Logarithm MCQ | Multiple Choice Questions -

Last Updated on: 3rd January 2025, 12:40 pm

Logarithm MCQ

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1. Which one is correct?

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

(b) log (m – n) = log m – log n

(d) None of these.

log_ba x log_a b = log_aa x log_b b = 1 x 1 = 1. So, option (c) is correct

2. Find the value of log _ba . log_cb . log_a c

(a) abc

(b) 0

(d) 4.

log_b a. log_c b. log_a c = (log_ca. log_b b). log_a c = (log_c a x 1) log_a c

= log_c a x log_a c = log_a a. log_c c = 1 x 1 = 1.

So, option (c) is correct

3. If log 2 = 0.30103, then the value of log 200 is

(a) 18.0103

(b) 6.30103

(d) 2.30103

log 200 = log (2 x 100) = log 2 + log 100 = log 2 + 2 log 10

= 0.30103 + 2 X 1 = 2.30103 (we know log 2 = 0.30103, log 10 = 1).

So, option (d) is correct

4. log 144 is equal to

(a) 2 log 4 + 2 log 2

(b) 4 log 2 + 2 log 3

(d) None of these.

2 log 4 + 2 log 2 = log 4² + log 2² = log 16 + log 4 = log (16 x 4) = log 64
4 log 2 + 2 log 3 = log 2⁴ + log 3² = log 16 + log 9 = log (16 x 9) = log 144
3 log 2 + 4 log 3 = log 2³ + log 3⁴ = log 8 + log 81 = log (8 x 81) = log 648

So, option (b) is correct

5. The relation log_e x + log_e (1 + x) = 0 is equal to

(a) x²+ x + e = 2

(b) x²+ x – e = 3

(d) x²+ x + 4 = 0.

log_e x + log_e (1 + x) = 0, Or, log_e x(1 + x) = 0, Or, x(1 + x) = e⁰,

Or, x + x² = 1, Or, x²+ x – 1 = 0.

So, option (c) is correct

Find value of (½)log10 (25) – 2 log10 (3) +log10 (18)

_{6. Find value of (½)log10 (25) – 2 log10 (3) +log10 (18)}

(a) 0

(b) 1

(d) 2.

$\displaystyle \frac{1}{2}$ log₁₀25-2log₁₀3+log₁₀18=log₁₀25^1/2-log₁₀3²+log₁₀18=log₁₀5-log₁₀9+log₁₀18

(log₁₀5-log₁₀9)+log₁₀18=log₁₀( $\displaystyle \frac{5}{9}$ )+log₁₀18+=log₁₀ $\displaystyle \left[ {\frac{{5\times 18}}{9}} \right]$ =log₁₀10=1

So, option (b) is correct

7. If log₁₀2 = 0.3010, the value of log₁₀ 80 is

(a) 1.9030

(b) 1.6020

(d) 4.9030.

Log₁₀ 80=log₁₀ (8 X 10) = log₁₀ 8 + log₁₀ 10 = log₁₀ 2³ + 1 = 3₁₀ log 2 + 1

= 3 x 0.3010 + 1 = 0.9030 + 1 = 1.9030. So, option (a) is correct

8. Find the value of log₃ (27)³

(a) 27

(b) 9

(d) 81.

log₃ (27)³ = 3 log₃ (27)= 3 log₃3³= (3 X 3) log₃3= 9 log₃ 3= 9 X1 = 9 (we know log₃ 3= 1).

So, option (b) is correct

9. If log₃(log₂x) = 1, Find the value of x

(a) 10

(b) 9

(d) 8

log₃(log₂x) = 1, or Now, log₂x = 3¹= 3, Or, x = 2³= 8, So, option (d) is correct

10. If log₄ (x² + x) – log₄ (x + 1) = 2, then:

(a) x = 8

(b) x = 4

(d) x = 16.

log₄ (x² + x) – log₄ (x + 1) = 2, or, log₄ $\displaystyle \left[ {\frac{{\left( {{{x}^{2}}+x} \right)}}{{\left( {x+1} \right)}}} \right]$ =2.

