53?

a.  4

b.  5

c.  6

d. 7

Solution: (option c is the correct answer)

log 343 = log 7³ = 3 log 7 = 2.53 (given).  So, log 7 = (2.53) / 3 …. (i)

now, 7ⁿ > 10⁵    or, log (7ⁿ) > log (10⁵),   or, n. log 7 >  5. log 10

or, n. (2.53) / 3 > 5, (as log₁₀ 10 = 1) or, n > (15 / 2.53), or, n > 5.93

So, the smallest integral value of n is 6.    => option (c)

Problem 2.  If a, b, c are three distinct positive numbers each different from 1, such that,

(log_ba. log_ca – log_aa) + (log_ab. log_cb – log_bb) + (log_ac. log_bc – log_cc) = 0, then the value of (abc)^1/3 is,

a. 1

b. 2

c. ½

d. none of these.

Solution: (option a is the correct answer)

Given, [(log a / log b). (log a / log c) – 1] + [(log b / log a). (log b/ log c) -1] + [(log c / log a). (log c / log b) – 1] = 0,

or, [(x / y). (x / z) – 1] + [(y / x). (y / z) – 1] + [(z /x). (z / y) – 1] = 0, [where, x = log a, y = log b, and z = log c]

or, [(x² / yz) – 1] + [(y² / zx) – 1] + [(z² / xy) – 1] = 0

or, [x² / yz] + [y² / zx ]+ [z² / xy] = 3, or, (x³ + y³ + z³) / (xyz) = 3,

or, x³ + y³ + z³ = 3xyz, or, x³ + y³ + z³ – 3xyz = 0,

or, (x + y + z). (x² + y² +z² -xy – yz -zx) = 0,

or, ½. (x + y + z). [(x – y)² + (y – z)² + (z – x)²] = 0,

• x + y + z = 0, (as x, y, z, are unequal) => log a + log b + log c = 0,
• log(abc) = 0 = log 1 => abc = 1 => (abc)^1/3 = (1)^1/3 = 1. => opt a

Problem 3.  log₂₀2 lies between

a. 1/5 and 1/4

b. 1/4 and 1/3

c. 1/3 and 1/2

d. 1/6 and 1/5.

Solution: (option a is the correct answer)

[Let us first understand the approach. Let, log (base20) 2 = 1/y. So, 20^1/y = 2, => 20 = 2^y. Also, we know, 2⁴ = 16, 2⁵ = 32]

We know, 16 < 20 < 32   => 2⁴ < 20 < 2⁵

=> log (2⁴) < log 20 < log 2⁵, (all the logarithms are taken to the base 20)

=> 4 log 2 < log 20 < 5 log 2 => 4 log 2 < 1 < 5 log 2 (since log₂₀20 = 1)

=> 4 log 2 < 1 and 1 < 5 log 2 => log 2 < 1/4 and 1/5 < log 2

=> 1/5 < log ₂₀2 < 1/4 => (opt a)

Problem 4. Find the value of log_0.5 √[8. √{8. √(8 …. to∞)}]

a. – 1

b. 2

c. – 3

d. 4/3

Solution: (option c is the correct answer)

Let, √[8. √{8. √(8 …. to ….. ∞)}] = x,  => 8. √{8. √(8 …. to ….. ∞)} = x² (on squaring).

[Note: From infinite number of 8, if one 8 comes out of the square root sign, the number of 8 inside the square root remains infinite]

Given problem = log _(1/2)8 = p (say) => (1/2)^p = 8, => (2^-1)^p = 2³, 2^-p = 2³ => p = -3 => (opt c)

Problem 5. The value of 7.log (16/15) + 5.log (25/24) + 3.log(81/80) is

a. log 3

b. 1

c. log 5

d. 1 – log 5

Solution: (option d is the correct answer)

[Let us solve this problem with two different ways.]

Method – 1

Given expression

= 7. log [(2⁴) / (3×5)] + 5. log [(5²) / (3×2³)] + 3. log [(3⁴) / (2⁴x5)]

= 7 (log 2⁴ – log 3 – log 5) + 5 (log 5² – log 3 – log 2³) + 3 (log 3⁴ – log2⁴ – log 5)

= 7 (4 log 2 – log 3 – log 5) + 5 (2 log 5 – log 3 – 3 log 2) + 3 (4 log 3 – 4log 2 – log 5)

= 28 log 2 – 7 log 3 – 7 log 5 + 10 log 5 – 5 log 3 – 15 log 2 + 12 log 3 – 12 log 2 – 3 log 5

= log 2. (28 – 27) + log 3. (12 – 12) + log 5. (10 – 10)

= log 2 = log (10/5) = log 10 – log 5 = 1 – log 5 => (opt d)

Method – 2

Given expression

= 7. log [(2⁴) / (3×5)] + 5. log [(5²) / (3×2³)] + 3. log [(3⁴) / (2⁴x5)]

= log [(2⁴) / (3×5)]⁷ + log [(5²) / (3×2³)]⁵ + log [(3⁴) / (2⁴x5)]³

= log [(2²⁸) / (3 ⁷x 5 ⁷)] + log [(5¹⁰) / (3⁵ x 2¹⁵)] + log [(3¹²) / (2¹² x 5³)]

= log [(2²⁸ x 5¹⁰ x 3¹²) / (3⁷ x 5⁷ x 3⁵ x 2¹⁵ x 2¹² x 5³)

= log [(2²⁸ x 5¹⁰ x 3¹²) / (2²⁷ x 3¹² x 5¹⁰)]

= log 2 = log (10/5) = log 10 – log 5 = 1 – log 5 => (opt d)

