Calculating Napier's logarithm

We’ll show that

  \[ \frac{x}{10^7} = e^{-\left(\frac{y}{10^7}\right)}, \]    

The point $Q$ moves at a constant speed of $10^7$, so we have $dy/dt = 10^7.$ Since $P$ moves at a speed that is proportional to the distance $x$ left to travel we have $dx/dt = -x.$ From this we see that

  \[ dy/dx = -\frac{10^7}{x}, \]    

which gives

  \[ y = -10^7 \ln {cx}, \]    

for some constant $c.$

We can work out the value of $c$ using our initial conditions. At the start, the point $P$ still needs to travel the whole length of the line segment $AB$, which is $10^7$. Therefore $x_0 = 10^7$. The point $Q$ hasn’t gone anywhere yet, so $y_0 = 0.$

Plugging this into the expression above gives

  \[ 0 = -10^7 \ln {c10^7}, \]    

so

  \[ e^0 = 1 = c10^7. \]    

Therefore,

  \[ c = \frac{1}{10^7}, \]    

so

  \[ y = -10^7 \ln {\frac{x}{10^7}}, \]    

and

  \[ \frac{x}{10^7} = e^{-\left(\frac{y}{10^7}\right)}, \]    

which is what we wanted to show.

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