Or, $\displaystyle \left[ {\frac{{\left( {{{x}^{2}}+x} \right)}}{{\left( {x+1} \right)}}} \right]$ = 4² = 16

Or, x²+ x = 16(x + 1), Or, x²+ x = 16x + 16,

Or, x²+ x – 16x – 16 = 0 , Or, x(x +1) – 16(x +1) = 0

Or, (x +1) (x -16), Either x = -1 or, x = 16,

So, option (d) is correct

11. Value of (log₁₀ 125 / log₁₀ 25) = ?

a. 5

b. $\displaystyle \frac{3}{2}.$

c. $\displaystyle \frac{1}{2}$

d. $\displaystyle \frac{1}{6}$

(log₁₀ 125 / log₁₀ 25)= (log₁₀ 5³ / log₁₀ 5²)

= 3 log₁₀ $\displaystyle \frac{5}{2}$ log₁₀ 5 = $\displaystyle \frac{3}{2}$ .

So, option (b) is correct

12. If log₈ m + log₈ 2= $\displaystyle \frac{2}{3}$ , then find the value of m

(a) 1

(b) 3

(d) 4

log₈ m + log₈ 2= $\displaystyle \frac{2}{3}$ , or, log₈ (m X 2) + =2/3, or, log₈ 2m = $\displaystyle \frac{2}{3}$ ,

So, 8 ⁽⁾ = 2m.

or, {2³}⁽⁾ = 2m.

Or, 2² = 2m, Or, m = 2.

So, option (c) is correct

13. Value of 7 log₁₀( $\displaystyle \frac{{16}}{{15}}$ )+ 5 log₁₀( $\displaystyle \frac{{25}}{{24}}$ )+ 3 log₁₀( $\displaystyle \frac{{81}}{{80}}$ )

(a) 2

(b) log₁₀2

(d) log₁₀ 5

7 log₁₀( $\displaystyle \frac{{16}}{{15}}$ )+ 5 log₁₀( $\displaystyle \frac{{25}}{{24}}$ )+ 3 log₁₀( $\displaystyle \frac{{81}}{{80}}$ )

= 7 {log₁₀16 – log₁₀15)} + 5 {log₁₀25 – log₁₀ 24} + 3 [{ (log₁₀ 3⁴) – log₁₀ (80}]

= 7 {log₁₀2⁴ – log₁₀(5 X 3)} + 5 {log₁₀5² – log₁₀(8X3)} } + 3 { log₁₀(3⁴)- log₁₀ (2⁴ X 5)}

= 7 { 4log₁₀2 – ( log₁₀5 + log₁₀3)} + 5 {2 log₁₀5- (log₁₀2³ + log₁₀3) } + 3 { 4log₁₀ 3– (4log₁₀ 2 + log₁₀ 5)}

= 28 log₁₀2 – 7 log₁₀5- 7 log₁₀ 3 +10 log₁₀ 5 – 15 log₁₀ 2 – 5 log₁₀ 3 + 12 log₁₀ 3 – 12 log ₁₀2- 3 log ₁₀ 5

= log ₁₀2. So, option (b) is correct

14. If log₁₀ p + log₁₀q = log₁₀ (p + q), which one of the following relations is true?

(a) p= $\displaystyle \frac{q}{{q-1}}$

(b) 3p=q=0

(d) p=q=2

log₁₀p + log₁₀q = log₁₀ (p + q), Or, log₁₀(pq) = log₁₀ (p + q),

Or, pq = p + q, Or, pq – p = q,

Or, p(q – 1) = q, or, p= $\displaystyle \frac{q}{{q-1}}$ .

