This Logarithms – CA Foundation Maths Study Material is designed strictly as per the latest syllabus and exam pattern.

## Logarithms – CA Foundation Maths Study Material

[Note : See You tube “Kailash Thakur” for better understanding]

If ab = c Where a ≠ 1 and a; c > o (positive)

Then b is said to be the logarithm of the number c to the base “a” and expressed as

Log_{a}c = b; Where a ≠ 1.

**Types of Logarithm**

(i) Natural Logarithm:

The Logarithm of a number to base “e” is called Natural Logarithm.

i.e. Log_{e}x

where x = a number

e = 2.7183

(ii) Common Logarithm:

Logarithm of a number to the base 10 is called common Logarithm.

i.e. Log_{10}x

where x = A number

Note: If base is not given then in arithmetical or commercial work; base is always taken as 10.

**Remember Some Formulae**

- If a
^{b}= c ⇔ Log_{a}c = b; Where a ≠ 1. - a
^{xlogab}= b^{x} - log
_{a}a = 1 - log
_{a}1 = 0 - log
_{b}a = \frac{1}{\log _a b} ⇒ log_{a}a.log_{a}b = 1 - log
_{a}a = log_{a}xlog_{x}a =log_{x}alog_{b}x - log
_{b}a = log_{x}a.log_{y}xlog_{z}y….log_{b}k - log
_{b}a = log_{b}x. log_{x}y.log_{y}z…. log_{k}a - log
_{b}a = \(\frac{\log _x a}{\log _x b}\) - log
_{b}a = \(\frac{\log _b x}{\log _a x}\) - If log
_{b}a = x Then- log
_{\(\frac{1}{b}\)}a = -x - log
_{b}\(\frac{1}{a}\) = -x - log
_{\(\frac{1}{b}\)}\(\frac{1}{a}\) = +x

- log
- log
_{a}(mn) = log_{a}m + log_{a}n - log
_{a}(mnr….) = log_{a}m + log_{a}n + log_{a}r + ………. - log
_{a}\(\left(\frac{m}{n}\right)\) = log_{a}m – log_{a}n - (i) log
_{a}b(m^{n}) = log_{a}m - log a = n log
_{a}m. - log
_{\(\frac{1}{b}\)}m = \(\frac{n}{b}\)log_{a}m - If log
_{a}m = log_{b}m ⇒ a = b - If log
_{a}m = log_{a}n ⇒ m = n

**Previous Year Exam Questions**

Question 1.

7log\(\left(\frac{16}{15}\right)\) + 5log\(\left(\frac{25}{24}\right)\) + 3log\(\left(\frac{81}{80}\right)\) is equal to

(a) 0

(b) 1

(c) log 2

(d) log 3

Answer:

Calculation Tricks

I Type i6 ÷ 15 × button Then push = button 6 time then push M+ button.

II Type 25 ÷ 24 × button Then push = button 4 times

III Then Push × button Then MRC button 2 time Then Push M+ button

IV Type 81 ÷ 80 × button Then = button 2 times

V Then Push × button and Then MRC button 2 times Then = button

We get; it is approx 2

So. Value = log 2

(c) is Correct

Question 2.

The value of the expression a^{logab.logbc.logcd.logdt}. [1 Mark, Feb. 2007]

(a) t

(b) abcdt

(c) (a + b + c + d + t)

(d) None.

Answer:

a^{logab.logbc.logcd.logdt}

= a^{1logat} = t^{1} = t

(a) is correct

Question 3.

If log_{10000}x = \(\frac{-1}{4}\), then x is given by: [1 Mark, Feb. 2007]

(a) 1/100

(b) 1/10

(c) 1/20

(d) None of these

Answer:

log_{10000}x = \(\frac{-1}{4}\)

∴ (10000)^{\(\frac{-1}{4}\)} = x

or (10^{4})^{\(\frac{-1}{4}\)} = x

or x = 10^{-1} = \(\frac{1}{10}\)

(b) is correct

Question 4.

