0

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

Nikunj
  • 6,160
user329048
  • 1
  • It is equivalent to $x \le e^{x-1}$, which is clearly true when $x=1$. You could then use derivatives – Henry Apr 06 '16 at 06:35
  • Hi, welcome to Math.SE. Please indicate what you have tried and where you are stuck. This will help people better tailor their answer to your background and situation. It will also demonstrate that you are interested in your question and not just looking for someone to do your homework for you - Math.SE is not a homework site. – Ian Miller Apr 06 '16 at 06:36

2 Answers2

0

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 $2$.

Anurag A
  • 41,067
  • Well it's clear how to prove $ex \le e^x$, using the formula for subtracting logs, in the domain where $\log x$ is defined. However ,$ex\le e^x$ seems to have only one solution $x=1$. A python plot also prooves this. – Alexander Cska Feb 11 '18 at 12:15
-1

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.

Holie Chen
  • 21
  • I guess that the last inequality is the only non-trivial thing in the question. –  Apr 06 '16 at 07:02
  • okay, there're two ways to understand this. First, you can draw the graphs of both side on an x-y plane, then you will get it. Second, I am not sure if you are familiar with calculus. If so, let $ f(x) = x - log(x) -1 $, take its derivative $ f'(x) = 1- \frac{1}{x} = 0 $, the root is $ x = 1 $, which implies that the minimum of $f(x)$ is $f(1) = 0$ – Holie Chen Apr 06 '16 at 07:07