use systematic trial and error to evaluate log50.
By trial and error you have to find $\displaystyle x$ (approximatly) such that:
$\displaystyle 10^x=50$
or dividing by $\displaystyle 10$:
$\displaystyle 10^{x-1}=5$
start with $\displaystyle x=1$, then $\displaystyle 10^{x-1}=1$, which is too small,
next try $\displaystyle x=2$, then $\displaystyle 10^{x-1}=10$, which is too large
So for our next try we take $\displaystyle x=1.5$, then $\displaystyle 10^{x-1}=\sqrt{10}\approx 3.16$ which is too small
Now we know that the required $\displaystyle x$ is between $\displaystyle 1.5$ and $\displaystyle 2$, so for our next trial we use $\displaystyle x=1.75 $and carry on as above refining the interval containing the required solution.
CB