Math (3)_text.pdf

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SSS^Sl
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WÊÈÊÈ
03
Scientific
LOGARITHMS
Notation
(move
décimal
point
five
places
the
right)
to
A
number
written
in
the
form
n
axl0
,wherel<a<10and
n
is
is
an
integer,
Change
each
of
the
following
calied
the
scientific
notation.
numbers
from
Write
each
of
the
following
(i)
scientific
notation
to
ordinary
notation.
6.35
x
10
6
(ii)
7.61
xlO
-4
ordinary
(i)
numbers
in
scientific
notation
(ii)
30600
30600
=
0.000058
4
Solution
(i)
6.35x10*
(move
=
6350000
Solution
(i)
3.06
x
10
the
décimal
point
six
places
(move
décimal
point
four
places
the
left)
(ii)
to
to
the
right)
(ii)
7.61
x
10T
4
=
0.000761
0.000058
=
5.8
x
10“
5
(move
the
décimal
point
four
places
to
the
left)
Exercise
3.1
Ql.
Express
each
of
the
following
Sol:
83,000
=
8.3xl0
4
(move
décimal
left)
numbers
in
i)
scientific
notation.
vi)
point
four
places
to
5700
5700
0.00643
0.00643
Sol:
=
5.7xl0
(move
3
décimal
Sol:
=
6.43
xlO
-1
(move
point
three
places
to
left)
ii)
décimal
point
three
places
to
right)
vii)
49,800,000
0.0074
Sol:
49,800,000
=
4.98x1
7
(move
left)
Sol:
0.0074
=
7.4
xlO"
3
(move
décimal
décimal
point
seven
places
to
iii)
point
three
places
to
right)
viii)
96,000,000
96,000,000
60,000,000
Sol:
=
9.6xl0
7
(move
left)
Sol:
60,000,000
=
6.0xl0
(move
left)
7
décimal
point
seven
places
to
iv)
décimal
point
seven
places
to
ix)
416.9
416.9
=
4.169xl0
2
0.00000000395
0.00000000395
Sol:
(move
décimal
left)
Sol:
=
3.95
xlO
-9
point
two
places
to
y)
(move
décimal
point
nine
places
right)
to
83,000
0
0
"
275,000
X)
0.0025
275,000
0.0025
base
4.
)
Find
log
4
2,
i.e.,
find
log
of
2
to
the
2.75X10
2.5x1
5
(
'
move
dcc
imn
poinl
fi
vc
p
aces
to
le
Cl
1
)
0'
3
f
move
deci
nul
point
thicc
places
to
ngiu
Tel
}
iog
4
2
its
x
is
Then
i.e.,
cxponenetial
form
7x
X
4
=
2
Q2.
Express
the
followtng
mmiberç
in
ordinary
notation.
2
=
2
1
2x~j
i)
6x10
4
4
Soi:
6x10
=
0.0006
i0
(move
lefî)
décimai
point
four
places
to
ii)
Sincc
a
°
Since
a
}
5.06x1
Soi:
5.06x1
i0
=
50,600,000,000
ter.
•
If
the
base
of
logarithm
is
is
taken
as
(move
décimal
point
right)
iii)
places
to
10
then
logarithm
called
Commun
Logarithm.
-6
9.01
8x1
9.01
8x1
Sol:
0'
6
~
0.000009018
(move
left)
décimal
point
six
places
to
iv)
The
intégral
part
of
the
logarithm
of
any
number
is
called
the
characteristic.
