Author Topic: VCE Methods Question Thread! (Read 4814971 times) Tweet Share

S_R_K · « **Reply #17865 on:** April 17, 2019, 12:34:24 pm »

Quote from: Rameen on April 17, 2019, 11:41:24 am

Thanks!
For 12b, how do I make the range of f, [0,∞), fit into the domain of g, R/{1}, for g(f1(x)) to be defined ?
The range of f can be written as (-∞, 1) U (1,∞). So would I make sqrt 2-x less than 1 and greater than 1?

That's not quite right, but you're on the right track. You can't write the range of f as (-∞, 1) U (1,∞), since even over its maximal domain the range of f(x) is [0,∞). However, if you do solve the inequalities sqrt(2 – x) < 1 and sqrt(2 – x) > 1, that will give you the correct domain for the composite function. Just be careful when solving: when you square both sides you may be introducing extraneous solutions, so double check that your answers satisfy the initial equation.

Jimmmy · « **Reply #17866 on:** April 19, 2019, 05:44:01 pm »

Solve for x in: $8e^{-x} - e^x = 2$

I've tried a few things here, namely rearranging the equation, trying to create a common base and create a log, but I couldn't work it out! The solutions in the textbook move the 2 to the LHS, then transitions it to:

$8 - 3^{2x} - 2e^x = 0$
$\therefore (e^x)^2 + 2e^x - 8 = 0$
then factorising and coming to $e^x = 2 \therefore x = ln(2)$

Can anyone clarify the the transition between the first three steps? I understand it from there, but I'm confused as to how they've juggled the numbers to be able to factorise in the first place.

MB_ · « **Reply #17867 on:** April 19, 2019, 06:05:47 pm »

Quote from: Jimmmy on April 19, 2019, 05:44:01 pm

Solve for x in: $8e^{-x} - e^x = 2$

I've tried a few things here, namely rearranging the equation, trying to create a common base and create a log, but I couldn't work it out! The solutions in the textbook move the 2 to the LHS, then transitions it to:

$8 - 3^{2x} - 2e^x = 0$
$\therefore (e^x)^2 + 2e^x - 8 = 0$
then factorising and coming to $e^x = 2 \therefore x = ln(2)$

Can anyone clarify the the transition between the first three steps? I understand it from there, but I'm confused as to how they've juggled the numbers to be able to factorise in the first place.

After moving 2 to the LHS, the whole equation is multiplied by e^x and then multiplied by -1

Rameen · « **Reply #17868 on:** April 20, 2019, 01:41:36 pm »

Hi. I am having trouble with part (d)(i) of this short answer question (images are uploaded)
Please look at the pictures to see the question, graph and and the previous parts of the question.
Thanks a lot

AlphaZero · « **Reply #17869 on:** April 20, 2019, 02:01:22 pm »

Quote from: Rameen on April 20, 2019, 01:41:36 pm

Hi. I am having trouble with part (d)(i) of this short answer question (images are uploaded)
Please look at the pictures to see the question, graph and and the previous parts of the question.
Thanks a lot

Define $d(x)$ to be the rule that gives the vertical distance between the the railway line and the cross-sectional boundary curve of the mountain/valley.

That is, $d(x)=\dfrac{-1}{8}x+\dfrac12-f(x)=\dfrac{-3x^3}{64}+\dfrac{7x^2}{32}-\dfrac{x}{8}$.

Then, use calculus techniques to find the value of $x$ (ie. the $x$-coordinate of $E$) that gives the maximal distance, noting that we are interested in $2/3<x<4$.

Jimmmy · « **Reply #17870 on:** April 20, 2019, 05:30:33 pm »

Quote from: MB_ on April 19, 2019, 06:05:47 pm

After moving 2 to the LHS, the whole equation is multiplied by e^x and then multiplied by -1

Thanks MB!

I've stumbled across another question; Q) Solve $4^{3x+1} > 5$ .

I've reached $\log_4(5/4) / 3$ , but the solutions show the answer as $\log_2(5/4) / 6$

I can see how it could work, but am a little confused as to the working out steps to get there. I'm pretty sure I'm just missing something very obvious.

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And another;

Q)Solve $\log_5(x) = 16\log_x(5)$
I got to $\ln(x) = \pm4\ln(5)$ and had no idea where to do to next. I checked the solutions and they went from there to;
$x = e^{\pm\ln(625)}$ and again I'm really confused how they just got rid of the natural log on both sides.
--------------------------------------------------------------------
And another! I'm really on a roll tonight; Q) Find inverse of function $f(x) = 5e^{2x} - 3$ . I find it pretty easily; $1/2(\ln((x+3)/5))$ , but I'm wondering how you would identify that x = -3 asymtote just from looking at the inverse, when the log includes the translation as being divided by 5 inclusive of the dilation in the y-axis? If it even possible we'd be asked such a question in Methods? I'm just trying to overrprepare, and wondering how I'd identify it without having the pre-known knowledge that the asymtotes simply switch with inverses.

