A quick query about working out on DE question from VCAA 2010 Exam 1 Q7. +1

[PB]

Hey guys,
Could anyone please enlighten me as to why the mod is taken out from the the log by the end of the answer?? I got loge (|1+x/1-x|)...
Also, is it true that "ln" is no longer allowed in VCE Spesh? only loge?

Thanks!

[b^3]

We have the restriction , so we need to look at what values can take with these values of .
Substituting in we get , and when as , . Now just to check the shape of the curve I'll expand it out.


Now we can see that it is a part of that hyperbola, so if you do a little sketch you'll see that for , and as you increase , it will increase and go off to infinity (as we found before), now this means (by looking at the sketch) that for those values of , , that is it is always positive, so the modulus can be removed.

Also, is it true that "ln" is no longer allowed in VCE Spesh? only loge?
Thanks!

I've heard this too, I'd stick with just to be safe.

EDIT: Wrote log instead of mod, fixed.

[lzxnl]

If you look carefully at the domain, -1<x<1
Now (1+x)/(1-x) is always positive in that domain. 1+x is always greater than zero and so is 1-x. If you want, (1+x)/(1-x) = (1-x^2)/(1-x)^2
From domain restriction abs x<1, so x^2<1. The top is always positive. The bottom is a perfect square. Therefore the argument of the log is always positive.

As for whether we can use ln, I can't say for sure, but IMO it would be stupid if it were forbidden.

[PB]

Thanks both of you. I think I might stick with b^3's method of sketching though...it seems clearer to me
Btw, is it really necessary to get rid of the mod? because technically it still isn't wrong right?

[lzxnl]

You can't be penalized for putting it in, but sometimes you need to remove the mod, such as when finding the inverse of this expression or something.

[b^3]

If you look carefully at the domain, -1<x<1
Now (1+x)/(1-x) is always positive in that domain. 1+x is always greater than zero and so is 1-x. If you want, (1+x)/(1-x) = (1-x^2)/(1-x)^2
From domain restriction abs x<1, so x^2<1. The top is always positive. The bottom is a perfect square. Therefore the argument of the log is always positive.

As for whether we can use ln, I can't say for sure, but IMO it would be stupid if it were forbidden.

I actually like this way better, it's the way I initially went to do it, but couldn't get it to work. Now I realise what I did, apparently a few hours ago I thought 1-(-1)=-2 .... sigh -.- (it confused the hell out of me when it didn't seem to work even though I knew it was meant to come out positive....)

It's probably advantageous to learn/understand the above method, can help you get out of tight spots quickly with less effort than the other method.

[PB]

ah thanks, B^3 and Nliu! Sorry for the late reply, I've been really busy what with the UMAT and stuff