7.865x10*
7.865
x
1
0*
Sol:
=
786,500,000
(move
décima)
point
eight
places
to
right)
The
characteristic
of
the
logarithn
of
a
number
greater
t'nan
1
is
always
one
less
than
the
number
of
digits
of
îhe
number.
in
the
Logarithm
of
a
Real
Nuniber
x
If
a
-
y
then
x
is
called
logarithm
of
y
to
the
base
vvritten
as log„
y
integra!
part
the
is
When
a
number
b
is
scientific
notation,
1
written
:n
the
in
the
V
and
0,
i.e.,
form
b
=
ax
i.e.,
=
x,
where
a
>
a
ï
10"
where
1
<
a
<
?.e
10,
the
power
of
10
and
y
>
0
i.e.,
n
the
logarithm
of
a
the
will
«i
characteristic
of
log
h.
the
base
‘a’
is
number
y
index
x
of
the
power
to
get
that
to
to
mm
Number
j
Scientific
J
Characteristic
1
which
a
must
he
raised
y-
]
number
Notation
of
the
l
a
-
y
and
log
a
y
=
x
are
équivalent.
When
one
relation
is
given,
relations
it
The
T02
|
1.02
i
x
x
x
10°
!
Logarithm
“
"Ô
—
i
!
99.6
|
9.96
1.092
!0
r
1
ï
can.be
convcrted
into
the
other.
Th
us
102
,
=
y
<=»
10H
3
j
2
3
~
loga
V
=À
1.662
A
|
1
.6624
x
10
Characteristic
of
Logarithm
of
(iii)
Again
proceeding
horizontally
the
till
mean
différence
column
The
ot
a
characteristic
of
the
logarithm
1,
is
corresponding
to
5
intersects
this
number
less
than
more
than
the
always
négative
row,
we
get
the
number
5
at
the
and
one
number
of
zeroes
(iv)
intersection.
immediately
after
the
décimal
point
of
the
Adding
5
the
two
numbers
6355
and
number.
we
get
.6360
as
the
mantissa
of
the
logarithm
of
43.25.
Write
the
characteristic
of
the
log
of
following
in
scientific
numbers
by
expressing
them
Find
the
mantissa
of
the
logarithm
of
0.002347
notation
and
noting
the
power
of
10.
Solution
0.872,
0.02,
0.00345
Here
also,
we
consider
only
wr(\
the
Number
Scientific
Characteristic
four
significant
digits
2347
Notation
of
the
We
first
locate
the
row
corresponding
to
23
in
the
logarithm
tables
Logarithm
0.872
|
8.72
x
10'
2
1
-1
and
proceed
as before.
~
0.02
2.0
>T
10'
-2
-3
Along
intersection
the
saine
row
to
its
0.00345
3.45
x
1(F
with
the
column
is
corresponding
to
4
the
resulting
number
Mantissa
3692.
The
number
fractional
at
the
intersection
of
The
iogaritf
ii
part
is
of
called
the
this
row
and
the
mean
différence
column
of
a
number
is
the
corresponding
to
7
is
13.
Hence
the
sum
of
mantissa.
Mantissa
IIM
always
positive
3692
and
13
gives
the
mantissa
of
the
logarithm
of
0.0023476
as
0.3705
ffli
Find
the
mantissa
of
the
logarithm
of
43.254
Find
(i)
(ii)
log
278.23
log
0.07058
Solution
Rounding
off
43.254
we
consider
only
the
four
significant
digits
4325.
(i)
Solution
(i)
278.23
can be
rounded
off
as
278.2
We
first
locate
to
the
in
row
log
corresponding
tables
and
(ii)
43
the
The
characteristic
is
is
2
and
the
mantissa,
using
log
tables,
.4443
Procecd
horizontal)
y
lill
we
reach
the
column
corresponding
to
2.
log
278.23
(ii)
=
2.4443
0.07058
is
The
characteristic
of
log
is
The
number
6355.
at
the
intersection
is
-2
which
written
as
2
by
convention.
Using
log
tables
the
mantissa
that
is
.8487,
so
corresponding
to
7
is
3.
Adding
2109
and
3
we
get
2112.
Log
0.07058
l.vamnle
=
2.8487
Since
the
characteristic
increased
by
1
is
1,
it
is
(because
there
should
be
the
intégral
part)
Find
the
numbers
whose
iogarithms
are
(i)
two
two
digits
in
and
therefore
the
décimal
point
is
fixed
after
1.3247
(ii)
2.1324
digits
front
left
in
2112.