If log (2a – 3b) = log a – log b, then a = ? [1 Mark, May 2007]

(a) \(\frac{3 b^2}{2 b-1}\)

(b) \(\frac{3 b}{2 b-1}\)

(c) \(\frac{b^2}{2 b+1}\)

(d) \(\frac{3 b^2}{2 b+1}\)

Answer:

log(2a – 3b) = log\(\frac{a}{b}\)

or 2a – 3b = a

or 2ab – 3b^{2} =a

or 2ab – a = 3b^{2}

or a(2b – 1) = 3b^{2}

or a = \(\frac{3 b^2}{2 b-1}\)

(a) is correct

Question 5.

\(\frac{1}{\log _{a b}(a b c)}+\frac{1}{\log _{b c}(a b c)}+\frac{1}{\log _{c a}(a b c)}\) is equal to: [1 Mark, Aug. 2007]

(a) 0

(b) 1

(c) 2

(d) -1

Answer:

\(\frac{1}{\log _{a b}(a b c)}+\frac{1}{\log _{b c}(a b c)}+\frac{1}{\log _{c a}(a b c)}\)

= log_{abc} (abJ)c.ca) = log_{abc} (abc)2

= 2 log_{abc} (abc) = 2 × 1 = 2

∴ (c) is correct

Question 6.

Number of digits in the numeral for 2^{64} [Given log 2 = 0.30103] : [1 Mark, Aug. 2007]

(a) 18 digits

(b) 19 digits

(c) 20 digits

(d) 21 digits

Answer:

(c) Let x = 2^{64}

log x= 64 log 2

= 64 × 0.30103

= 19.26592

x = AL (19.26592) characteristics = 19

∴ No. of digits in the number = 19 + 1 = 20

∴ (c) is correct

Question 7.

The value \(\frac{\log _3 8}{\log _9 16 . \log _4 10}\) is: [1 Mark, Nov. 2007]

(a) 3log_{10}2

(b) 7log_{10}3

(c) 3log_{e}z

(d) None

Answer:

(a) is correct

Question 8.

If x = \(\) then the value of n is: [1 Mark, Feb. 2008]

(a) \(\frac{1}{2}\)log_{e}\(\frac{1+x}{1-x}\)

(b) log_{e}\(\frac{1+x}{1-x}\)

(c) log_{e}\(\frac{1-x}{1+x}\)

(d) log_{e}\(\frac{1-x}{1+x}\)

Answer:

Question 9.

log 144 is equal to:

(a) 2 log 4 + 2 log 2

(b) 4 log 2 + 2 log 3

(c) 3 log 2 + 4 log 3

(d) 3 log 2 -4 log 3

Answer:

log 144 = log(16 × 9)

= log 16 + log 9

= log 2^{4} +log 3^{2}

= 4 log 2 + 2 log 3

(b) is correct

Tricks:- Go by choices.

Question 10.

If log_{2} [log_{3}(log_{2} x)] = 1, then x equals ; [1 Mark, June 2008]

(a) 128

(b) 256

(c) 512

(d) None

Answer:

log_{2}[log_{3}(log_{2}x)] = 1

Tricks:- 2^{321} = x

or x = 2^{321} = 2^{3×3} = 2^{9} = 512

(c) is correct

Detail Method log_{2} [log_{3} (log_{2} x)] = 1

or log_{3} (log_{2} x) = 2^{1} = 2

or log_{2}x = 3^{2}

or log_{2} x = 9

or x = 2^{9} = 512

(c) is correct

Question 11.

If log\(\left(\frac{a+b}{4}\right)=\frac{1}{2}\) (log a + log b) then: \(\frac{a}{b}+\frac{b}{a}\)

(a) 12

(b) 14

(c) 16

(d) 8

Answer:

log\(\left(\frac{a+b}{4}\right)=\frac{1}{2}\) (log a + log b)

or log\(\left(\frac{a+b}{4}\right)\) = log(ab)^{1/2}

or \(\frac{a+b}{4}=\sqrt{a b}\)

or a + b = 4\(\sqrt{a b}\)

Squaring on both sides; we get (a + b)^{2} = 16ab

or a^{2} + b^{2} + 2ab = 16ab

or a^{2} + b^{2} = 14ab

or \(\frac{a^2}{a b}+\frac{b^2}{a b}=\frac{14 a b}{a b}\) [Dividing by ab on both sides]

or \(\frac{a}{b}+\frac{b}{a}\) = 14

(b) is correct

Question 12.

