Logarithm - Arithmetic Aptitude Questions and Answers

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If log 2 = 0.3010 and log 3 = 0.4771, the value of log₅ 512 is:

A.

2.870

B.

2.967

C.

3.876

D.

3.912

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Answer : Option C

Explanation :

= $\frac { log2^9 } {log(10/2) }$

= $\frac {9 log 2} {log10-log2 }$

= $\frac { (9*0.30100} { 1-0.3010 }$

= $\frac { 2.709 } { 0.699 }$

= $\frac { 2.709 } { 699 }$

= 3.876

log₅ 512

=

$\frac { log512 } {log^5 }$

$\frac { log\sqrt{8} } { log8 }$ is equal to:

A.

$\frac { 1 } { \sqrt{8} }$

B.

$\frac { 1 } { 4 }$

C.

$\frac { 1 } {4 }$

D.

$\frac { 1 } { 8}$

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Answer : Option C

Explanation :

$\frac { log\sqrt{8} } { log8 }$ = $\frac {log(8)^{1/2} } { log8 }$ = $€‹€‹ \frac { \frac{1}{2} log 8 } { log 8 }$ = $\frac { 1 } { 2 }$

If log 27 = 1.431, then the value of log 9 is:

A.

0.934

B.

0.945

C.

0.954

D.

0.958

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If log $\frac {a} {b }$ +log $\frac {b} {a }$ =log (a + b), then

A.

a + b = 1

B.

a - b = 1

C.

a = b

D.

a² - b² = 1

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Answer : Option A

Explanation :

log $\frac {a} {b }$ +log $\frac {b} {a }$ =log (a + b)

log (a + b) = log[ $\frac {a} {b } \times$ $\frac {b} {a }$ ]=log1

So, a + b = 1.

If log₁₀ 7 = a, then log_{10 $[ \frac { 1 } { 70 }]$}is equal to:

A.

- (1 + a)

B.

(1 + a)^-1

C.

$\frac { a } { 10 }$

D.

$\frac { 1 } { 10a }$

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Answer : Option A

Explanation :

log₁_{_{$[ \frac { 1 } { 70 }]$}=} log₁₀ 1 - log₁₀ 70
	= - log₁₀ (7 x 10)
	= - (log₁₀ 7 + log₁₀ 10)
	= - (a + 1).

If log₁₀ 2 = 0.3010, then log₂ 10 is equal to:

A.

$\frac { 699 } { 301 }$

B.

$\frac { 1000 } { 301 }$

C.

0.3010

D.

0.6990

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Answer : Option B

Explanation :

log₂ 10 = $\frac { 1 } {log10^2 }$ = $\frac { 1 } {0.3010 }$ = $\frac { 10000 } {3010 }$ = $\frac { 1000 } { 301 }$

If log₁₀ 2 = 0.3010, the value of log₁₀ 80 is:

A.

1.6020

B.

1.9030

C.

3.9030

D.

None of these

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If log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + 1, then x is equal to:

A.

1

B.

3

C.

5

D.

10

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The value of^{$[\frac { 1 } { log^360 }+€‹€‹$ $[\frac { 1 } { log^460 }+€‹€‹$}^{$[\frac { 1 } { log^560 }]$ is}

A.

0

B.

1

C.

5

D.

60

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If log 2 = 0.30103, the number of digits in 2⁶⁴ is:

A.

18

B.

19

C.

20

D.

21

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If log_{x $[\frac { 9 } {16 } ]=$ $-[\frac { 1 } {2 } ],$}then x is equal to:

A.

$-\frac { 3} {4 }$

B.

$\frac { 3} {4 }$

C.

$\frac {8 1 } { 256 }$

D.

$\frac {256 } { 81}$

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Answer : Option D

Explanation :

If log_{x_{$[\frac { 9 } {16 } ]=$}_{$-[\frac { 1 } {2 } ]$}}

$\frac { X^-1/2 } { }$ = $\frac {9 } {16}$

$\frac { 1 } { \sqrt{X}}$ = $\frac {9 } {16}$

${ \sqrt{X} } { }$ = $\frac {16 } {9}$

x =[ $\frac {16 } {9}$ ]²

x = $\frac {256 } { 81}$

If a^x = b^y, then:

A.

log $\frac {a } {b }$ = $\frac {X } {Y }$

B.

$\frac { loga } { logb }$ = $\frac {X } {Y }$

C.

$\frac { loga } { logb }$ = $\frac {Y } {X }$

D.

None of these

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Answer : Option C

Explanation :

a^x = b^y

log a^x = log b^y

x log a = y log b

$\frac { loga } { logb }$ = $\frac {Y } {X }$

If log_x y = 100 and log₂ x = 10, then the value of y is:

A.

2¹⁰

B.

2¹⁰⁰

C.

2¹⁰⁰⁰

D.

2¹⁰⁰⁰⁰

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The value of log₂ 16 is:

A.

$\frac { 1 } { 8 }$

B.

4

C.

8

D.

16