Solution
(i)
Hence
(ii)
antilog
of
1.3247
is
21.12.
1.3247
2.1324
Proceeding
as
in
(i)
the
signifîcant
Reading
corresponding
0.3247),
this
alotig
the
row
to
.32
(as
mantissa
=
of
figures
corresponding
to
the
mantissa
is
1.
we
get
2109
at
the
intersection
0.1324
are
1356.
Since
the
characteristic
row
with
the
colutnn
corresponding
to
intersection
of
this
2
,
its
numerical
value
2
there
will
be
is
decreased
by
4.
The
number
at
the
row
and
the
mean
Hence
one
zéro
after
the
différence
coiumn
décimal
point.
Hence
antilog
of
2
.1324
is
0.01356.
Ql.
Find
the
comuton
logarithm
of
Hence
log
0.3206
Q2.
If
log
31.09
=
1.5060
the
foliowing
numbers.
1)
=
1.4926,
find
the
232.92
232.92 can
be
rounded
off
as
232.9
Characteristic
values
of
foliowing:
i)
log
3.109
log
3.109
Characteristic
=
=3
2
.3672
Sol:
Mantissa
a
0
Hence
log
232.92
H)
*
=
2.3672
Mantissa
29326
29.326
can be
rounded
off
as
29.33
Characteristic
SS
So
log
3.109
ü)
Sol:
=
.4926
=
0.4926
log
310.9
log
3
10.9
Characteristic
1
Mantissa
35
.4673
=
2
.4926
Hence
log
29.326
üi)
.
=
-
=
—
1.4673
Mantissa
=
=
0.00032
Characteristic
So
4
.5051
ni)
log
310.9
2.4926
log
0.003109
log
0.003109
Characteristic
Mantissa
Sol:
Hence
log
0.0032
iv)
4.5051
0.3206
Characteristic
ï
Mantissa
=
=
=
3
.4926
Soiog
0.003109
iv)
3.4926
Mantissa
=
.5060
log
0.3109
Sol:
log
0.3109
Characteristic
1
Mantissa
=
=
a
ï
2
=6
Squaring
both
side
.4926
î
So
Q3.
log
0.3109
=
numbers
are:
.4926
Find
tbe
whose
Ml
üi)
!
=(*)
comnion
logarithms
i)
3.5621
log,
n
let
the
number
be
x
=
2
form
cl
log
x
=
3.5621
In
exponential
Characteristic
Mantissa
x
x
=
antilog
ss
3.5621
=
=
=
required
3
.5621
5
=n
3648
iv)
10
p
=40
form
3648
is
the
In
logarithmic
Hence
3648
ü)
number
or
Ï.7427
Log
jo
40
log
40
Characteristic
=
P
=P
=
1
Let
the
number
be
x
Log
x
=
1.7427
Mantissa
=
=
.6021
Characteristic
Mantissa
=
=
=
the
So,
ï
P
Evaluatc
log,
1.6021
.7427
Q5,
i)
=
antilog
x
=
0.5530
x
ï.7427
0.5530
128
is
Hence
number
Q4.
0.5530
required
Let
x
-
Log,
—
62
128
What
replacement
for
the
In
exponential
form
i)
unknown
in
each
of
following
will
make
the
staiement
true?
logj81
=
L
In
exponential
r-JL
128
2*
X
_ J_
2
7
form
2
3
3
=>
|L
l
1
=81
=
2'
7
'
=
3
4
Bases
are
=>
equal
.y
=
—7
=
4|
so
ii)
log
5
12
to
the
base
2yfl
log
2
exponents
are
equal
ii)
Sol:
^512
=
log
log,
6
=
0.5
form
„0.S
Let
x
512
form
In
exponential
In
exponential
a
C
=6

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