So, option (a) is correct

15. If log $\displaystyle \frac{{\left( {x+y} \right)}}{5}$ = $\displaystyle \frac{1}{2}$ (logx + log y), find value of ( $\displaystyle \frac{x}{y}$ ) + $\displaystyle \frac{y}{x}$ )

(a) 20

(b) 23

(d) 18.

log $\displaystyle \frac{{x+y}}{5}$ = ½ (logx + log y), or, log $\displaystyle \frac{{x+y}}{5}$ = ½ log (xy), or, log $\displaystyle \frac{{x+y}}{5}$ = $\displaystyle \frac{1}{2}$ log (xy),

log $\displaystyle \frac{{x+y}}{5}$ = log [(xy) ^½].

Or, $\displaystyle \frac{{x+y}}{5}$ = (xy) ^½,

or {(x+y)/5}²= (xy), or, (x² + y² + 2xy) / 25=xy

or, (x² + y² + 2xy) =25xy, x² + y²=23xy, or, $\displaystyle \frac{{\left( {{{x}^{2}}+{{y}^{2}}} \right)}}{{xy}}$ =23,

Now, ( $\displaystyle \frac{x}{y}$ ) + $\displaystyle \frac{y}{x}$ ) = $\displaystyle \frac{{{{x}^{2}}+{{y}^{2}}}}{{xy}}$ = 23.

So, option (b) is correct

16. If log (2a – 3b) = log a – log b, then a =

(a) $\displaystyle \frac{{3{{b}^{2}}}}{{\left( {2b-1} \right)}}$

(b) $\displaystyle \frac{{3{{b}^{3}}}}{{\left( {2b-1} \right)}}$

(d) None of the above

log (2a – 3b) = log a – log b, or, log (2a – 3b) = log ( $\displaystyle \frac{a}{b}$ ). or, 2a-3b= $\displaystyle \frac{a}{b}$ , Or, 2ab – 3b² = a

Or, 2ab – a = 3b², Or, a(2b – 1) = 3b², or a= $\displaystyle \frac{{3{{b}^{2}}}}{{\left( {2b-1} \right)}}$ .

So, option (a) is correct

17. Find value of log ( $\displaystyle \frac{{{{a}^{2}}}}{{bc}}$ ) + log ( $\displaystyle \frac{{{{b}^{2}}}}{{ac}}$ ) + log ( $\displaystyle \frac{{{{c}^{2}}}}{{ab}}$ )

(a) 1

(b) 0

(d) log a

log ( $\displaystyle \frac{{{{a}^{2}}}}{{bc}}$ ) + log ( $\displaystyle \frac{{{{b}^{2}}}}{{ac}}$ ) + log ( $\displaystyle \frac{{{{c}^{2}}}}{{ab}}$ )

= (log a²– log bc) + (log b²– log ac) + (log c²– log ab)

= log a²+ log b²+ log c²– (log bc + log ac + log ab)

= log a²b²c²– log (bc. .ab)

= log a²b²c²– log a²b²c² = 0. So, option (b) is correct

18. If log ( $\displaystyle \frac{m}{n}$ ) + log ( $\displaystyle \frac{n}{m}$ ) = log (m+n) then find true statement

(a) m+n=1

(b) $\displaystyle \frac{m}{n}$ =2

(d) m² – n² = 4

log (m/n) + log (n/m) = log (m+n), or log {( $\displaystyle \frac{m}{n}$ ) X ( $\displaystyle \frac{n}{m}$ )} = log (m+n), or log1=log (m+n). or m+n=1.

So, option (a) is correct

19. If log_a3= $\displaystyle \frac{1}{3}$ , then find value of a

(a) 27

(b) 81

(d) 24.

log_a3= $\displaystyle \frac{1}{3}$ , So, a^1/3=3, or a=(3)³ = 27. So, option (a) is correct

20. Find the value of log₂ ( $\displaystyle \frac{1}{{64}}$ )

(a) 6

(b) -6

(d) 4.

log₂ ( $\displaystyle \frac{1}{{64}}$ ) = log₂ ( $\displaystyle \frac{1}{{{{2}^{6}}}}$ ) = log₂ 1 – 6log₂2 = 0 – (6X 1) = -6 (as log₂2= 1).

