How to prove $\log\left(\frac e{2t}\right) \approx \log\left(\frac1t\right)$ for $0 < t < \frac 12$?

Question

How do you prove $\log\left(\dfrac e{2t}\right) \approx \log\left(\dfrac1t\right)$ for $0 < t < \dfrac 12$?

I've simplified this to proving just $\log(2) \approx 1$. How do I prove this now without using a calculator?

Answer 1

In fact you have $$\log\left(\dfrac e{2t}\right)-\log\left(\dfrac 1t\right)=\log\left(\dfrac e2 \right)=1-\log2$$ You assume that $1-\log2\approx0\implies e\approx 2$ which is not a very good approximation.