log (m+n) = log m+ log n, m can be expressed as : [1 Mark, June 2009]

(a) m = \(\frac{n}{n-1}\)

(b) m = \(\frac{n}{n+1}\)

(c) m = \(\frac{n+1}{n}\)

(d) m = \(\frac{n+1}{n-1}\)

Answer:

log(m + n) = log m + log n

or log (m + n) = log(mn)

or m + n = mn

or m – mn = -n

or m (1 – n) = -n

or m = \(\frac{-n}{1-n}=\frac{n}{n-1}\)

∴ (a) is correct

Tricks Go by choices.

Question 13.

log_{4}(x^{2} + x) – log_{4}(x + 1) = 2. find x. [1 Mark, June 2009]

(a) 16

(b) 0

(c) -1

(d) None of these

Answer:

log_{4}(x^{2} + x) – log_{4}(x + 1) = 2

or log \(\left(\frac{x^2+x}{x+1}\right)\) = 2

or \(\frac{x(x+1)}{x+1}\) = 4^{2}

or x= 16

∴ (a) is correct

Tricks: Go by choices.

Question 14.

Find the value of [log_{10} \(\sqrt{25}\) – log_{10}(2^{3}) + log_{10}(4)^{2}]^{x}: [1 Mark, Dec. 2009]

(a) x

(b) 10

(c) 1

(d) none

Answer:

[log_{10} \(\sqrt{25}\) – log_{10}(2^{3}) + log_{10}(4)^{2}]^{x} = [log_{10}(\(\frac{5}{8}\) × 16)]^{x}

= (log_{10}10)^{x} = 1^{x} = 1

∴ (c) is correct

Question 15.

If log_{a} b + log_{a} c = 0 then: [1 Mark, June 2010]

(a) b = c

(b) b = -c

(c) b = c = 1

(d) b and c are reciprocals.

Answer:

log_{a}b + log_{a}c = 0

or log_{a}(bc) = log_{a}1

∴ bc = 1

b = \(\frac{1}{c}\)

∴ (d) is correct

Question 16.

The value of 2log x + 2 log x^{2} + 2 log x^{3} + ………. + 2 log x^{n} will be: [1 Mark, Dec. 2010]

(a) \(\frac{n(n+1) \log x}{2}\)

(b) n(n + 1)log x

(c) n^{2} log x

(d) None of these

Answer:

Detail Method:

2 log x + 2 log x^{2} + 2logx^{3} + ………….. + 2 log x^{n}

=2 log x + 2.2 logx + 2.3logx + ……….. + 2.n.logx

= 2 logx. [1 + 2 + 3 + ………. + n]

= 2 log x. \(\frac{n(n+1)}{2}\) = n(n+1)

= (b) is correct

Tricks Put n = 2 in options directly.

This should be equal to sum of 1st 2 terms = 2 logx + 2.2logx = 6 logx

Which gives option (b)

∴ (b) is correct.

Question 17.

Solve: \(\frac{\log _x 10-3}{2}+\frac{11-\log _x 10}{3}\) = 2. [1 Mark, Dec. 2010, June 2011]

(a) 10^{-1}

(b) 10^{2}

(c) 10

(d) 10^{3}

Answer:

\(\frac{\log _x 10-3}{2}+\frac{11-\log _x 10}{3}\) = 2

Tricks: Go by choice [Do Mentally]

For (a) x = 10^{-1}

L.H.S = \(\frac{\log _{\left(10^{-1}\right)} 10-3}{2}+\frac{11-\log _{10^{-1}} 10}{3}\)

= \(\frac{\log _{\left(10^{-1}\right)} 10-3}{2}+\frac{11-\log _{10^{-1}} 10}{3}\)

= -2 + 4 = 2 = (R.H.S)

∴ (a) is correct

Question 18.