So, option (b) is correct

21. log_x( $\displaystyle \frac{{16}}{{25}}$ ) = – $\displaystyle \frac{1}{2}$ , Find value of x

(a) $\displaystyle \frac{{625}}{{256}}$

(b) $\displaystyle \frac{{256}}{{625}}$

(d) $\displaystyle \frac{{32}}{{50}}$

log_x( $\displaystyle \frac{{16}}{{25}}$ )=- $\displaystyle \frac{1}{2}$ . Or x^(-1/2) = $\displaystyle \frac{{16}}{{25}}$ , or 1/{x^(1/2)} = $\displaystyle \frac{{16}}{{25}}$ , or x^(1/2)} = $\displaystyle \frac{{25}}{{16}}$ , or x= $\displaystyle \frac{{{{{25}}^{2}}}}{{{{{16}}^{2}}}}$ = $\displaystyle \frac{{625}}{{256}}$ .

So, option (a) is correct

22. If log_x10000 = -1/4, find x

(a) $\displaystyle \frac{1}{{10}}$

(b) $\displaystyle \frac{1}{{100}}$

(d) $\displaystyle \frac{1}{{1000}}$

log_x10000 = $\displaystyle -\frac{1}{4}$ , so, (10000)^-1/4= x,

or $\displaystyle \frac{1}{{{{{\left( {10000} \right)}}^{{\frac{1}{4}}}}}}$ = x,

or x= 1/{(10⁴)}^1/4= $\displaystyle \frac{1}{{10}}$ .

So, option (a) is correct

23. Find value of log₁₀ (0.00001)

(a) – 5

(b) $\displaystyle \frac{1}{5}$

(d) 5

log₁₀ (0.00001) = log₁₀ ( $\displaystyle \frac{1}{{100000}}$ ) = log₁₀ $\displaystyle \left( {\frac{1}{{{{{10}}^{5}}}}} \right)$ = log₁₀ 10^-5) = 5 log₁₀= -5 $\displaystyle \times$ 1= – 5 .

So, option (a) is correct

24. If log₁₀ x + log₁₀y = z, then find value of x

(a) $\displaystyle \frac{z}{y}$

(b) $\displaystyle \frac{{10}}{{3z}}$

(d) yz

log₁₀ x + log₁₀ y = z, or Or, log₁₀ (xy) = z,

10^z = xy, x= $\displaystyle \frac{{{{{10}}^{z}}}}{y}$ .

So, option (c) is correct

25. Find value of log₁₀ 5 assuming log₁₀ 2 = 0.3010

(a) .4010

(b) 0.6911

(d) .6021.

log₁₀ 5 = log₁₀ ( $\displaystyle \frac{{10}}{2}$ ) = log₁₀ 10 – log₁₀ 2 = 1- 0.3010 =0.6990.

So, option (c) is correct

26. Find value of (log₅3) x (log₃625)

(a) 1

(b) 2

(d) 4.

log₅3 x log₃ 625 = log₃3 x log₅ 625 (by change of base)

= 1 x log₅ 5⁴ = log₅ 5⁴ = 4 log₅ 5 = 4 $\displaystyle \times$ 1=4.

So, option (d) is correct

27. If log(x² – 6x + 10) = 0, then the value of x is

(a) 4

(b) 7

(d) 5.

log(x² – 6x + 10) = 0, Or, log( x² – 6x + 10) = log 1, x² – 6x + 10=1, or x² – 6x + 10 – 1 = 0

Or, x² – 6x + 9 = 0, Or, x² – 3x – 3x + 9 = 0, Or, x (x – 3) – 3(x – 3) = 0, Or, (x – 3) (x – 3) = 0, or x = 3. So, option (c) is correct

28. Find x, if log₄3 + log₄(x + 2) = 2

(a) $\displaystyle \frac{{10}}{3}$

(b) $\displaystyle \frac{4}{3}$

(d) $\displaystyle \frac{{12}}{5}$

log₄3 + log₄ (x + 2) = 2, log₄{3 X (x+2)} =2, or, log₄{3x+6)} =2,

Or, 4² = 3x + 6 Or, 3x+ 6 = 16 Or, 3x = 16- 6 = 10

Or, x = $\displaystyle \frac{{10}}{3}$ .

