# Logarithms – CA Foundation Maths Study Material

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
Logac = b; Where a ≠ 1.

Types of Logarithm
(i) Natural Logarithm:
The Logarithm of a number to base “e” is called Natural Logarithm.
i.e. Logex
where x = a number
e = 2.7183

(ii) Common Logarithm:
Logarithm of a number to the base 10 is called common Logarithm.
i.e. Log10x
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 ab = c ⇔ Logac = b; Where a ≠ 1.
• axlogab = bx
• loga a = 1
• loga 1 = 0
• logb a = \frac{1}{\log _a b} ⇒ loga a.loga b = 1
• loga a = loga xlogx a =logx alogbx
• logb a = logx a.logy xlogz y….logb k
• logba = logbx. logx y.log y z…. logk a
• logba = $$\frac{\log _x a}{\log _x b}$$
• logba = $$\frac{\log _b x}{\log _a x}$$
• If logba = x Then
• log$$\frac{1}{b}$$a = -x
• logb$$\frac{1}{a}$$ = -x
• log$$\frac{1}{b}$$$$\frac{1}{a}$$ = +x
• loga(mn) = logam + logan
• loga(mnr….) = logam + logan + logar + ……….
• loga$$\left(\frac{m}{n}\right)$$ = logam – logan
• (i) logab(mn) = logam
• log a = n logam.
• log$$\frac{1}{b}$$ m = $$\frac{n}{b}$$logam
• If logam = logbm ⇒ a = b
• If logam = logan ⇒ 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

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 alogab.logbc.logcd.logdt. [1 Mark, Feb. 2007]
(a) t
(b) abcdt
(c) (a + b + c + d + t)
(d) None.
alogab.logbc.logcd.logdt
= a1logat = t1 = t
(a) is correct

Question 3.
If log10000x = $$\frac{-1}{4}$$, then x is given by: [1 Mark, Feb. 2007]
(a) 1/100
(b) 1/10
(c) 1/20
(d) None of these
log10000x = $$\frac{-1}{4}$$
∴ (10000)$$\frac{-1}{4}$$ = x
or (104)$$\frac{-1}{4}$$ = 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) $$\frac{3 b^2}{2 b+1}$$
log(2a – 3b) = log$$\frac{a}{b}$$
or 2a – 3b = a
or 2ab – 3b2 =a
or 2ab – a = 3b2
or a(2b – 1) = 3b2
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
$$\frac{1}{\log _{a b}(a b c)}+\frac{1}{\log _{b c}(a b c)}+\frac{1}{\log _{c a}(a b c)}$$
= logabc (abJ)c.ca) = logabc (abc)2
= 2 logabc (abc) = 2 × 1 = 2
∴ (c) is correct

Question 6.
Number of digits in the numeral for 264 [Given log 2 = 0.30103] : [1 Mark, Aug. 2007]
(a) 18 digits
(b) 19 digits
(c) 20 digits
(d) 21 digits
(c) Let x = 264
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) 3log102
(b) 7log103
(c) 3logez
(d) None

(a) is correct

Question 8.
If x =  then the value of n is: [1 Mark, Feb. 2008]
(a) $$\frac{1}{2}$$loge$$\frac{1+x}{1-x}$$
(b) loge$$\frac{1+x}{1-x}$$
(c) loge$$\frac{1-x}{1+x}$$
(d) loge$$\frac{1-x}{1+x}$$

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
log 144 = log(16 × 9)
= log 16 + log 9
= log 24 +log 32
= 4 log 2 + 2 log 3
(b) is correct
Tricks:- Go by choices.

Question 10.
If log2 [log3(log2 x)] = 1, then x equals ; [1 Mark, June 2008]
(a) 128
(b) 256
(c) 512
(d) None
log2[log3(log2x)] = 1
Tricks:- 2321 = x
or x = 2321 = 23×3 = 29 = 512
(c) is correct

Detail Method log2 [log3 (log2 x)] = 1
or log3 (log2 x) = 21 = 2
or log2x = 32
or log2 x = 9
or x = 29 = 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
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 a2 + b2 + 2ab = 16ab
or a2 + b2 = 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 = $$\frac{n+1}{n-1}$$
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.
log4(x2 + x) – log4(x + 1) = 2. find x. [1 Mark, June 2009]
(a) 16
(b) 0
(c) -1
(d) None of these
log4(x2 + x) – log4(x + 1) = 2
or log $$\left(\frac{x^2+x}{x+1}\right)$$ = 2
or $$\frac{x(x+1)}{x+1}$$ = 42
or x= 16
∴ (a) is correct
Tricks: Go by choices.

