I want check my answer

**r-soy** · November 12th 2009, 08:43 AM

Hi
I want check my answer
Question 1

Question 2

**TKHunny** · November 12th 2009, 07:02 PM

$x^{log(x)}\;=\;100 \cdot x$

You started out nicely

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

Then you rather wandered off.

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

This is tantalizingly familiar.

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

It's Algebra I from here. Don't forget the ORIGINAL Domain.

**r-soy** · November 12th 2009, 09:10 PM

I want exaplin to first step :

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

and last step

why you say - 2 why you don't say - 100
$[log(x)]^{2}\;-\;log(x)\;-\;2\;=\;0$

**Bacterius** · November 13th 2009, 01:35 AM

Because $log_{10}(100) = 2$ , I guess.

**TKHunny** · November 13th 2009, 02:35 PM

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.

**r-soy** · November 14th 2009, 07:46 AM

thanx a lot

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