Logarithm
- The population of a town at the beginning of the year 1986 was 2,65,000. If the rate of increase be 52 per thousand of the population. Find the population at the beginning of the year 1991. ?
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We have r = Rate of increase
= 52/1000 x 100
= 5.2, n = 5, P0 = 265000
∵ P = 265000(1 + 5.2 / 100)5
⇒ log P = log 265000 + 5(log 105.2 - log 100)
Correct Option: B
We have r = Rate of increase
= 52/1000 x 100
= 5.2, n = 5, P0 = 265000
∵ P = 265000(1 + 5.2 / 100)5
⇒ log P = log 265000 + 5(log 105.2 - log 100)
= 5.4232 + 5(2.0220 - 2)
= 5.4232 + 0.1100
= 5.5332
∴ P = antilog(5.5332) = 341400
- What rate per cent per annum compound interest will Rs. 3,000 in 3 years if the interest is reckoned half yearly ?
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∵ 3000 = 2000(1 + r/200)6
⇒ 3/2 = (1 + r/200)6
⇒ 1 + r/200 = (3/2)1/6
⇒ log(1 + r/200)= 1/6(log 3 - log 2 )
⇒ log(1 + r/200)= 1/6(0.4771-.3010)
Correct Option: C
∵ 3000 = 2000(1 + r/200)6
⇒ 3/2 = (1 + r/200)6
⇒ 1 + r/200 = (3/2)1/6
⇒ log(1 + r/200)= 1/6(log 3 - log 2 )
⇒ log(1 + r/200)= 1/6(0.4771-.3010)
= 0.02935
⇒ (1 + r/200) = antilog(.02935)
⇒ 1 + r/200 = 1.070 = 1 + 7/100
∴ r = 14%