If n = m! where (‘m’ is a positive integer > 2) then the value of ∴ \(\frac{1}{\log _2 n}+\frac{1}{\log _3 n}+\frac{1}{\log _4 n}+\ldots \ldots \ldots \ldots+\frac{1}{\log _m n}\). [1 Mark, June 2011]

(a) 1

(b) 0

(c) -1

(d) 2

Answer:

Given n = m !

\(\frac{1}{\log _2 n}+\frac{1}{\log _3 n}+\frac{1}{\log _4 n}+\ldots \ldots \ldots \ldots+\frac{1}{\log _m n}\)

= log_{n}2 + log_{n}3 + log_{n}4 + ………… + log_{n}m

= log_{n}(2. 3.4 ……………….. m)

= log_{n} (1 . 2. 3. 4 …………….. m) = log(ml) (ml) = 1

∴ (a) is correct

Question 19.

If log_{2}x + log_{4}x = 6, then the value of x is: [1 Mark, 2011 Dec.]

(a) 16

(b) 32

(c) 64

(d) 128

Answer:

(a) is correct

Tricks: Go by choices for (a) if x = 16

L.H.S = log_{2}16 + log_{4} 16 = 4 + 2 = 6(R.H.S)

∴ (a) is correct

Detail Method log_{2}x + log_{4}x = 6

or log_{2}x + log_{2}2 x = 6

or log_{2}x + \(\frac{1}{2}\)log_{2}x = 6

or (1 + \(\frac{1}{2}\))log_{2}x = 16

or log_{2}x = \(\frac{6 \times 2}{3}\) = 4

∴ x = 2^{4} = 16

Question 20.

If log_{x}Y= 100 and log_{2} x = 10, then the value of‘Y’

(a) 2^{10}

(b) 2^{100}

(c) 2^{1,000}

(d) 2^{10,000}

Answer:

(c) log 2 x = 10 ∴ x = 210

Now logxy = 100 y = ;

c100 y = (210)100 =21000 (c) is correct

Question 21.

Which of the following is true. If \(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{1}{a b c}\). [1 Mark, Dec.2012]

(a) log (ab + bc +ca) =abc

(b) log \(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) =abc

(c) log (abc) = 0

(d) log(a+b + c) = 0

Answer:

(a) is correct \(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}=\frac{1}{a b c}\)

Multiplying both sides by abc; \(\frac{a b c}{a b}+\frac{a b c}{b c}+\frac{a b c}{c a}=\frac{a b c}{a b c}\)

or; c + a + b = 1

or; a + b + c= 1

Taking log on both sides ; we get

log (a + b + c) = log 1 = 0

Question 22.

If (log_{\(\sqrt{x}\) } 2)^{2} = log_{x} 2 then x = [1 Mark, June 2013]

(a) 16

(b) 32

(c) 8

(d) 4

Answer:

(a) is correct (log_{\(\sqrt{x}\) } 2)^{2} = log_{x}2

or (log_{x1/2} 2)^{2} = log_{x} 2

or (\(\frac{1}{\frac{1}{2}}\)log_{x}2)^{2} = log_{x}2

or 4(log_{x}2)^{2} – log_{x}2 = 0

or 4(log_{x}2)2 – log_{x}2 = 0

or log_{x}2[4log_{x} – 1] = 0

If log_{x}2 = 0 (Invalid) 4log_{x}2 – 1 = 0

or 4log_{x}2 = 1

or log_{x}2 = \(\frac{1}{4}\)

or x^{1/4} = 2 ⇒ x = 2^{4} = 16

Tricks Go by choices

For (a) LHS = (log^ 2)^{2} = (log4 2)^{2}

RHS = log_{16} 2 = log_{24} 2 = \(\frac{1}{4}\)log_{2}2 = \(\frac{1}{4}\)

(a) is correct

Note Never write; check mentally.

Question 23.

Find Value of [log_{y}x.log_{z}y.log_{x} z]^{3} =

(a) 0

(b) -1

(c) 1

(d) 3

Answer:

(c) is correct [log_{y}x.log_{z}y.log_{x} z]^{3}

= [log_{x}x] = [1]^{3} = 1

Question 24.

