1. logs question

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

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

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

$512 = x^2$

$\sqrt{512} = x$

so x = 22.6 , is this correct?

appreciate any input.

thanks

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

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

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

$512 = x^2$

$\sqrt{512} = x$

so x = 22.6 , is this correct?

appreciate any input.

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

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

$log_2{x} = \pm 3$

$x = 2^{\pm 3}$

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

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

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

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

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

You want

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

which is where the

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