# I want check my answer

• Nov 12th 2009, 08:43 AM
r-soy
Hi
Question 1
Attachment 13826
Question 2
Attachment 13827
• Nov 12th 2009, 07:02 PM
TKHunny
$\displaystyle x^{log(x)}\;=\;100 \cdot x$

You started out nicely

$\displaystyle log(x) \cdot log(x)\;=\;log(100 \cdot x)$

Then you rather wandered off.

$\displaystyle [log(x)]^{2}\;=\;log(x) + log(100)$

This is tantalizingly familiar.

$\displaystyle [log(x)]^{2}\;-\;log(x)\;-\;2\;=\;0$

It's Algebra I from here. Don't forget the ORIGINAL Domain.
• Nov 12th 2009, 09:10 PM
r-soy
I want exaplin to first step :

$\displaystyle log(x) \cdot log(x)\;=\;log(100 \cdot x)$

and last step

why you say - 2 why you don't say - 100
$\displaystyle [log(x)]^{2}\;-\;log(x)\;-\;2\;=\;0$
• Nov 13th 2009, 01:35 AM
Bacterius
Because $\displaystyle log_{10}(100) = 2$, I guess.
• Nov 13th 2009, 02:35 PM
TKHunny
It appeared to be a more interesting problem if these were logs base 10, so I assumed that they were and just threw that step in without discussion to see if you were paying attention. You were! Good work.

log(100x) = log(100) + log(x)

After that, why would I say just "100"? We're stuck with the logarithm function until it is treated properly.
• Nov 14th 2009, 07:46 AM
r-soy
thanx a lot