Logarithm Questions and Answers

Q1 :

If log (a+b3)=12(log a+log b)Big(frac{a + b}{3}Big) = frac{1}{2} left(text{log } a + text{log } b right)

A
  

a2+b2=aba^2 + b^2 = ab

B
  

a2+b2=7aba^2 + b^2 = 7ab

C
  

a2+b2=−3aba^2 + b^2 = -3ab

D
  

a2+b2=3aba^2 + b^2 = 3ab

View Answer
Correct Answer: B

a2+b2=7aba^2 + b^2 = 7ab

Description:

log (a+b3)=12(log a+log b)text{log } Big(frac{a + b}{3}Big) = frac{1}{2} (text{log } a + text{log } b)

∴log (a+b3)= log (ab)12 or a+b3=(ab)12therefore text{log } Big(frac{a + b}{3}Big) = text{ log } (ab)^frac{1}{2} text{ or } frac{a + b}{3} = (ab)^frac{1}{2}

 or (a+b)2=9ab⇒a2+b2=7abtext{ or } (a +b )^2 = 9ab Rightarrow a^2 + b^2 = 7ab

Q2 :

The expression  log 115+log143−log 2215text{ log } frac{11}{5} + text{log} frac{14}{3} - text{log } frac{22}{15 } is equal to

A
  

log 2

B
  

log 3

C
  

log 5

D
  

log 7

View Answer
Correct Answer: D

log 7

Description:

 log 11− log 5+ log 2+ log 7− log 3 − log 2− log 11+ log 5+ log 3begin{aligned} & text{ log } 11 - text{ log } 5 + text{ log } 2 + text{ log } 7 - text{ log } 3 \ &- text{ log } 2 - text{ log } 11 + text{ log } 5 + text{ log }3 end{aligned}

∴ log 7therefore text{ log } 7

Q3 :

If  logba× logyx= log catext{ log}_b a times text{ log}_y x = text{ log }_c a, then the value of xx and yy are respectively

A
  

b and cb text{ and } c

B
  

c and bc text{ and } b

C
  

c and ac text{ and } a

D
  

b and ab text{ and } a

View Answer
Correct Answer: A

b and cb text{ and } c

Description:

 logba× logyx= log catext{ log}_b a times text{ log}_y x = text{ log }_c a

 logba× logcb= log catext{ log}_b a times text{ log}_c b = text{ log }_c a

∴x=b,y=ctherefore x = b, y = c

Q4 :

If log⁡107=alog_{10} 7 = a, then log⁡10(170)log_{10}Big(frac{1}{70}Big) is equal to 

A
  

a10frac{a}{10}

B
  

110afrac{1}{10a}

C
  

(1+a)−1(1 + a)^{-1}

D
  

−(1+a)-(1 + a)

View Answer
Correct Answer: D

−(1+a)-(1 + a)

Description:

log⁡170=log⁡1−log⁡70 =−log⁡70 =−log⁡(7×10) =−(log⁡7+log⁡10) =−log⁡7−log⁡10 =−a−1=−(1+a) [∵log⁡7=a]begin{aligned} &log frac{1}{70} = log 1 - log 70 &quadquad = - log 70 &quadquad = - log (7 times 10) &quadquad = - (log 7 + log 10) &quadquad = - log 7 - log 10 &quadquad = - a - 1 = - (1 + a) &quadquad [because log 7 = a] end{aligned}

Q5 :

If antilog (0.0385) = 1092 and log⁡x=3ˉ.0385log x = bar 3.0385, then xx is equal to

A
  

.1092

B
  

.01092

C
  

.001092

D
  

.0001092

View Answer
Correct Answer: C

.001092

Description:

log⁡x=3ˉ.0385=−3.0385 =−6+3.0385 =−6log⁡10+3.0385 =log⁡10−6+log⁡(1092) =log⁡(1092×10−6) =log⁡(0.001092) ⇒x=0.001092begin{aligned} log x &= bar 3.0385 = - 3.0385 quadquad; &= - 6 + 3.0385 quadquad; &= - 6 log 10 + 3.0385 quadquad; &=log 10^{-6} + log (1092) quadquad; &=log (1092 times 10^{-6}) quadquad; &= log (0.001092) Rightarrow x &= 0.001092 end{aligned}

Q6 :

