VCE Methods Question Thread!

[Bhootnike]

With logs, say if you had:
ln(x)+ln(3x-1)=1

that is the same as :
ln(x)+ln(3x-1)=ln(e)

so x(3x-1) = e

if you just thought, hey, doesnt x= e, and x = 1/3, what are you doing wrong mathematically?
because, with questions and their answers, we're meant to expand, rearrange and use q'formula to find x.

[b^3]

With logs, say if you had:
ln(x)+ln(3x-1)=1

that is the same as :
ln(x)+ln(3x-1)=ln(e)

so x(3x-1) = e

if you just thought, hey, doesnt x= e, and x = 1/3, what are you doing wrong mathematically?
because, with questions and their answers, we're meant to expand, rearrange and use q'formula to find x.

You can only apply the null factor law when you have one side equalling 0. Otherwise above you are saying that x=e and 3x-1=0, so you are saying that e*0=e.

So what you have to do is as said above, get everything onto the one side, then solve for the x.




Now because anything inside a log has to be greater than 0, we have a restrcition of x>0 and 3x-1>0, hence x>0.
The solution with the minus in it will give a negative answer, so x<0 for that answer, so the solution will only be the positive solution.
i.e.

[Bhootnike]

With logs, say if you had:
ln(x)+ln(3x-1)=1

that is the same as :
ln(x)+ln(3x-1)=ln(e)

so x(3x-1) = e

if you just thought, hey, doesnt x= e, and x = 1/3, what are you doing wrong mathematically?
because, with questions and their answers, we're meant to expand, rearrange and use q'formula to find x.

You can only apply the null factor law when you have one side equalling 0. Here you are saying that at the same time x=e and 3x-1=0, so you are saying that e*0=e.

So what you have to do is as said above, get everything onto the one side, then solve for the x.




Now because anything inside a log has to be greater than 0, we have a restrcition of x>0 and 3x-1>0, hence x>0.
The solution with the minus in it will give a negative answer, so x>0 for that answer, so the solution will only be the positive solution.
i.e.

thank you very much. i think thats something i forgot from yr 9 math!
haha

ok another qtn - for those using the classpad,
when we sketch graphs and we want the e.g. point at which x = 3
we get a value of like 4.46357365
the question states, all answers must be in exact form.
when you copy that above value for y and go to main - paste-execute
you get something like
is there a way to get a more simpler answer cause like, when we have to sketch graphs and we have to label them, putting in such a fraction seems silly.

[Bhootnike]

ok another qtn - for those using the classpad,
when we sketch graphs and we want the e.g. point at which x = 3
we get a value of like 4.46357365
the question states, all answers must be in exact form.
when you copy that above value for y and go to main - paste-execute
you get something like
is there a way to get a more simpler answer cause like, when we have to sketch graphs and we have to label them, putting in such a fraction seems silly.

^^ the above question, and :

is the equation
subbing a few points in you find that

the next line of working to find a is as follows:

                             --- 1
 
              ------2

therefore,         --------3

can someone please explain how to get from step 2 to 3 ?

[b^3]

When you substitute what you found for b into you'll end up with (bringing the power from the front up).
Now remember that if
So that means that what is inside the log will be multipled by what was left.



This next bit isn't required, but just to show that is true (as I don't remember if the book gives this or not, but it should help you understand anyway)
Let
If we take the natural log of both sides

(Bringing the power down)
Now we know that a log with the same base as what is in the bracket is 1, i.e.
So we end up with (as

So that means that

[pi]

Remember than e and ln(x) are inverses of each other





edit: beaten cubed

[#1procrastinator]

For x^9 > 9, why is it ±x > 3 not x > ±3

[pi]

Do you mean x^2 > 9 ?

[link125]

I'm attempting the Exam 1 for 2011, question 3b

Struggling at the moment, could someone show me how to work it out?

