If $5^{a_{n+1} - a_n} = 1+\frac{1}{n+\frac 23} $, and $a_1=1$ then prove that $[a_{207} ] =3$ (Greatest integer function) I am going to skip through the exhausting computation part to get my result
$$a_{n+1} -a_1= \log_{5} (\frac{5+3n}{5})$$
Using $n+1 \to n$
$$\implies a_n-1 =\log_5 \frac { 2+3n}{5}$$
So $a_{207}$ is
$$a_{207} = 1+\log_5 \frac{623}{5}$$
It would be a shame to see that I came this far but got stuck at the final part of answer, ie. calculation of the logarithm, but here we are. I am not supposed to have a calculator or a log table at my disposal, so how can I guess the value?

You just use the fact that $5^3 < 623 < 5^4$, so that $5^3/5 < 623/5 < 5^4/5$, and hence
$$\log_5  5^3/5 < \log_5 623/5 < \log_5 5^4/5$$
In other words, $2 < \log_5 623/5 < 3$. Thus
$$\lfloor a_{207} \rfloor = \lfloor 1 + \log_5 623/5 \rfloor$$
$$= 3$$