The value of,13log⁡10125−2log⁡104+log⁡1032frac{1}{3} log_{10} 125 - 2 log_{10} 4 + log_{10} 32 is

A
  

0

B
  

1

C
  

2

D
  

3

View Answer
Correct Answer: B

1

Description:

13log⁡10125−2log⁡104+log⁡1032 =log⁡10(125)1/3−log⁡1042+log⁡1032 [∵log⁡mn=nlog⁡m] =log⁡105−log⁡1016+log⁡1032 =log⁡105×3216 =log⁡1010 =1 [∵log⁡m×n=log⁡m+log⁡n] [log⁡mn=log⁡m−log⁡n]begin{aligned} &frac{1}{3} log_{10} 125 - 2 log_{10} 4 + log_{10} 32 \ &= log_{10} (125)^{1/3} - log_{10} 4 ^2 + log_{10}^{32} \ &Big[because log m^n = n log mBig] \ &=log_{10} 5 - log_{10} 16 + log_{10} 32 \ &=log_{10} frac{5 times 32}{16} \ &=log_{10} 10 \ &= 1 \ &Big[ because log m times n = log m + log nBig] \ &Big[log frac{m}{n} = log m - log n Big] &end{aligned}

Q7 :

For 0 < a <1, the numbers 1afrac{1}{a}, a,a2log⁡a, a^2 log a when arranged in ascending order of their values, are:

A
  

log⁡a,a2,a,1alog a, a^2, a, frac{1}{a}

B
  

a,1a,log⁡a,a2a, frac{1}{a}, log a, a^2

C
  

a,a2,log⁡a,1aa, a^2, log a, frac{1}{a}

D
  

log⁡a,1a,a,a2log a, frac{1}{a}, a, a^2

View Answer
Correct Answer: A

log⁡a,a2,a,1alog a, a^2, a, frac{1}{a}

Description:

Given , 0 < a < 1

Let a=12=0.5a = frac{1}{2} = 0.5

a2=14=0.25a^2 = frac{1}{4} = 0.25

1a=2frac{1}{a} = 2

log⁡a=log⁡12=log⁡1−log⁡2=0−0.30= negativelog a = log frac{1}{2} = log_1 - log_2 = 0 - 0.30 = text{ negative}

∴therefore In ascending order of their values are

log⁡a,a2,a,1alog a, a^2, a, frac{1}{a}

Q8 :

The value of log⁡10125log_{10} 125, (given log⁡102=0.30log_{10} 2 = 0.30), is

A
  

2.05

B
  

2.10

C
  

2.15

D
  

2.21

View Answer
Correct Answer: B

2.10

Description:

log⁡10125=log⁡10(5)3 =3log⁡105[∵log⁡mn=nlog⁡m] =3log⁡10(102) =3[log⁡1010−log⁡102] =3[1−0.30] =3×0.70 =2.10begin{aligned} log_{10}125 &= log_{10}(5)^3 &= 3 log_{10}5 [because log m^n = n log m] &= 3 log_{10} Big(frac{10}{2}Big) &= 3[log_{10}10 - log_{10}2] &= 3 [1 - 0.30] &= 3 times 0.70 &= 2.10 end{aligned}

Q9 :

What is the value of log100_{100} 0.1 ?

A
  

1/2

B
  

-1/2

C
  

-2

D
  

2

View Answer
Correct Answer: B

-1/2

Description:

log1000.1=log102(110)=12log10(110)=12log10(10)−1=−12log1010=−12begin{aligned} & text{log}_{100} hspace{0.2cm} 0.1 = text{log}_{10^2} bigg( frac{1}{10} bigg) & =frac{1}{2} text{log}_{10} bigg(frac{1}{10} bigg) = frac{1}{2} text{log}_{10} hspace{0.2cm} (10)^{-1} & = -frac{1}{2} text{log}_{10} hspace{0.2cm} 10 = -frac{1}{2} end{aligned}

Q10 :

The value of 3 log 3 + 2 log 2 is

A
  

log 108

B
  

log 106

C
  

log 109

D
  

None of these

View Answer
Correct Answer: A

log 108

Description:

3 log 3 + 2 log2
= log 33^3 + log 22^2
= log 27 + log 4
= log (27 ×times 4)
= log 108

Q11 :

