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Napier's Logarithm's interpreted with differential equations.

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(as in Struik's Notes to NAPIER.)

Let TS = A=10^{7}.

Let dS = y.

The position of g at time t is A-y from its starting
point, I will let P=A-y.

The position of the token a at time t is bc and is called
*x.*

Let the initial velocity of g be denoted v0.

Then dP/dt = - dy/dt.

Napier's statement is that " a geometrically moving point
approaching a fixed one has its velocities proportionate to its distances
from the fixed one."
Thus at time t, dy/dt = k y where k is a constant for
proportionality. Napier is using k= -1.

Thus y = v0 e^{-t} . Solving this equation
for t using the natural logarithm gives t=ln(v0/y)= ln(v0)-ln(y)

Now since the token a is moving at a constant rate, namely
v0, we have *x* = v0*t. *x *is Napier's log for the y value
at the same time t. Hence Napier's log for y, which I will denote by NOG(y)
has NOG(y)=v0*(ln(v0/y)) = v0*(ln(v0)-ln(y)).

Notice that for y = A, NOG(y) = 0, so v0=A.

Notice that the computation using NOG is more complicated
then those using the modern logarithms (when A is not 1).

Multiplication Note:

Find the product of two numbers *x and y.*

Find NOG(*xy*) and then use the table
to find *xy*. This is the same as how ln would be used. However the
formula used to compute NOG(*xy*) is not as simple as ln(*xy*)=
ln(*x*) + ln(*y*).
NOG (*xy*) = A* (ln(A) - ln(*x*) - ln(*y*))

= NOG(*x*) -A*ln(*y*)

= NOG(*x*) + NOG(*y*) - A*ln(A).

Using A = 10000000, ln(A) is approximately
16.118095651

Notice that if a/b=c/d then

NOG(a)- NOG(b)=v0*(ln(b)-ln(a)) = v0*(ln(d)-ln(c)=NOG(c)-NOG(d)

thus

NOG(c)= NOG(a) + NOG(d) - NOG(b)