Tricky Logarithmic inequality I have tried proving this logarithmic inequality but I did not succeed. I tried to put every term on one side, I expanded and tried to use one of the properties of logarithms but the proof does not come out clearly mathematically. 

$$\log_2(x) \leq (x-1)\log_2(e)$$

Thanks for your response
Hint

For $x \geq 1$ we have $ex \leq e^x$ (can you see why?). Then use monotonicity of loagrithm to take log of both sides with base $2$.
Actually this is easy.

Notice that we have $$ \log_2(x) = \frac{\log(x)}{\log(2)} $$

Plug it in both side and we have
$$
\frac{\log(x)}{\log(2)} \leq \frac{x-1}{\log(2)} \\
\log(x) \leq x-1
$$
which is trivial.