1. ## I want check my answer

Hi
Question 1

Question 2

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

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.

3. 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
$\displaystyle [log(x)]^{2}\;-\;log(x)\;-\;2\;=\;0$

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

5. 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.

6. thanx a lot