ATAR Notes: Forum
Archived Discussion => 2009 => Endofyear exams => Exam Discussion => Victoria => Mathematical Methods (and CAS) => Topic started by: almostatrap on November 06, 2009, 12:00:34 pm

so x=1 is not valid. and 2ln(x) does not always = ln(x^2)
I just went through the year 11 and 12 essentials textbooks, and nowhere does it mention exceptions to log laws. Even searching the internet i couldn't find it. And i have never been taught it (my teacher is awesome).
So we're just meant to know intuitively that log laws are not valid in this situation?
I declare question 9 ambiguous! marks for everyone!
I doubt this will be the case, but yeah.
what do people think?

I said the domain is the intersection of the two functions.
For y = 2ln(x) the domain is x > 0
For y = ln(x+3) the domain is x>3
Thus the domain is x>0
Since 1<0
We only take

I assumed that the negative answer I got couldn't be right because then you would be taking the log of a negative.
BUT, the negative answer I got was not 1 because I made a silly mistake.
Still, same logic.

I figured that rather than putting the 2 above the x, say you divided the other logs by 2 instead. You are left with loge(1), therefore it is undefined.

You have to see whether the solutions are defined in the original unaltered expression.

all good points! oh well

so x=1 is not valid. and 2ln(x) does not always = ln(x^2)
I just went through the year 11 and 12 essentials textbooks, and nowhere does it mention exceptions to log laws. Even searching the internet i couldn't find it. And i have never been taught it (my teacher is awesome).
So we're just meant to know intuitively that log laws are not valid in this situation?
I declare question 9 ambiguous! marks for everyone!
I doubt this will be the case, but yeah.
what do people think?
Also remember squaring a number leads to redundant solutions.
However if you had you'd only be left with the positive answer .
Same logic applies, squaring the x leads to the redundant solution of x = 1.

It's a good habit to define domains (at least in your head) before you attempt a question.

do you get marks off if you forgot to simplfy things?
also, i forgot to include units for rate of change and i accidently left the ve value forlog...is that 1 mark each??

In spesh the assessment reports say that you must simplify. But I haven't seen that in a methods one so maybe not simplifying it completely will not be penalized :)
The units for rate of change, the question specifically asked you to put units so not sure. And the ve value for log is an incorrect answer

you can only take x>3/2 since it wasnt an absolute function, also taking the square of 1 will give you + answers therefore you must reject the negative... i would think that this was just common sense since bodmas clearly applies within the brackets. But i guess it could have been a bit ambiguous.

In spesh the assessment reports say that you must simplify. But I haven't seen that in a methods one so maybe not simplifying it completely will not be penalized :)
The units for rate of change, the question specifically asked you to put units so not sure. And the ve value for log is an incorrect answer
well i accidently forgot to disregard the ve.. i still had the 3/2.. would i get one mark off for including that x=1?
so annoyed, i made sososososo many stupid mistakes