S_R_K · « **Reply #17871 on:** April 21, 2019, 10:31:41 am »

Quote from: Jimmmy on April 20, 2019, 05:30:33 pm

Thanks MB!

I've stumbled across another question; Q) Solve $4^{3x+1} > 5$ .

I've reached $\log_4(5/4) / 3$ , but the solutions show the answer as $\log_2(5/4) / 6$

I can see how it could work, but am a little confused as to the working out steps to get there. I'm pretty sure I'm just missing something very obvious.

Two ways you might get the answer using the base-2 logarithm:

Method 1: Rewrite the initial inequality in the following form:

$(2^2)^{3x+1}>5$

Given what you've written about the answer you got, you should be able to proceed from there to the answer using the base-2 logarithm

Method 2: Convert your answer to the base-2 logarithm using the change of base formulae:

$\textrm{log}_4(\frac{5}{4}) = \frac{\textrm{log}_2(\frac{5}{4})}{\textrm{log}_2(4)} = \frac{\textrm{log}_2(\frac{5}{4})}{2}$

Then we have, and the result follows immediately:

$3x > \frac{1}{2}\textrm{log}_2(\frac{5}{4})$

Quote

--------------------------------------------------------------------
And another; Q)Solve $\log_5(x) = 16\log_x(5)$
I got to $\ln(x) = \pm4\ln(5)$ and had no idea where to do to next. I checked the solutions and they went from there to;
$x = e^{\pm\ln(625)}$ and again I'm really confused how they just got rid of the natural log on both sides.

$4\textrm{log}_e(5) = \textrm{log}_e(5^4) = \textrm{log}_e(625)$

Then make both sides the exponent of e. This is the inverse function of the natural logarithm, hence they "cancel out".

Quote

--------------------------------------------------------------------
And another! I'm really on a roll tonight; Q) Find inverse of function $f(x) = 5e^{2x} - 3$ . I find it pretty easily; $1/2(\ln((x+3)/5))$ , but I'm wondering how you would identify that x = -3 asymtote just from looking at the inverse, when the log includes the translation as being divided by 5 inclusive of the dilation in the y-axis? If it even possible we'd be asked such a question in Methods? I'm just trying to overrprepare, and wondering how I'd identify it without having the pre-known knowledge that the asymtotes simply switch with inverses.

If you transform the plane by reflecting all points around the line y = x, then horizontal lines become vertical lines and vice versa. Hence, if the graph of an invertible function has a horizontal asymptote y = k, the graph of its inverse has a vertical asymptote x = h (and vice versa).

Alternatively, the vertical asymptote for a graph of y = ln(ax + b) can always be found by solving ax + b = 0.

Alternatively, if you work out the sequence of transformations mapping ln(x) to ln((x+3)5), you'll get x' = 5x – 3. You are correct to be suspicious about why the vertical asymptote remains at x = –3, because in general if you apply a dilation before a translation, you can't just find the asymptote of the image by applying only the translation to the asymptote of the pre-image (consider, eg. applying x' = 5x – 3 to ln(x –1)). But for y = ln(x) the vertical asymptote is the y-axis, so the dilation has no effect on points on this line; the only transformation that moves the asymptote is the translation.

Jimmmy · « **Reply #17872 on:** April 21, 2019, 11:04:48 am »

Quote from: S_R_K on April 21, 2019, 10:31:41 am

Two ways you might get the answer using the base-2 logarithm:

Method 1: Rewrite the initial inequality in the following form:

$(2^2)^{3x+1}>5$

Given what you've written about the answer you got, you should be able to proceed from there to the answer using the base-2 logarithm

Method 2: Convert your answer to the base-2 logarithm using the change of base formulae:

$\frac{\textrm{log}_4(\frac{5}{4})}{\textrm{log}_2(4)}=\frac{1}{2}\textrm{log}_2(\frac{5}{4})$

Then we have, and the result follows immediately:

$3x > \frac{1}{2}\textrm{log}_2(\frac{5}{4})$

$4\textrm{log}_e(5) = \textrm{log}_e(5^4) = \textrm{log}_e(625)$

Then make both sides the exponent of e. This is the inverse function of the natural logarithm, hence they "cancel out".

If you transform the plane by reflecting all points around the line y = x, then horizontal lines become vertical lines and vice versa. Hence, if the graph of an invertible function has a horizontal asymptote y = k, the graph of its inverse has a vertical asymptote x = h (and vice versa).

