? A = 12,000(1 + 12/100)10
= 12000(28/25)10
? log A = log 12000 + 10[log 28 - log 25]
? log A = 4.0792 + 10(1.4472 - 1.3979)
= 4.0792 + 0.493
= 4.5722
? A = antilog 4.5722 = 37342
C.I. = 37342 - 12000 = 25342
=25350
? loga, logb, logc are in A.P. Then,
? logb - loga = logc - logb
? log b/a = log c/b
? b/a = c/b
? b2 = ac
? a, b, c are in G.P.
? log5[(x2 + x ) / x] = 2
? log10(x + 1) = 2
? x + 1 = 25
? x = 24
Since, 0.3274 gives characteristic 1. Therefore value of log (0.3274) = 1.5150
? log tan 45 = log 1 = 0
Hence, Whole Expression = (something) x zero
Given Exp.
= log (57 x 100 / 100) + 3log(0.57) + 1/2log(0.57)
= log (0.57) + log 102 + 3log(0.57) +1/2log (0.57)
= (1 + 3 + 1/2) log(0.57) + 2 [? log102 = 2]
= (4.5 x 1.756) + 2 = 4.5 x (-1 + 0.756) + 2
= 3.402 - 4.5 + 2
= 0.902
? x = 264
? log x = log 264
? log x = 64 log 2
= 64 x .3010 = 19.264
? No.of digits = 19 + 1 = 20
Let N = 312 x 28
? log N = 12 x log3 + 8 x log 2
? log N = 12 x 0.47712 + 8 x 0.30103
? log N = 8.13368
? No.of digits = 8 + 1 = 9
Given exp. = 1/(log2 ?) + 1/(log6 ?)
= log? 2 + log? 6
= log? (2 x 6)
= log?12
Since 12 > ? so the value of given expression is more than 1.
? N = (875)16
? log N = log (875)16
? log N = 16 log (875)
= 16 x (2.942)
= 47.072
? No.of digits = 47 + 1 = 48
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
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