Ex 2.3 class 9 math | new book
Ex2.3 class 9 math new book will be completely solved.
Q1 Find characteristic of the following number:
(i) 5287 Ch = 3
(ii) 59.28 Ch = 1
(iii) 0.0567 Ch = -2 0r $$Characteristic\;=\;\overline2$$
(iv) 234.7 Ch = 2
(v) 0.000049 Ch = -5 or $$Characteristic\;=\;\overline5$$
(vi) 145000 Ch = 5
Q2 Find Logarithm of the following numbers:
(i) 43
Characteristic = 1 Mantissa = 0.6335
Log 43 = 1.6335
(ii) 579
Characteristic = 2 Mantissa = 0.7627
Log 579 = 2.7627
(iii) 1.982
Characteristic = 0 Mantissa = 0.2971
Log 1.982 = 0.2971
(iv) 0.0876
$$Charactersitic\;=\;-\;2\;=\;\overline2$$ Mantissa = 0.9425
$$Log\;0.0876=\;\overline2.9428$$
(v) 0.047
$$Charactersitic\;=\;-\;2\;=\;\overline2$$ Mantissa = 0.6628
$$Log\;0.047=\;\overline2.6628$$
(vi) 0.000354
$$Characteristic\;=\;-4\;=\;\overline4$$
Mantissa = 0.5490
$$Log\;0.000354\;=\;\overline4\;.\;5490$$
Q3 If Log 3.177 = 0.5019, then find
(i) Log 3177
$$Log\;3177\\=\;Log\;(\;3.177\;\times\;10^3\;)\\=\;Log\;3.177\;+\;Log\;10^3$$
lets apply power rule of log and Log 10 = 1 and Log 3.177 = 0.5019 ( Given )
Log 3177 = Log 3.177 + 3 log 10
= 0.5019 + 3 ( 1 ) = 0.5019 + 3 = 3.5019
(ii) Log 31.77
$$Log\;31.77\\=\;Log\;(\;3.177\;\times\;10^1\;)\\=\;Log\;3.177\;+\;Log\;10^1$$
lets apply power rule of log and Log 10 = 1 and Log 3.177 = 0.5019 ( Given )
Log 31.77 = Log 3.177 + (1) log 10
= 0.5019 + 1 ( 1 ) = 0.5019 + 1 = 1.5019
(iii) Log 0.03177
$$Log\;0.03177\\=\;Log\;(\;3.177\;\times\;10^{-2}\;)\\=\;Log\;3.177\;+\;Log\;10^{-2}$$
lets apply power rule of log and Log 10 = 1 and Log 3.177 = 0.5019 ( Given )
Log 31.77 = Log 3.177 + (-2) log 10
= 0.5019 + (-2) ( 1 ) = 0.5019 – 2 = -1.4981
Because Mantissa will never be negative, so we will change it into positive
-1.4981 +2 – 2 = -2 + 0.5019 = $$\overline2\;.\;5019$$
Q4 Find the value of x
(i) Log x = 0.0065
Log x = 0.0065
x = Antilog 0.0065
= 1.015
(ii) Log x = 1.192
= Antilog 1.192
x = 15.56
(iii) Log x = -3.434
x = Antilog ( -3.434)
Mantissa will never be negative.
x = Antilog ( -3.434 +4-4 )
= Antilog ( -4 + 0.566 )
$$Anti\log\;(\;\overline4.566\;)\;=\;0.003681$$
(iv) Log x = -1.5726
x = Antilog ( -1.5726)
x = Antilog ( -1.5726 +2-2 )
= Antilog ( -2 + 0.4274 )
$$Anti\log\;(\;\overline2.4274\;)\;=\;0.02675$$
(v) Log x = 4.3561
x = Antilog 4.3561 = 22710
(vi) Log ( -2.0184 )
x = Antilog ( -2.0184)
x = Antilog ( -2.0184 +3-3 )
= Antilog ( -3 + 0.9816 )
$$Anti\log\;(\;\overline3.9816\;)\;=\;0.009585$$
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