The formula,
log n – log (n – 1) = 1/n + 1/2n

^{2}+ 1/3n^{3 }+ 1/4n^{4 }+…., can be used to generate a simple infinite series expression for every number.
Basically
if n is of the form t/(t – 1), then log n – log (n – 1) is of the form,

log t/(t – 1) – log (1/t – 1) = log t.

So, for example to find log 2 we set t = 2.

Therefore log (2/1) – log (1/1) = log 2.

Thus with n in this case = 2/1,

log 2 = 1/1.2
+ 1/2.2

^{2 }+ 1/3.2^{3 }+ 1/4.2^{4 }+…
So summing the first four terms of the series we obtain .68229…,
which already approximates well to the correct answer .693147…

Likewise to find the corresponding expression for log 3 we
set t = 3, with n thereby = 3/2.

Thus log 3 = 2/1.3 + 2

^{2}/2.3^{2 }+ 2^{3 }/3.3^{3 }+ 2^{4}/4.3^{4 }+…
So again by summing the first four terms we obtain 1.03703…
as against the correct answer, 1.09861…

This indicates that in this case the answer converges more
slowly to the actual result and this will be increasingly the case as n becomes
larger!.

Indeed, in principle we can use the same formula i.e..

log n – log
(n – 1) = 1/n + 1/2n

^{2}+ 1/3n^{3 }+ 1/4n^{4 }+…,
to find an unlimited number of infinite expressions for the
log of any number.

For example, 4 = 2

^{2 }and 8 = 2^{3}, so we equally express log 2 in terms of any number that is a power of 2 and by extension the log of any number in terms of a number that is a corresponding power of that number..
So log 2 = (log
4)/2 and (log 8)/3 respectively!

Thus these
two alternative infinite series expressions for log 2 are,

(3/1.4 + 3

^{2}/2.4^{2 }+ 3^{3 }/3.4^{3 }+ 3^{4}/4.4^{4 }+… )/2 and
(7/1.8
+ 7

^{2}/2.8^{2 }+ 7^{3 }/3.8^{3 }+ 7^{4}/4.8^{4 }+… )/3 respectively.
However, a significant practical problem is that they
converge ever more slowly to the actual result.

This problem can however be solved in an interesting manner!

For example to find a quickly converging series for log 2,
we can set

t = 2 raised
to a small fractional power!

For example,
for t = 2

^{1/10}, n = t/(t – 1) = 2^{1/10}/(2^{1/10 }– 1)
Therefore

log 2 = 10{(2

^{1/10 }– 1)/2^{1/10 }+ (2^{1/10 }– 1)^{2}/2. 2^{2/10 }+ …….}
= 10(.066967…
+ .002242… + ….)

= .69209…

So, the result has already converged quite close to the actual
result (i.e. .693147…)

However by raising here 2 to a sufficiently small power, we
can approximate the log 2 as closely as we wish to the actual result through
sole consideration of the first term.

So in general if a is the number whose log we wish to estimate,
and 1/t is the power to which it is raised then the approximation = t(a

^{1/t }– 1)/a^{1/t}
Therefore for example, when a = 2 and t = 1,000,000 the
approximation for log 2 is given as,

1,000,000{(2

^{1/1,000,000 }– 1)/2^{1/1,000,000}}^{ }
= 1,000,000} (1.0000006931474… – 1)/1.0000006931474},

= 1,000,000(.0000006931469…) = .6931469…

So this answer is already correct to 6 significant figures
with respect to the actual result.

Indeed as the denominator a/

^{1/t }approximates ever more closely to 1 for very large t, the log for any number a can be approximated closely by the even more simple expression,
t(a

^{1/t }– 1) where t is sufficiently small.
Again for
example, log 145 = 4.97673374242…

So in this
case setting t = 1,000,000,000,

a

^{1/t }– 1 = 1.0000000049767337548… – 1 = .0000000049767337548…
Therefore
log 145 ~ 1,000,000,000(.0000000049767337548…)

=
4.9767337548…

As we can
see this approximation is already correct to 7 significant figures.

And we can continually improve this approximation by making t progressively larger.

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