Find the value of Log_{4}9.Log_{3}2 = [1 Mark, Dec. 2013]

(a) 3

(b) 9

(c) 2

(d) 1

Answer:

(d) is correct Log_{4}9.Log_{3}2 = l0g_{(2)2} (3^{2}).log_{3}2

= \(\frac{2}{2}\)log_{2}3.log_{3}2

= 1 × 1 = 1

Question 25.

If X = log_{24}12; y = log_{36} 24; z = log_{48} 36 then xyz + 1 = ? [1 Mark, June 2014]

(a) 2xy

(b) 2zx

(c) 2yz

(d) 2

Answer:

(c) is correct xyz + 1 = log_{24}12. log_{36} 24. log_{48} 36+1

= log_{48}12 + log_{48} 48

= log_{48} (12 × 48)

= log_{48}(12 × 2)^{2}

= 2log_{48} 24

= 2 log_{36} 24. log_{48} 36

= 2yz

Question 26.

If x2 + y2 = 7xy then log \(\frac{1}{3}\)(x + y) = [1 Mark, June 2014]

(a) log x + log y

(b) \(\frac{1}{2}\)(log x + log y)

(c) \(\frac{1}{3}\)(log x + log y)

(d) \(\frac{1}{3}\)(log x. log y)

Answer:

Question 27.

If log x= a – b; log y = a + b then log\(\left(\frac{10 x}{y^2}\right)\)

(a) 1- a – 3b

(b) a – 1 + 3b

(c) a + 3b – 1

(d) 1 – b + 3a

Answer:

(a) is correct

log x = a+b ; log y = a-b.

log\(\left(\frac{10 x}{y^2}\right)\) = log_{10}10 + log x – log y^{2}

= 1 + a + b – 2log y = 1 + a + b – 2 (a – b)

= 1 + a + b – 2a + 2b = 1 – a + 3b

Question 28.

If x = 1 + log_{p} qr, y = 1 + log_{q} rp and z = 1 + log_{r} pq ; then the value of \(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}\) = ……….

(a) 0

(b) 1

(c) -1

(d) 3

Answer:

(A) is correct 1 1 1

Tricks:- Cyclic order

So, \(\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}+\frac{1}{\mathrm{z}}\) = 1

(See Quicker BMLRS)

Question 29.

If log x = m + n; log y = m – n then log\(\left(\frac{10 x}{y^2}\right)\) =

(a) 1 – m + 3n

(b) m – 1 + 3n

(c) m + 3n + 1

(d) None

Answer:

(a) If log x = m + n; log y = m – n

Then log\(\left(\frac{10 x}{y^2}\right)\)

= log 10 + log x – log y^{2}

= 1 + log x – 2 log y

= 1 + (m + n) – 2(m – n)

= 1 + m + n – 2m + 2n

= 1 – m + 3n

∴ (a) is correct.

Question 30.

log_{3} 5. × log_{5} 4 × log_{2} 3: [1 Mark, Dec. 2015]

(a) 2

(b) 5

(c) -2

(d) None of these

Answer:

(a) is correct log_{3} 5. log_{5} 4. log_{2} 3

= log_{3} 4. log_{2} 3

= log_{2} 4 = 2

Question 31.

The integral part of a logarithm is called __________, and the decimal part of a logarithm is called __________. [1 Mark, June 2016]

(a) Mantissa, Characteristic

(b) Characteristic, Mantissa

(c) Whole, Decimal

(d) None of these

Answer:

(b) is correct.

Question 32.

The value of \(\frac{1}{\log _3 60}+\frac{1}{\log _4 60}+\frac{1}{\log _5 60}\) = __________. [1 Mark, June 2016]

(a) 0

(b) 1

(c) 5

(d) 60

Answer:

(b) is correct. log_{60} 3 + log_{60} 4 + log_{60} 5

= log_{60}(3 × 4 × 5) + log_{2} 60 = 1

Question 33.

If log_{4}(x^{2} + x) – log_{4}(x + 1) = 2 then the value of x is

(a) 2

(b) 3

(c) 16

(d) 8

Answer:

(c) is correct.

or log_{4}x – 2 ⇒ x = 4^{2} = 16

Question 34.