So, option (a) is correct

29. If a2 + b2 = 23ab, evaluate log $\displaystyle \frac{{\left( {a+b} \right)}}{5}$

(a) 5

(b) log(ab) /2

(d) log(a+b) /4

a² + b² = 23ab, Or, a² + b² + 2ab = 23ab + 2ab,

Or, (a + b)² = 25ab, {(a+b)/5)}²=ab, or $\displaystyle \frac{{a+b}}{5}$ =(ab)^½

or log { $\displaystyle \frac{{a+b}}{5}$ }=log {(ab)^½ } = ½ log ab = {log(ab)} /2.

So, option (b) is correct

30.Evaluate {1/log_ab(abc)} + {1/log_bc(abc)} + {1/log_ca(abc)}

(a) 0

(b) 1

(d) 3

{1/log_ab(abc)} + {1/log_bc(abc)} + {1/log_ca(abc)}

= log _abc (ab) + log _abc (bc) + log _abc (ca) = log _abc (ab . bc . ca), = log _abc (abc)²

= 2 log _abc abc = 2.1 = 2. So, option (c) is correct

31. If log (2a – 3b) = log a – log b, find value of a

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

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

(d) $\displaystyle \frac{{2{{b}^{2}}}}{{2b-1}}$

log (2a – 3b) = log a – log b, Or, log (2a – 36) = log ( $\displaystyle \frac{a}{b}$ ),

Or, 2a – 3b = $\displaystyle \frac{a}{b}$

Or, 2ab – 3b² = a, Or, 2ab – a = 3b², Or, a(2b – 1) = 3b²,

or a= $\displaystyle \frac{{3{{b}^{2}}}}{{2b-1}}$ .

So, option (a) is correct

32. If log $\displaystyle \frac{{a+b}}{4}=\frac{1}{2}$ (log a + log b), then find value of $\displaystyle \left( {\frac{a}{b}+\frac{b}{a}} \right)$

(a) 18

(b) 14

(d) 10

log $\displaystyle \frac{{a+b}}{4}=\frac{1}{2}$ (log a + log b),

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

or $\displaystyle \frac{{{{{\left( {a+b} \right)}}^{2}}}}{{16}}$ = ab,

Or, a² + 2ab + b² = 16ab, or, a² + b²

= 14ab, (a² + b²) / (ab) = 14, $\displaystyle \left( {\frac{a}{b}+\frac{b}{a}} \right)$ =14.

So, option (b) is correct

33. Which of the following statements is not correct?

log₁₀ 10 = 1
log (2 + 3) = log (2 x 3)
log₁₀ 1 = 0
log (1 + 2 + 3) = log 1 + log 2 + log 3

(a) log_a a = 1, so log₁₀ 10 = 1.

(b) log (2 + 3) = log 5, log (2 x 3) = log 6 = log 2 + log 3

So, log (2 + 3) log (2 x 3)

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, option (b) is correct

34. $\displaystyle \frac{{\log \sqrt{8}}}{{\log 8}}$ is equal to

$\displaystyle \frac{1}{{\sqrt{8}}}$
¼
½
$\displaystyle {\sqrt{8}}$

$\displaystyle \frac{{\log \sqrt{8}}}{{\log 8}}$ = log 8 ^½ / log 8 = (½ log 8) / log 8 = $\displaystyle \frac{1}{2}$
So, option (c) is correct