Problem 6. If a² + b² = 7ab, and log [(a + b)/3] = k. (log a + log b), then the value of k is

a. 1/5

b. 3

c. 1/2

d. 4

Solution: (option c is the correct answer)

Given, a² + b² = 7ab, or, a² + b² + 2ab = 7ab + 2ab => (a +b)² = 9ab

⇒(a + b) = 3. (ab)^1/2,

⇒(a + b)/3 = (ab)^1/2 => log [(a + b)/3] = ½. log(ab)

⇒log [(a + b)/3] = 1/2. (log a + log b)

⇒k = ½ [comparing with log {(a + b)/3} = k. (log a + log b)](opt c)

Problem 7. What is the least value of the expression, 2 log₁₀x – log_x0.01, for x > 1

a. 2

b. 4

c. 8

d. none of these

Solution: (option b is the correct answer)

[All the logarithms are taken to the base 10]

Let p = 2 log x – (log 0.01) / (log x) [using log_nm = (log m) / (log n)]

= 2 a – [log (10)^-2 / a]      [where, log x = a]

= 2 a + (2 log 10) / a = 2 a + 2/a = 2 (a + 1/a) …. (i)

[since, x > 1, so, log x > 0, that is, a > 0.

We know that for any positive number a, (a + 1/a) ≥ 2 (very important and useful for CAT)]

So, from (i), p ≥ 2×2 => p ≥ 4

So, the least value of the expression is 4 => (opt b)

Problem 8. If (x)^log₁₀^x = 100 x, then number of possible values of x is/are:

a. 0

b. 1

c. 2

d. none of these

Solution: (option c is the correct answer)

[All the logarithms are taken to the base 10]

x ^{log x} = 100 x     => log (x ^{log x}) = log 100 + log x

=> log x. log x = log 10² + log x

=> a² = 2 log 10 + a [where a = log x]   => a² = 2 + a   

=> a² – a – 2 = 0, => a² + a – 2a – 2 = 0   => a(a +1) – 2(a +1) = 0,

=> (a + 1) (a – 2) = 0    => a = -1, 2

When a = -1, log₁₀x = – 1    => x = 10^-1 = 1/10

When a = 2, log₁₀x = 2       => x = 10² = 100

So, x = 1/10, 100     => 2 values of x are possible.  => (opt c)

Problem 9. If log₃₀ 3 = p and log₃₀ 5 = q, then, how is log₃₀ 8 expressed in terms of p and q

a. 3 – 3(p + q)

b. 5p + 2q – 1

c. p + q -2

d. none of these

Solution: (option a is the correct answer)

[At first let us understand the approach. The base of the logarithm is 30. We have to express log 8 in terms of log 3 and log 5. Now, 8 = 2³ and 2 = 30/15 = 30 / (3×5)]

[All the logarithms are taken to the base 30]

log 8 = log 2³ = 3 log 2 = 3 log (30/15) = 3 [log 30 – log(3×5)]

= 3[1 – log 3 – log 5]           (since, log₃₀ 30 = 1)

= 3[1 – p – q] = 3 – 3(p + q) => (opt a)

Problem 10.  The value of log _{cos A} √[1 + (sec² A / cosec² A)] is

a. 1

b. 0

c. log 2

d. -1

Solution: (option d is the correct answer)

sec²A / cosec² A = (1/cos²A) / (1/sin²A) = sin²A / cos²A = tan²A

So, the given expression = log (base cos A) √(1 + tan² A)

= log (base cos A) √[sec²A] = log (base cos A) (sec A)

= log _{cos A} (cos A)^-1 = -1. log _{cos A} (cos A) = – 1. => (opt d)

Problem 11. If log_0.3 (x – 1) < log_0.09 (x – 1), then x lies in the interval

a. (2, ∞)

b. (- ∞, 2)

c. (1, 2)

d. none of these

Solution: (option a is the correct answer)

log 0.09 = log (0.3)² = 2 log 0.3

Now, log_0.09 (x – 1) = log (x – 1) / log 0.09 = log (x – 1) / 2 log 0.3

= ½. iog_0.3 (x – 1)        [or, we can use log _a^p (b) = 1/p. log _a b]

Given, log _0.3 (x – 1) < ½. log _0.3 (x – 1)

=> 2 log _0.3 (x – 1) < log _0.3 (x – 1) => log _0.3 (x – 1) < 0

=> (x – 1) > (0.3)⁰

[since the base is less than 1, so the inequality sign gets reversed]

⇒(x – 1) > 1 => x > 2

⇒So, the interval is (2, ∞) => (opt a)

 In this post you have learnt various types of important concept building problems on logarithm involving inequality, range of values, trigonometry and others.

We will carry on from here to the next set of questions on logarithm in the coming post, which will give you an exposure to more varieties of problems much needed for cracking CAT.

This article was contributed by Sandip Gupta.

CAT Exam Questions based on Logarithms

Q1: If log (2^a × 3^b × 5^c) is the arithmetic mean of log (2^2 × 3^3 × 5), log (2^6 × 3 × 5^7), and log(2 × 3^2 × 5^4), then a equals
Check answer of logarithm Q1 from CAT

Q2: If x is a real number such that log(base 3)5 = log(base 5)(2 + x), then which of the following is true?
Check answer of logarithm Q2 from CAT

Q3: The value of log (base 0.008) √5 + log (base√3) 81 – 7 is equal to
Check answer of logarithm Q3 from CAT

Q4: Suppose, log(base3)x = log(base12)y = a, where x, y are positive numbers. If G is the geometric mean of x and y, and log(base6)G is equal to
Check answer of logarithm Q4 from CAT

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