1. ## logs question

$\displaystyle \frac{ log_2{16} + log_2{32} }{log_2{x}} = log_2{x}$

$\displaystyle \frac{log_2{512}}{log_2{x}} = log_2{x}$

$\displaystyle log_2{512} = log_2{x^2}$

$\displaystyle 512 = x^2$

$\displaystyle \sqrt{512} = x$

so x = 22.6 , is this correct?

appreciate any input.

thanks

2. Originally Posted by Tweety
$\displaystyle \frac{ log_2{16} + log_2{32} }{log_2{x}} = log_2{x}$

$\displaystyle \frac{log_2{512}}{log_2{x}} = log_2{x}$

$\displaystyle log_2{512} = log_2{x^2}$

$\displaystyle 512 = x^2$

$\displaystyle \sqrt{512} = x$

so x = 22.6 , is this correct?

appreciate any input.

thanks
$\displaystyle 16 = 2^4$
$\displaystyle 32 = 2^5$

$\displaystyle 9 = (log_2{x})^2$

$\displaystyle log_2{x} = \pm 3$

$\displaystyle x = 2^{\pm 3}$

Therefore $\displaystyle x = 8 \: , \: x = \frac{1}{8}$

3. Originally Posted by Tweety
$\displaystyle \frac{ log_2{16} + log_2{32} }{log_2{x}} = log_2{x}$

$\displaystyle \frac{log_2{512}}{log_2{x}} = log_2{x}$

$\displaystyle log_2{512} = log_2{x^2}$
PROBLEM is here, as

$\displaystyle (log_2{x})*(log_2{x})\not=log_2{x^2}$

You want

$\displaystyle log_2{512} = (log_2{x})^2$ ,

which is where the

$\displaystyle 9 = (log_2{x})^2$ written in the above post, comes from.