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).
Find the product of two numbers x and 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
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)