Question 14.
Find the value of [log10 $$\sqrt{25}$$ – log10(23) + log10(4)2]x: [1 Mark, Dec. 2009]
(a) x
(b) 10
(c) 1
(d) none
[log10 $$\sqrt{25}$$ – log10(23) + log10(4)2]x = [log10($$\frac{5}{8}$$ × 16)]x
= (log1010)x = 1x = 1
∴ (c) is correct

Question 15.
If loga b + loga c = 0 then: [1 Mark, June 2010]
(a) b = c
(b) b = -c
(c) b = c = 1
(d) b and c are reciprocals.
logab + logac = 0
or loga(bc) = loga1
∴ bc = 1
b = $$\frac{1}{c}$$
∴ (d) is correct

Question 16.
The value of 2log x + 2 log x2 + 2 log x3 + ………. + 2 log xn will be: [1 Mark, Dec. 2010]
(a) $$\frac{n(n+1) \log x}{2}$$
(b) n(n + 1)log x
(c) n2 log x
(d) None of these
Detail Method:
2 log x + 2 log x2 + 2logx3 + ………….. + 2 log xn
=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) 102
(c) 10
(d) 103
$$\frac{\log _x 10-3}{2}+\frac{11-\log _x 10}{3}$$ = 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
Given n = m !
$$\frac{1}{\log _2 n}+\frac{1}{\log _3 n}+\frac{1}{\log _4 n}+\ldots \ldots \ldots \ldots+\frac{1}{\log _m n}$$
= logn2 + logn3 + logn4 + ………… + lognm
= logn(2. 3.4 ……………….. m)
= logn (1 . 2. 3. 4 …………….. m) = log(ml) (ml) = 1
∴ (a) is correct

Question 19.
If log2x + log4x = 6, then the value of x is: [1 Mark, 2011 Dec.]
(a) 16
(b) 32
(c) 64
(d) 128
(a) is correct
Tricks: Go by choices for (a) if x = 16
L.H.S = log216 + log4 16 = 4 + 2 = 6(R.H.S)
∴ (a) is correct

Detail Method log2x + log4x = 6
or log2x + log22 x = 6
or log2x + $$\frac{1}{2}$$log2x = 6
or (1 + $$\frac{1}{2}$$)log2x = 16
or log2x = $$\frac{6 \times 2}{3}$$ = 4
∴ x = 24 = 16

Question 20.
If logxY= 100 and log2 x = 10, then the value of‘Y’
(a) 210
(b) 2100
(c) 21,000
(d) 210,000
(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
(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 = logx 2 then x = [1 Mark, June 2013]
(a) 16
(b) 32
(c) 8
(d) 4
(a) is correct (log$$\sqrt{x}$$ 2)2 = logx2
or (logx1/2 2)2 = logx 2
or ($$\frac{1}{\frac{1}{2}}$$logx2)2 = logx2
or 4(logx2)2 – logx2 = 0
or 4(logx2)2 – logx2 = 0
or logx2[4logx – 1] = 0
If logx2 = 0 (Invalid) 4logx2 – 1 = 0
or 4logx2 = 1
or logx2 = $$\frac{1}{4}$$
or x1/4 = 2 ⇒ x = 24 = 16

Tricks Go by choices
For (a) LHS = (log^ 2)2 = (log4 2)2
RHS = log16 2 = log24 2 = $$\frac{1}{4}$$log22 = $$\frac{1}{4}$$
(a) is correct
Note Never write; check mentally.

