Mathematics Question Thread

[12070]

How do you work this style of question out? I feel like it should be easy I don't know which laws to use.

Which expression is equivalent to 4 + logx2 ?
(A) log2(2x)
(B) log2(16 + x)
(C) 4log2(2x)
(D) log2(16x)

Sorry for the copy and paste issue but it's from the 2016 HSC paper; Question 10

[RuiAce]

How do you work this style of question out? I feel like it should be easy I don't know which laws to use.

Which expression is equivalent to 4 + logx2 ?
(A) log2(2x)
(B) log2(16 + x)
(C) 4log2(2x)
(D) log2(16x)

Sorry for the copy and paste issue but it's from the 2016 HSC paper; Question 10

Addressed in posts #1990 and #1991

[MisterNeo]

How do you work this style of question out? I feel like it should be easy I don't know which laws to use.

Which expression is equivalent to 4 + logx2 ?
(A) log2(2x)
(B) log2(16 + x)
(C) 4log2(2x)
(D) log2(16x)

Sorry for the copy and paste issue but it's from the 2016 HSC paper; Question 10

Here's the proof for the question.
Note: You wrote logx2 when it was log2x

Looking at the options of multiple choice can help identify which log laws you need to apply.
Hope this helps

[12070]

Here's the proof for the question.
Note: You wrote logx2 when it was log2x
(Image removed from quote.)
Looking at the options of multiple choice can help identify which log laws you need to apply.
Hope this helps

Thanks for that. I guess the hard part for me at least was realising that 4= log16 base 2. Tricky

[IronBark]

Thank you so much for your guys help.

I need help with this question: As dry air moves upward, it expands and cools. The temp at a height of 1 km = 10(deg) and temp at 3 km (-5(deg))

1. Assuming there is a constant rate of change of temp with respect to height, express the temp in terms of the height h
2. What is the meaning of the slope?
3. What is the meaning of the horizontal intercept of the line

Can you please show working.... 

for 1 would you assume something like T= H(10).. I dont know

[MisterNeo]

Thank you so much for your guys help.

I need help with this question: As dry air moves upward, it expands and cools. The temp at a height of 1 km = 10(deg) and temp at 3 km (-5(deg))

1. Assuming there is a constant rate of change of temp with respect to height, express the temp in terms of the height h
2. What is the meaning of the slope?
3. What is the meaning of the horizontal intercept of the line

Can you please show working.... 

for 1 would you assume something like T= H(10).. I dont know

I haven't done this topic, but I think this is it. Never heard of "meaning of.."   

 

[jamonwindeyer]

Thank you so much for your guys help.

I need help with this question: As dry air moves upward, it expands and cools. The temp at a height of 1 km = 10(deg) and temp at 3 km (-5(deg))

1. Assuming there is a constant rate of change of temp with respect to height, express the temp in terms of the height h
2. What is the meaning of the slope?
3. What is the meaning of the horizontal intercept of the line

Can you please show working.... 

for 1 would you assume something like T= H(10).. I dont know

Just to help with 2 and 3, I think it wants more physical interpretations:

- The slope of the line represents the change in temperature with temperature, how fast the temperature drops with altitude.
- The horizontal intercept represents the altitude where the temperature reaches zero!

[IronBark]

I haven't done this topic, but I think this is it. Never heard of "meaning of.."   
(Image removed from quote.)
 

Your a legend Cheers. Sorry with this exam, it does not include answers so i dont know If I am getting these right or not. These are some of the questions, Which i am not sure if are correct.

1. (integral sign)4xcos(x^2+1)dx
So I put that it = ((4x)^2)/2 x sin((x^2+1)^2)/2 x 2x
                       = 2x^3sin(x^2+1)/2 + c

2. Anti derivative of f(x)?

 (f(x)^(n+1))/n

3. f(x) =4(root)x+(1)/((2x)^2), find f'(x)
f(x)= (4x)^1/2 + 2x^-2
f'(x)= ((2x)^-1/2)-4x^-3
      = (2/(root)x) - 4/x^3

4. Explain what the function notation g: [0,1] (-1,1) means:

As 0,1 approach the x and y axis, they intercept the boundaries of -1,1 though not being limited to

[MisterNeo]

Your a legend Cheers. Sorry with this exam, it does not include answers so i dont know If I am getting these right or not. These are some of the questions, Which i am not sure if are correct.

