*Saturday, November 7th, 2020*

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,

- Write down all the formulae and try to remember them,
- 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.

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

Note: In the following section,

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

**Problem 1. **Find the least integer n, such that 7^{n }> 10^{5}, given log_{10}343 = 2.53?

a. 4

b. 5

c. 6

d. 7

**Solution: **(option c is the correct answer)

log 343 = log 7^{3} = 3 log 7 = 2.53 (given). So, log 7 = (2.53) / 3 …. (i)

now, 7^{n} > 10^{5 } or, log (7^{n}) > log (10^{5}), or, n. log 7 > 5. log 10

or, n. (2.53) / 3 > 5, (as log_{10 }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_{b}a. log_{c}a – log_{a}a) + (log_{a}b. log_{c}b – log_{b}b) + (log_{a}c. log_{b}c – log_{c}c) = 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, [(x^{2} / yz) – 1] + [(y^{2} / zx) – 1] + [(z^{2 }/ xy) – 1] = 0

or, [x^{2} / yz] + [y^{2} / zx ]+ [z^{2 }/ xy] = 3, or, (x^{3} + y^{3} + z^{3}) / (xyz) = 3,

or, x^{3} + y^{3} + z^{3} = 3xyz, or, x^{3 }+ y^{3 }+ z^{3 }– 3xyz = 0,

or, (x + y + z). (x^{2 }+ y^{2 }+z^{2 }-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. **log_{20}2 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, 2^{4 }= 16, 2^{5} = 32]

We know, 16 < 20 < 32 => 2^{4} < 20 < 2^{5}

=> log (2^{4}) < log 20 < log 2^{5}, (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 log_{20}20 = 1)

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

=> 1/5 < log _{20}2 < 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 ….. **∞**)} = x^{2} (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} = 2^{3}, 2^{-p} = 2^{3}**

**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 [(2^{4}) / (3×5)] + 5. log [(5^{2}) / (3×2^{3})] + 3. log [(3^{4}) / (2^{4}x5)]

= 7 (log 2^{4} – log 3 – log 5) + 5 (log 5^{2} – log 3 – log 2^{3}) + 3 (log 3^{4} – log2^{4} – 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 [(2^{4}) / (3×5)] + 5. log [(5^{2}) / (3×2^{3})] + 3. log [(3^{4}) / (2^{4}x5)]

= log [(2^{4}) / (3×5)]^{7} + log [(5^{2}) / (3×2^{3})]^{5 }+ log [(3^{4}) / (2^{4}x5)]^{3 }

= log [(2^{28}) / (3 ^{7}x 5 ^{7})] + log [(5^{10}) / (3^{5 }x 2^{15})] + log [(3^{12}) / (2^{12 }x 5^{3})]

= log [(2^{28 }x 5^{10 }x 3^{12}) / (3^{7 }x 5^{7 }x 3^{5 }x 2^{15 }x 2^{12 }x 5^{3})

= log [(2^{28 }x 5^{10 }x 3^{12}) / (2^{27 }x 3^{12 }x 5^{10})]

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

**Problem 6****. **If a^{2} + b^{2} = 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, a^{2 }+ b^{2} = 7ab, or, a^{2} + b^{2 }+ 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 log_{10}x – log_{x}0.01, 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_{n}m = (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)^{log}_{10}^{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 10^{2} + log x

=> a^{2} = 2 log 10 + a [where a = log x] => a^{2} = 2 + a ** **

=> a^{2 }– a – 2 = 0, => a^{2 }+ a – 2a – 2 = 0 => a(a +1) – 2(a +1) = 0,

=> (a + 1) (a – 2) = 0 => a = -1, 2

When a = -1, log_{10}x = – 1 => x = 10^{-1 }= 1/10

When a = 2, log_{10}x = 2 => x = 10^{2} = 100

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

**Problem 9****. **If log_{30 }3 = p and log_{30 }5 = q, then, how is log_{30 }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 = 2^{3} and 2 = 30/15 = 30 / (3×5)]

[All the logarithms are taken to the base 30]

log 8 = log 2^{3 }= 3 log 2 = 3 log (30/15) = 3 [log 30 – log(3×5)]

= 3[1 – log 3 – log 5] (since, log_{30 }30 = 1)

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

**Problem 10. ** The value of log _{cos A} √[1 + (sec^{2 }A / cosec^{2} A)] is

a. 1

b. 0

c. log 2

d. -1

**Solution: **(option d is the correct answer)

sec^{2}A / cosec^{2} A = (1/cos^{2}A) / (1/sin^{2}A) = sin^{2}A / cos^{2}A = tan^{2}A

So, the given expression = log (base cos A) √(1 + tan^{2 }A)

= log (base cos A) √[sec^{2}A] = 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.

**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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