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

Title: log question, posssibly considered ambigous?
Post 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?
Title: Re: log question, posssibly considered ambigous?
Post by: TrueTears on November 06, 2009, 12:01:58 pm: 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 $\frac{3}{2}$
Title: Re: log question, posssibly considered ambigous?
Post by: simpak on November 06, 2009, 12:02:41 pm: 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.
Title: Re: log question, posssibly considered ambigous?
Post by: nala on November 06, 2009, 12:06:13 pm: 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.
Title: Re: log question, posssibly considered ambigous?
Post by: m@tty on November 06, 2009, 12:07:28 pm: You have to see whether the solutions are defined in the original unaltered expression.
Title: Re: log question, posssibly considered ambigous?
Post by: almostatrap on November 06, 2009, 12:08:20 pm: all good points! oh well
Title: Re: log question, posssibly considered ambigous?
Post by: TrueTears on November 06, 2009, 12:09:33 pm: Quote from: 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?
Also remember squaring a number leads to redundant solutions.

$x^2 = 4$

$x = \pm 2$

However if you had $x = \sqrt{4}$ you'd only be left with the positive answer $x = 2$ .

Same logic applies, squaring the x leads to the redundant solution of x = 1.
Title: Re: log question, posssibly considered ambigous?
Post by: NE2000 on November 06, 2009, 12:38:54 pm: It's a good habit to define domains (at least in your head) before you attempt a question.
Title: Re: log question, posssibly considered ambigous?
Post by: lilyrose on November 06, 2009, 01:10:30 pm: 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??
Title: Re: log question, posssibly considered ambigous?
Post by: NE2000 on November 06, 2009, 01:11:40 pm: 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
Title: Re: log question, posssibly considered ambigous?
Post by: tomygun_123 on November 06, 2009, 01:20:25 pm: 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.
Title: Re: log question, posssibly considered ambigous?
Post by: lilyrose on November 06, 2009, 01:25:36 pm: Quote from: NE2000 on November 06, 2009, 01:11:40 pm
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