35. If log (a/b) + log (b/a) = log (a+b), then

a + b = 1
a – b = 1
a = b
a² – b² = 1

LHS=log (a/b) + log (b/a) = log (a+b)
RHS=log (a+b) = log (a/b X b/a) = 1
So, a + b = 1.
So, option (a) is correct

36. If log₁₀ 7 = a, then log₁₀( $\displaystyle \frac{1}{{70}}$ ) is equal to :

-(1 + a)
(1 + a)^-1
(1 + a)
None of the above

log₁₀ ( $\displaystyle \frac{1}{{70}}$ ) = log₁₀ 1 – log₁₀ 70
= 0 – log₁₀ (7 x 10)
= – (log₁₀ 7 + log₁₀ 10)
= – (a + 1)
So, option (a) is correct

37. The value of 1/log₃60 +1/log₄60 + 1/log₅60

1/log₃60 +1/log₄60 + 1/log₅60
=log₆₀ 3+ log₆₀ 4+ log₆₀ 5
= log₆₀ (3 x 4 x 5) = = log₆₀ 60 =1
So, option (b) is correct

38 If a^x = b^y, then:

log $\displaystyle \frac{a}{b}=\frac{x}{y}$
$\displaystyle \frac{{\log a}}{{\log b}}=\frac{x}{y}$
$\displaystyle \frac{{\log a}}{{\log b}}=\frac{y}{x}$
None of the above

a^x = b^y
or, log a^x = log b
or, x log a = y log b
or $\displaystyle \frac{{\log a}}{{\log b}}=\frac{y}{x}$
So, option (c) is correct

39. The Value of log₄32 is

$\displaystyle \frac{1}{2}$
$\displaystyle \frac{5}{2}$
16
8

log₄32 = x, or 4^x = 32, or, (2²)^x = 2x2x2x2x2
or, 2^2x = 2⁵, So, 2x=5 or x= $\displaystyle \frac{5}{2}$
So, option (b) is correct

40. If log_am=n, value of a^n-1 is

(a) $\displaystyle \frac{m}{a}$

(b) $\displaystyle \frac{a}{m}$

(d) $\displaystyle \frac{{{{n}^{a}}}}{m}$

log_am=n, or aⁿ=m, or $\displaystyle \frac{{{{a}^{n}}}}{a}=\frac{m}{a}$
0r, a^n-1= $\displaystyle \frac{m}{a}$
So, option (a) is correct

41. The value of ln(1+5x/1+3x).
a) 2x – 8x² + 152x³/3 – …
b) x² + 7x²/2–12x³/5 + …
c) x – 15x²/2+163x³/4 – …
d) 1 – x²/2+x⁴/4 – …

ln(1+5x/1+3x) = ln(1+5x) – ln(1+3x) = (5x – (5x)²/2+(5x)³/3 – …) – (3x – (3x)²/2+(3x)³/3 – …)
= 2x – 8x² + 152x^3/3 –
So, option (a) is correct

42. If log₁₀(x + 5) + log₁₀10 = 4, then the value of x is
(a) 720
(b) 995
(c) 729
(d) 825

log₁₀(x + 5) + log₁₀ 10 = 4,
Or, log₁₀10 x (x + 5) = 4,
Or, 10 (x + 5) = 10⁴,
Or, (x + 5) = 10³ = 1000
Or, x = 1000 – 5 = 995
So, option (b) is correct

43. The value of log_ba x log_cb x log_ac is equal to
(a) Log₁₀a
(b) 0
(c) 2
(d) Log₁₀10

Log_ba x log_c b x log_a c = (Log_ba x log_b c) x log_c b
=Log_ba x log_c b x log_a c
= (Log_ba x log_b c) x log_c b
=(Log_aa x log_b c) x log_c b (By changes of base)
=1 X log_b c X log_c b
= log_cc X log_b b (By changes of base)
=1 X 1=1=log₁₀10

So, option (d) is correct

Logarithm MCQ | Multiple Choice Questions

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