Hi

I want check my answer

Question 1

Attachment 13826

Question 2

Attachment 13827

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- Nov 12th 2009, 08:43 AMr-soyI want check my answer
Hi

I want check my answer

Question 1

Attachment 13826

Question 2

Attachment 13827 - Nov 12th 2009, 07:02 PMTKHunny
$\displaystyle x^{log(x)}\;=\;100 \cdot x$

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 PMr-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 AMBacterius
Because $\displaystyle log_{10}(100) = 2$, I guess.

- Nov 13th 2009, 02:35 PMTKHunny
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 AMr-soy
thanx a lot