Arithmetic Aptitude :: Logarithm

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


A0.934

B0.945

C0.954

D0.958

ENone of these

Answer: Option C

Explanation:

Log 3^3 =1.431

3 log 3=1.431

Log 3= 0.477

 Log 9 => Log (3)^2

2(log 3 )

2(0.477)= 0.954

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If log 125 + log 8 = x, then x is equal to


A3-Jan

B2

C0.064

D3

Answer: Option D

Explanation:

log 125 + log 8 = log (125 * 8)
=> log (1000) = log (10^3)
=> 3 log 10 = 3

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If log5 + log(5x+1) = log(x+5) + 1, then x is equal to


A1

B3

C5

D10

Answer: Option B

Explanation:

If log5   log(5x 1) = log(x 5)   1

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If log 0.318=0.3364 and log 0.317=0.33320 then log 0.319 =?


A0.33365

B0.3368

C0.3396

D0.3369

Answer: Option C

Explanation:

log 0.319=log0.318+(log0.318-log0.317) = 0.3396

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If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?


A0.3396

B0.4396

C0.5396

DNone of these

Answer: Option A

Explanation:

log 0.317=0.3332 and log 0.318=0.3364
then
log 0.319=log0.318+(log0.318-log0.317)
=0.3396

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If log(base 10) 5 + log(base 10)(5x + 1) = log(base 10)(x + 5) + 1, then x is equal to:


A9

B3

C1

D4

Answer: Option B

Explanation:

Given log(base 10) 5 + log(base 10)(5x + 1) = log(base 10)(x + 5) + 1
=> log(base 10) 5 + log(base 10)(5x + 1) = log(base 10)(x + 5) + log(base 10) 10
Hint: log (base Z)Z = 1
=> log(base 10)[5 (5x + 1)] = log(base 10) [10(x + 5)] ------ (Point to Remember.)
=> 5(5x + 1) = 10(x + 5)
=> 5x + 1 = 2x + 10
=> x = 3

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When numbers are written in base b, we have 15*22 = 414, the value of b is


A8

B7

C6

DNone of these

Answer: Option C

Explanation:

When-numbers-are-written-in-base-b

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Choose the correct option.

The Value of log8(64) - log64(4096) is


A0

B2

C1

D6

Answer: Option A

Explanation:

log8 (64)-log64(4096)
log8 (8^2)-log 64(64^2) = 2 - 2 = 0

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Choose the correct option.

log base 12 144p=3 then p= ?


A12

B13

C15

D11

Answer: Option A

Explanation:

log base 12 144p=3
=> 144p=12^3
=> 144p=12^2 *12
=> 144p=144*12
=> p=12

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Change 101010 two to base ten.


A42

B3

C84

D30

Answer: Option A

Explanation:

(101010) base2 = (1*2^5 + 0*2^4 + 1*2^3 +0*2^2 + 1*2^1 + 0*2^0) 10 = (42) base10

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