# Logarithm

The solution $x(t)$ of the initial value problem

• $\frac{dx}{dt}=\frac{1}{t}$ for $t>1$
• $x(1)=0$,

is the logarithm $x(t)=\log (t)$, which is computed by time stepping $x = x + \frac{dt}{t}$ and extended to $t>0$ by time stepping backwards in time t.

The logarithm $x=\log(t)$ is the inverse of the exponential $t=\exp(x)$.

To do:

1. Play with the code changing into $dx=C\frac{dt}{t}$ with C a given number.

2. Show that by construction

• $\log(a*b)=\log(a) +\log(b)$.

Hint: Show that if $y(t) = \log(a*t)-\log(a)$, then $\frac{dy}{dt}=\frac{a}{a*t}=\frac{dt}{t}$ (set $s = a*t$ and note that $ds = a*dt$) and $y(t)=0$.  Conclude that $y(t)=\log(t)$, that is $\log(a*t)-\log(a)=\log(t)$ and so $\log(a*b)=\log(a)+\log(b)$, choosing $t = b$.

3. Show that log and exp are inverse functions (use that $dx=\frac{dt}{t}$ is the same as $dt=tdx$).