If log2_2 2=16sqrt{2} = frac{1}{6} , then the value of aa is

A
  

(2)6(sqrt{2})^6

B
  

(6)12(6)^{frac{1}{2}}

C
  

3

D
  

-6

View Answer
Correct Answer: A

(2)6(sqrt{2})^6

Description:

∵loga2=16⇒a1/6=2∴a=(2)6begin{aligned} & because hspace{0.2cm} text{log}_a hspace{0.2cm} sqrt{2} = frac{1}{6} & Rightarrow a^{1/6} = sqrt{2} & therefore hspace{0.2cm} a= (sqrt{2})^6 end{aligned}

Q12 :

Find the logarithm of 1728 to the base 232sqrt{3}

A
  

3.124

B
  

3.1732

C
  

6

D
  

5

View Answer
Correct Answer: C

6

Description:

Let log23_{2sqrt{3}} 1728 = xx
⇒(23)x=1728∵1728=26(3)6=(23)6(23)x=(23)6 On comparing both sides, we get x=6begin{aligned} & Rightarrow (2sqrt{3})^x = 1728 \ & because 1728 = 2^6(sqrt{3})^6 = (2sqrt{3})^6 \ & (2sqrt{3})^x = (2sqrt{3})^6 \ & text{ On comparing both sides, we get } x = 6 end{aligned}

Q13 :

What is the value of 12frac{1}{2} log10_{10} 25 - 2log10_{10} 3 + log10_{10} 18 ?

A
  

2

B
  

3

C
  

1

D
  

0

View Answer
Correct Answer: C

1

Description:

12log1025−2log103+log1018=log10251/2−log1032+log1018=log105−log109+log1018=log105×189=log10909=log1010=1begin{aligned} & frac{1}{2} text{log}_{10} hspace{0.2cm} 25 - 2 hspace{0.2cm} text{log}_{10} hspace{0.2cm} 3 + text{log}_{10} hspace{0.2cm} 18 \ & = text{log}_{10} hspace{0.2cm} 25^{1/2} - text{log}_{10} hspace{0.2cm} 3^2 + text{log}_{10} hspace{0.2cm} 18 \ & = text{log}_{10} hspace{0.2cm} 5 - text{log}_{10} hspace{0.2cm} 9 + text{log}_{10} hspace{0.2cm} 18 \ & = text{log}_{10} frac{5 times 18}{9} = text{log}_{10} frac{90}{9} \ & = text{log}_{10} hspace{0.2cm} 10 = 1 end{aligned}

Q14 :

What is the value of [log10(5 log10100)]2[text{log}_{10} hspace{0.2cm} (5 text{ log}_{10} hspace{0.2cm}100)]^2 ?

A
  

4

B
  

3

C
  

2

D
  

1

View Answer
Correct Answer: D

1

Description:

[log10(5 log10100)]2=[log10(5 log10102)]2=[log10(10 log1010)]2=[log1010]2=[∵log1010=1]=12=1begin{aligned} & [text{log}_{10} hspace{0.2cm} (5 text{ log}_{10} hspace{0.2cm}100)]^2 \ & = [text{log}_{10} hspace{0.2cm} (5 text{ log}_{10} hspace{0.2cm}10^2)] ^2 \ & = [text{log}_{10} hspace{0.2cm} (10 text{ log}_{10} hspace{0.2cm}10)] ^2 \ & = [text{log}_{10} hspace{0.2cm} 10]^2 \ & =[because text{log}_{10} hspace{0.2cm} 10=1] \ & =1^2 = 1 end{aligned}

Q15 :

The value of logyx⋅logzy⋅logxztext{log}_y hspace{0.2cm} x cdot text{log}_z hspace{0.2cm} y cdot text{log}_x hspace{0.2cm} z is

A
  

log xyz

B
  

xyz

C
  

1

D
  

0

View Answer
Correct Answer: C

1

Description:

logyx⋅logzy⋅logxz=log xlog y×log ylog z×log zlog x=1[∵log ab=log blog a]begin{aligned} & text{log}_y hspace{0.2cm} x cdot text{log}_z hspace{0.2cm} y cdot text{log}_x hspace{0.2cm} z \ & = frac{text{log } x}{text{log } y} times frac{text{log } y}{text{log } z} times frac{text{log } z}{text{log } x} = 1 \ & bigg[because text{log }_a hspace{0.2cm} b = frac{text{log } b} {text{log } a} bigg] end{aligned}