# Thread: [SOLVED] Log identity problem

1. ## [SOLVED] Log identity problem

Prove that :

(Log(base 2)X -1)^ 2 / Log(base 2)X = Log(base 2)X + Log(base x)2 –2

I have tried but I always end up with Log(base 2)X + Log(base 2)X^-2 + 1

All help appreciated

2. We have to prove that $\displaystyle \frac{\left(\log_2x-1\right)^2}{\log_2x}=\log_2x+\log_x2-2$ ($\displaystyle x>0$, $\displaystyle x\ne 1$).

$\displaystyle \frac{\left(\log_2x-1\right)^2}{\log_2x}=\log_2x+\frac{\log_22}{\log_2 x}-2\Leftrightarrow$ $\displaystyle \frac{\left(\log_2x-1\right)^2}{\log_2x}=\log_2x+\frac{1}{\log_2x}-2\Leftrightarrow$ $\displaystyle \left(\log_2x-1\right)^2=\log_2^2x+1-2\log_2x$ which is an identity. QED

3. Hello, MarcoMP!

We need this identity: .$\displaystyle \log_ba \:=\:\frac{1}{\log_ab}$

Prove that: .$\displaystyle \frac{(\log_2x - 1)^2}{\log_2x} \;=\;\log_2x + \log_x2 - 2$

On the left side we have:

. . $\displaystyle \frac{(\log_2x)^2 - 2\log_2x + 1}{\log_2x}$

. . $\displaystyle =\;\frac{(\log_2x)^2}{\log_2x} - \frac{2\log_2x}{\log_2x} + \frac{1}{\log_2x}$

. . $\displaystyle =\;\log_2x - 2 + \underbrace{\frac{1}{\log_2x}}_{\downarrow}$
. . $\displaystyle = \;\log_2x - 2 + \log_x2$