What is the proof of $x\ln x - x = x\ln\left(\frac{x}2\right)$?

Question

What is the proof of $x\ln x - x = x \ln\left(\frac{x}2\right)$ ?

i know that: $x\ln x - x$ = $\int \ln x \, dx $ (after a bit of math manipulation)

so $\int \ln x \, dx $ = $x\ln\left(\frac{x}2\right)$, how do i continue to complete it?

Answer 1

$$x\log(x)-x=x(\log(x)-1)=x(\log(x)-\log(e))=x\left(\log\left(\frac{x}{e}\right)\right) \neq x\left(\log\left(\frac{x}{2}\right)\right)$$ But: $$x\log_2(x)-x=x(\log_2(x)-1)=x(\log_2(x)-\log_2(2))=x\left(\log_2\left(\frac{x}{2}\right)\right)$$

Answer 2

You can't prove it because it is not true:$$x\ln\left(\frac x2\right)=x\left(\ln(x)-\ln(2)\right)=x\ln(x)-x\ln(2)$$which, in general, is not $x\ln(x)-x$.

Answer 3

That's not true.

You have $x\ln{x}-x=x(\ln{x}-1)=x(\ln{x}-\ln{e})=x\ln{\frac xe}$

Of course I have taken the classical logarithm function.

If you take $\log_2$, then yes, it holds true.

Answer 4

You can't prove it, as it's not true. For instance, substituting $e$ in for $x$ you have $e - e = 0$ on the left and $e \ln (e/2) \neq 0$ on the right.