log 125 / log 5 = x.
log 5³ / log 5 = x
x = 3 log5 / log5
= 3

log₂(1/64)
= log₂(1) – log₂(64)
= 0 – log₂(2⁶)
= -6log₂(2).
= −6

Here are some values to base ten represented as logarithms :

10^0.3010 = 2 ,
take log base 10 both sides ,
log₁₀10^0.3010 = log₁₀2
0.3010 = log₁₀2

Given :
log_0.125125 = log_125/1000125
= log_0.55 = 1 / log₅0.5
= 1 / -log₅2
= -log₁₀5 / log₁₀2
= -2.321 [ log5 = 0.6989 and log2 = 0.3010]

logX + logY = log (XY)
= log (a²/bc) + log (b²/ac) + log (c²/ab)
= log (a² b² c² / bcacab)
= log (a² b²c² / a²b²c²)
= log 1
= 0

Given:
= log75/16 – 2 log5/9 + log32/343
= log [(25 x 3)/(4 x 4)] – log (25/81) + log [(16 x 2)/(81 x 3)] = log(25 x 3) – log ( 4 x 4 ) – log(25) + log81 + log(16 x 2) – log (81 x 3)
= log 25 + log 3 – log 16 – log 25 + log 81 + log 16 + log 2 – log 81 – log 3
= log 2

Given:
= log [ { (9 / 8) / (27 / 32) } x 3 / 4) ] = log [ (9 / 8) x (3 / 4) x (32 / 27) ] = log 1
= 0

log(ab²) – log(ac) + log abc⁴ – 3log(bc)
= log ab² – logac + log abc⁴ – log b³c³
= log_a (ab²/ac) x (abc⁴/b³c³)
= log_aa = 1

log₂x = 10 ⇒ x = 2¹⁰.
log_xY= 100
⇒ y = x¹⁰⁰
⇒ y = (2¹⁰)¹⁰⁰ [put value of x] ⇒ y = 2¹⁰⁰⁰

Let required number of hours be x
More examiners, less hours(indirect proportion)
More days, less hours(indirect proportion)
More answer books, more hours(direct proportion)
Hence we can write as

• 9 * 12 * 2 * 5 = 4 * 30 * 1 * x

• x = 9 * 12 * 2 * 5 / 4 * 30 * 1

• = 9 * 3 * 2 * 5 / 30

= 9

• 36839/17 = 2167, Dividend contains 2 places of decimal.
  So, 368.39 / 17 = 21.67
• 17050/62 = 275, Dividend contains 2 places of decimal.
  So, 70.50/62 = 2.75
• 87565/83 = 1055, Dividend contains 2 places of decimal.
  So, 875.65/83 = 10.55

Divide 86700 by 3 = 28900
The square root of 28900 = √28900
= 170 is the perfect square

Subtract 0.000009 from 0.006898
= 0.006898 – 0.000009
= 0.006889
Square root of 0.006889 = 0.083
= 0.083 is a perfect square

Weight of 1 apple = 250 g
Weight of 6 apples = 250 * 6 = 1.5 kg
Initial weight = 14.5 – 1.5 = 13 kg
Number of apples in his carton initially :
= 13000/250
= 52 apples

Take the lcm of 3 , 4 , 5 , 6.
3 = 3×1
4 = 2×2
5 = 5×1
6 = 2×3
LCM = 2 × 2 × 3 × 5 = 60
Therefore 60 is the number that is completely divisible by these numbers. But we have to find the smallest number that leaves remainder 2.
Answer : 60 + 2 = 62

LCM of 18, 24 and 32:
–> 18 = 2 x 3 x 3
–> 24 = 2 x 2 x 2 x 3
–> 32 = 2 x 2 x 2 x 2 x 2
LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288.

The smallest four digit number that is divisible by 288:
288 x 3 = 864
288 x 4 = 1152

Since 288 is divisible by 18, 24 and 32, 1152 is also divisible by all these numbers. Therefore, 1152 is the smallest four digit number divisible by 18, 24 and 32.

Since the numbers are co-prime, they contain only 1 as the common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29.
So, First number is = 551/29 = 19.
Third number = 1073/29 = 37
So sum of these numbers is = (19 + 29 + 37) = 85.

LCM of 2 , 4 , 6 ,8 , 10 , 12 is 120 seconds, that’s 2 minutes.
Now 60 / 2 = 30
Adding one bell at the beginning it’ll be 30 + 1 = 31.

In this type of question, we need to calculate the HCF, trick is,
HCF (1661 – 10) and (2045 -13)
= HCF (1651, 2032)
= 127