Solve the equation:

sin(2x+pie/3)=1/2 for x [0,pie]

Thanks

[Aurelian]

I'm attempting the Exam 1 for 2011, question 3b

Struggling at the moment, could someone show me how to work it out?

Solve the equation:

sin(2x+pie/3)=1/2 for x [0,pie]

Thanks

hmm solve for x E [0, pie] you say? I dunno... maybe you're meant to eat it? :|

But srsly;

sin(2x + pi/3) = 1/2

Let 2x + pi/3 = @

sin(@) = 1/2

If you know your exact values, you'll know that the two base solutions for that are

@ = pi/6 and @ = 5pi/6

Now, each of these repeat every 2pi. So;

@ = ...pi/6 - 2pi, 5pi/6 - 2pi, pi/6, 5pi/6, pi/6 + 2pi, 5pi/6 + 2pi,...
@ = ...-11pi/6, -7pi/6, pi/6, 5pi/6, 13pi/6, 17pi/6...
@ = 2x + pi/3

Therefore;

2x + 2pi/6 = ...-11pi/6, -7pi/6, pi/6, 5pi/6, 13pi/6, 17pi/6...
2x = ...-13pi/6, -9pi/6, -pi/6, 3pi/6, 11pi/6, 15pi/6...
x = ...-13pi/12, -9pi/12, -pi/12, 3pi/12, 11pi/12, 15pi/12...

But x = [0, pi]

Therefore, x = 3pi/12, 11pi/12 only.

(or, simplified, x = pi/4, 11pi/12)

[TrueTears]

For x^9 > 9, why is it ±x > 3 not x > ±3

assuming x^2>9

+ or - x >3

case 1: x>3

case 2: -x>3 => x<-3

[link125]

I'm attempting the Exam 1 for 2011, question 3b

Struggling at the moment, could someone show me how to work it out?

Solve the equation:

sin(2x+pie/3)=1/2 for x [0,pie]

Thanks

hmm solve for x E [0, pie] you say? I dunno... maybe you're meant to eat it? :|

But srsly;

sin(2x + pi/3) = 1/2

Let 2x + pi/3 = @

sin(@) = 1/2

If you know your exact values, you'll know that the two base solutions for that are

@ = pi/6 and @ = 5pi/6

Now, each of these repeat every 2pi. So;

@ = ...pi/6 - 2pi, 5pi/6 - 2pi, pi/6, 5pi/6, pi/6 + 2pi, 5pi/6 + 2pi,...
@ = ...-11pi/6, -7pi/6, pi/6, 5pi/6, 13pi/6, 17pi/6...
@ = 2x + pi/3

Therefore;

2x + 2pi/6 = ...-11pi/6, -7pi/6, pi/6, 5pi/6, 13pi/6, 17pi/6...
2x = ...-13pi/6, -9pi/6, -pi/6, 3pi/6, 11pi/6, 15pi/6...
x = ...-13pi/12, -9pi/12, -pi/12, 3pi/12, 11pi/12, 15pi/12...

But x = [0, pi]

Therefore, x = 3pi/12, 11pi/12 only.

(or, simplified, x = pi/4, 11pi/12)

Thanks, helped a lot

Just wondering if you could explain why we use 2pi here:

"Now, each of these repeat every 2pi"

Would we use 2pi for each question for that part?

[pi]

It repeats very 2pi because that is the period of the basic sine function (ie. in this case sin(@))

[link125]

It repeats very 2pi because that is the period of the basic sine function (ie. in this case sin(@))

I don't understand entirely, isn't the period pi?
2pi/n=2pi/2 = pi

[Phy124]

It repeats very 2pi because that is the period of the basic sine function (ie. in this case sin(@))

I don't understand entirely, isn't the period pi?
2pi/n=2pi/2 = pi

The period of is before any dilation's occur (that's why he called it the basic sine function).

So the original sine function is equal to 1/2 for;





As an undilated sine function has a period of so the two values will reoccur every

But then the graph has been translated such that instead of x equalling these values, , does instead.





And then solve for x for the given domain as done above.