1. (integral sign)4xcos(x^2+1)dx
So I put that it = ((4x)^2)/2 x sin((x^2+1)^2)/2 x 2x
                       = 2x^3sin(x^2+1)/2 + c

2. Anti derivative of f(x)?

 (f(x)^(n+1))/n

3. f(x) =4(root)x+(1)/((2x)^2), find f'(x)
f(x)= (4x)^1/2 + 2x^-2
f'(x)= ((2x)^-1/2)-4x^-3
      = (2/(root)x) - 4/x^3

4. Explain what the function notation g: [0,1] (-1,1) means:

As 0,1 approach the x and y axis, they intercept the boundaries of -1,1 though not being limited to

Hmm...your post is a bit sketchy but I'll try to interpret it.

Not sure about Q4 though...

[Joseph41]

This is such a good thread, far out. Make sure y'all +1 the helpful responses!

Keep being awesome, ATAR Notes.

[RuiAce]

Your a legend Cheers. Sorry with this exam, it does not include answers so i dont know If I am getting these right or not. These are some of the questions, Which i am not sure if are correct.

1. (integral sign)4xcos(x^2+1)dx
So I put that it = ((4x)^2)/2 x sin((x^2+1)^2)/2 x 2x
                       = 2x^3sin(x^2+1)/2 + c

2. Anti derivative of f(x)?

 (f(x)^(n+1))/n

3. f(x) =4(root)x+(1)/((2x)^2), find f'(x)
f(x)= (4x)^1/2 + 2x^-2
f'(x)= ((2x)^-1/2)-4x^-3
      = (2/(root)x) - 4/x^3

4. Explain what the function notation g: [0,1] (-1,1) means:

As 0,1 approach the x and y axis, they intercept the boundaries of -1,1 though not being limited to

Whilst Q4 is in the VCE course it is NOT in the HSC course.

It means that the function takes values from the domain \(0\le x \le 1\) and maps them over to the codomain \(-1< x < 1\)

Asking a ton of questions especially when it's continuous can be quite offputting. Please relax the speed that you ask the questions. Additionally, if it's too hard to type please resort to simply posting images. (You could also try learning LaTeX, but that may be too time consuming and posting images would be more efficient)

[RuiAce]

Hmm...your post is a bit sketchy but I'll try to interpret it.
(Image removed from quote.)
Not sure about Q4 though...

By the way, regarding your Q1 I personally would skip line 2 That assumes that you could divide by 2x even when there was no x term there to begin with which is false. Jump straight to the answer, or write out a derivative and alternatively compare it to the integral.

Note: reversing the chain rule technically isn't required until Extension 2, but it's nice to know.

[jamonwindeyer]

Asking a ton of questions especially when it's continuous can be quite offputting. Please relax the speed that you ask the questions. Additionally, if it's too hard to type please resort to simply posting images. (You could also try learning LaTeX, but that may be too time consuming and posting images would be more efficient)

What Rui means to say is, if you ask a lot of questions, it might take a bit to get to them all. So if you're really struggling with a specific thing, posting that one thing (or perhaps one question to check your understanding) would probably get you an answer faster

But please ask as many questions as you like! Ultimately, if Rui doesn't answer, I will. Or MisterNeo will. Or jakesilove will. Or kiwiberry will. Or ellipse will. Or Shadowxo will. Point being, heaps of people around to answer questions (such is the awesomeness of ATAR Notes), we'll get to all of them eventually

[cxmplete]

hi,
how would you graph a x-2lnx graph, and are there particular steps to graphing it?

[jamonwindeyer]

hi,
how would you graph a x-2lnx graph, and are there particular steps to graphing it?

Hey! if you were looking for just a rough sketch (without calculus), then your best bet would be to use superposition. Draw a graph of \(y=x\), and then another of \(y=-2\ln{x}\) (that's just a regular looking log graph, flipped vertically). Adding the two function values at any given \(x\)-coordinate will give you the \(y-\)-coordinate, if that makes sense. Remember, if one of the functions doesn't exist (like the logarithm doesn't for negative values), your new one won't either!!

Otherwise, you'd apply the normal techniques based on differentiation (turning points, inflexions, etc.)