Question 23.
Find Value of [logyx.logzy.logx z]3 =
(a) 0
(b) -1
(c) 1
(d) 3
(c) is correct [logyx.logzy.logx z]3
= [logxx] = [1]3 = 1

Question 24.
Find the value of Log49.Log32 = [1 Mark, Dec. 2013]
(a) 3
(b) 9
(c) 2
(d) 1
(d) is correct Log49.Log32 = l0g(2)2 (32).log32
= $$\frac{2}{2}$$log23.log32
= 1 × 1 = 1

Question 25.
If X = log2412; y = log36 24; z = log48 36 then xyz + 1 = ? [1 Mark, June 2014]
(a) 2xy
(b) 2zx
(c) 2yz
(d) 2
(c) is correct xyz + 1 = log2412. log36 24. log48 36+1
= log4812 + log48 48
= log48 (12 × 48)
= log48(12 × 2)2
= 2log48 24
= 2 log36 24. log48 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)

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
(a) is correct
log x = a+b ; log y = a-b.
log$$\left(\frac{10 x}{y^2}\right)$$ = log1010 + log x – log y2
= 1 + a + b – 2log y = 1 + a + b – 2 (a – b)
= 1 + a + b – 2a + 2b = 1 – a + 3b

Question 28.
If x = 1 + logp qr, y = 1 + logq rp and z = 1 + logr 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

(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
(a) If log x = m + n; log y = m – n
Then log$$\left(\frac{10 x}{y^2}\right)$$
= log 10 + log x – log y2
= 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.
log3 5. × log5 4 × log2 3: [1 Mark, Dec. 2015]
(a) 2
(b) 5
(c) -2
(d) None of these
(a) is correct log3 5. log5 4. log2 3
= log3 4. log2 3
= log2 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
(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
(b) is correct. log60 3 + log60 4 + log60 5
= log60(3 × 4 × 5) + log2 60 = 1

Question 33.
If log4(x2 + x) – log4(x + 1) = 2 then the value of x is
(a) 2
(b) 3
(c) 16
(d) 8
(c) is correct.

or log4x – 2 ⇒ x = 42 = 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
(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 (13 + 23 + 33 + ………. + n3) = .[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
log (13 + 23 + 33 + …………… + n3)
= 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 log3[log4(logxx)] = 0 then X =
(a) 4
(b) 8
(c) 16
(d) 32
(c)
Tricks: GBC
for option (c) log3[log4(log2x)]
= log3[log4(log216)]
= log3(log44) = log31 = 0
∴ (c) is correct.

Question 37.
If log$$\left(\frac{x-y}{2}\right)=\frac{1}{2}$$(log x + log y) then x2 + y2 = _______. [1 Mark, Dec. 2017]
(a) 6 xy
(b) 2xy
(c) 3x2y2
(d) 4x2y2

or; x2 + y2 – 2xy = 4xy
or; x2 + y2 =6xy

Question 38.
If log2(3√2) = $$\frac{1}{15}$$ then x = . [1 Mark, June 2018]
(a) 2
(b) 8
(c) 16
(d) 32
(d) log2(3√2) = $$\frac{1}{15}$$
or x1/15 = 3√2 = 21/15
or x = (21/15)15 = 25 = 32

Question 39.
The value of the expression :
alogab.logb c. logcd. logdt. [1 Mark, Nov. 2018]
(a) t
(b) abcdt
(c) (a + b + c + d +1)
(d) None
alogab.logb c. logcd. logdt
= alogat = a1.logat = t1 = t
[Formula axlogab = bx]

Question 40.
log2 log2 log216 = ? [1 Mark, Nov. 2018]
(a) 0
(b) 3
(c) 1
(d) 2
(c) log2 log2 log216 = log2 log24 = log22 = 1

Question 41.
The value of
log5(1+$$\frac{1}{5}$$) + log5 (1 + $$\frac{1}{6}$$) + ……… + log5 (1 + $$\frac{1}{624}$$)
(a) 2
(b) 3
(c) 5
(d) 0

Question 42.
log2$$\sqrt{2}$$ (512): log3$$\sqrt{2}$$ 324 = .[1 Mark, June 2019]
(a) 128:61
(b) 2:3
(c) 3 :2
(d) None
log2$$\sqrt{2}$$ (512) = 5 + 1
Similarly For log3$$\sqrt{2}$$ 324
log3$$\sqrt{2}$$324 value = 3 + 1=4
So; log2$$\sqrt{2}$$512:log3$$\sqrt{2}$$324