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Logarithm as Integral
Oguz Kurt
OSUMath 152
Spring 2008
Oguz Kurt
(OSUMath 152)
Logarithm as Integral
Spring 2008
1 / 11
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Question:
For positive real numbers
a, b
, how can we ﬁnd a solution to
the equation
a
x
=
b
? Namely, Find
log
a
b
=?
Say we wish to solve
2
x
= 5
. Namely,
log
2
5 =?
We know
2
2
= 4
<
5
and
2
3
= 8
>
5
.So,
2
< x <
3
.
Try
2
2
.
5
= 5
.
65685424949
>
5
. So,
2
< x <
2
.
5
One way is to try the number in the middle of the last interval (for
example, trying
2
.
25
and so on) till you get close enough to the perfect
answer. However, ﬁnding the answer to our problem through this
method is very hard and timeconsuming.
Not a very efﬁcient method
Oguz Kurt
(OSUMath 152)
Logarithm as Integral
Spring 2008
2 / 11
Scottish Mathematician Napier used a different method while he
invented logarithm as the ﬁrst time in early 17th Century. He found the
integer powers of
1
.
001
for a set of reasons:
1
it’s easy to calculate its powers
1
.
001
2
=
1
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This note was uploaded on 04/07/2008 for the course MATH 152 taught by Professor Kurt during the Spring '08 term at Ohio State.
 Spring '08
 kurt
 Real Numbers

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