Given log 2 = 0.3010 and log 3 = 0.4771 then the value of log 24 [1 Mark, Dec. 2016]

(a) 1.3081

(b) 1.1038

(c) 1.3801

(d) 1.8301

Answer:

(c) is correct.

Calculator Tricks:

Type 24 then √ button 19 times – 1 x 227695 = button. We will get the required value of log 24.

Question 35.

log (1^{3} + 2^{3} + 3^{3} + ………. + n^{3}) = .[1 Mark, June 2017]

(a) 2 log n + 2 log (n + 1) – 2 log2

(b) log n + 2 log (n +1)-2 log2

(c) 2 log n + log (n +1)- 2 log 2

(d) None

Answer:

log (1^{3} + 2^{3} + 3^{3} + …………… + n^{3})

= log\(\left(\frac{n(n+1)}{2}\right)^2\) = 2log\(\frac{n(n+1)}{2}\)

= 2 [log n + log (n +1) – log 2]

= 2 log n + 2log(n+1) – 2log 2

So, (a) is correct

Tricks: Go by choices

Question 36.

If log_{3}[log_{4}(log_{x}x)] = 0 then X =

(a) 4

(b) 8

(c) 16

(d) 32

Answer:

(c)

Tricks: GBC

for option (c) log_{3}[log_{4}(log_{2}x)]

= log_{3}[log_{4}(log_{2}16)]

= log_{3}(log_{4}4) = log_{3}1 = 0

∴ (c) is correct.

Question 37.

If log\(\left(\frac{x-y}{2}\right)=\frac{1}{2}\)(log x + log y) then x^{2} + y^{2} = _______. [1 Mark, Dec. 2017]

(a) 6 xy

(b) 2xy

(c) 3x^{2}y^{2}

(d) 4x^{2}y^{2}

Answer:

or; x^{2} + y^{2} – 2xy = 4xy

or; x^{2} + y^{2} =6xy

Question 38.

If log_{2}(3√2) = \(\frac{1}{15}\) then x = . [1 Mark, June 2018]

(a) 2

(b) 8

(c) 16

(d) 32

Answer:

(d) log_{2}(3√2) = \(\frac{1}{15}\)

or x^{1/15} = 3√2 = 2^{1/15}

or x = (2^{1/15})^{15} = 2^{5} = 32

Question 39.

The value of the expression :

alog_{a}b.log_{b} c. log_{c}d. log_{d}t. [1 Mark, Nov. 2018]

(a) t

(b) abcdt

(c) (a + b + c + d +1)

(d) None

Answer:

alog_{a}b.log_{b} c. log_{c}d. log_{d}t

= a^{logat} = a^{1.logat} = t^{1} = t

[Formula a^{xlogab} = b^{x}]

Question 40.

log_{2} log_{2} log_{2}16 = ? [1 Mark, Nov. 2018]

(a) 0

(b) 3

(c) 1

(d) 2

Answer:

(c) log_{2} log_{2} log_{2}16 = log_{2} log_{2}4 = log_{2}2 = 1

Question 41.

The value of

log_{5}(1+\(\frac{1}{5}\)) + log_{5} (1 + \(\frac{1}{6}\)) + ……… + log_{5} (1 + \(\frac{1}{624}\))

(a) 2

(b) 3

(c) 5

(d) 0

Answer:

Question 42.

log_{2\(\sqrt{2}\) }(512): log_{3\(\sqrt{2}\)} 324 = .[1 Mark, June 2019]

(a) 128:61

(b) 2:3

(c) 3 :2

(d) None

Answer:

Calculator Tricks

log_{2\(\sqrt{2}\) }(512) = 5 + 1

Type 2 × 2√ button = button.

Then press x button then continue pressing = button until to get 512

Here = button has been pressed 5 times. So log value

= No. of = button pressings + 1

Similarly For log_{3\(\sqrt{2}\)} 324

Type 3 × 2√ button = button then

x = button 3 times ; we get

log_{3\(\sqrt{2}\)}324 value = 3 + 1=4

So; log_{2\(\sqrt{2}\)}512:log_{3\(\sqrt{2}\)}324

= 6: 4 = 3: 2