ATAR Notes: Forum
Archived Discussion => 2009 => End-of-year exams => Exam Discussion => Victoria => Mathematical Methods (and CAS) => Topic started by: almostatrap on November 06, 2009, 12:00:34 pm
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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?
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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
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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.
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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.
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You have to see whether the solutions are defined in the original unaltered expression.
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all good points! oh well
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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.
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It's a good habit to define domains (at least in your head) before you attempt a question.
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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??
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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
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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.
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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