Alternatively, the vertical asymptote for a graph of y = ln(ax + b) can always be found by solving ax + b = 0.

Alternatively, if you work out the sequence of transformations mapping ln(x) to ln((x+3)5), you'll get x' = 5x – 3. You are correct to be suspicious about why the vertical asymptote remains at x = –3, because in general if you apply a dilation before a translation, you can't just find the asymptote of the image by applying only the translation to the asymptote of the pre-image (consider, eg. applying x' = 5x – 3 to ln(x –1)). But for y = ln(x) the vertical asymptote is the y-axis, so the dilation has no effect on points on this line; the only transformation that moves the asymptote is the translation.

You're a star! I understand 3 perfectly, 2 pretty well and mostly the part of 1. Just with Q1, how do you get the 1/2 out front after the change of base formulae? I must be missing something obvious, can you literally just take a power out of the log base and divide it by the whole logarithm?

S_R_K · « **Reply #17873 on:** April 21, 2019, 11:34:11 am »

Quote from: Jimmmy on April 21, 2019, 11:04:48 am

You're a star! I understand 3 perfectly, 2 pretty well and mostly the part of 1. Just with Q1, how do you get the 1/2 out front after the change of base formulae? I must be missing something obvious, can you literally just take a power out of the log base and divide it by the whole logarithm?

Sorry! I actually made a mistake in the previous post – I'll edit it shortly to fix it up. That might help to answer your question. In any event, it's also good to see a derivation of the change of base formula, because that will help explain how I get the answer. You don't need to do this every time when using the change of base formulae, but it's worth seeing so that you understand how it works.

$\textrm{Let log}_4(\frac{5}{4})=y$

The idea here is that we're going to find an expression for y using the base-2 logarithm, but we do that using log laws. First we need to get rid of the base-4 logarithm so that we can then take the base-2 logarithm. We do that using the inverse function: making both sides the exponent of 4.

$\frac{5}{4}=4^y$

Now we can take the base-2 logarithm:

$\textrm{log}_2(\frac{5}{4})=\textrm{log}_2(4^y)= y\textrm{log}_2(4)$

In the second line, I'm using one of the log-laws. Now we just solve for y, and this gives:

$y= \frac{\textrm{log}_2(\frac{5}{4})}{\textrm{log}_2(4)}$

But the base-2 logarithm of 4 is 2, hence we get:

$y= \frac{1}{2}\textrm{log}_2(\frac{5}{4})$

And then substituting in for y gives us

$\textrm{log}_4(\frac{5}{4})= \frac{1}{2}\textrm{log}_2(\frac{5}{4})$

Jimmmy · « **Reply #17874 on:** April 21, 2019, 11:56:51 am »

Quote from: S_R_K on April 21, 2019, 11:34:11 am

Sorry! I actually made a mistake in the previous post – I'll edit it shortly to fix it up. That might help to answer your question. In any event, it's also good to see a derivation of the change of base formula, because that will help explain how I get the answer. You don't need to do this every time when using the change of base formulae, but it's worth seeing so that you understand how it works.

$\textrm{Let log}_4(\frac{5}{4})=y$

The idea here is that we're going to find an expression for y using the base-2 logarithm, but we do that using log laws. First we need to get rid of the base-4 logarithm so that we can then take the base-2 logarithm. We do that using the inverse function: making both sides the exponent of 4.

$\frac{5}{4}=4^y$

Now we can take the base-2 logarithm:

$\textrm{log}_2(\frac{5}{4})=\textrm{log}_2(4^y)= y\textrm{log}_2(4)$

In the second line, I'm using one of the log-laws. Now we just solve for y, and this gives:

$y= \frac{\textrm{log}_2(\frac{5}{4})}{\textrm{log}_2(4)}$

But the base-2 logarithm of 4 is 2, hence we get:

$y= \frac{1}{2}\textrm{log}_2(\frac{5}{4})$

And then substituting in for y gives us

$\textrm{log}_4(\frac{5}{4})= \frac{1}{2}\textrm{log}_2(\frac{5}{4})$

That makes perfect sense, and explains the change of base formulae so much better than I understood it in class. Massive thanks!

persistent_insomniac · « **Reply #17875 on:** April 21, 2019, 08:36:40 pm »

I don't know if I should post this here or on the spesh thread but can someone plz explain why for (d) they changed the modulus like that whereas for (e) they didn't?

aspiringantelope · « **Reply #17876 on:** April 21, 2019, 10:00:05 pm »

This sounded a little easy to me when I read it but I'm not just don't seem to know why?
For the circle $x^2+y^2-4x+6y=14$, find the equation of the diameter which passes through the origin.

I graphed it up but it the diameter doesn't go pass the origin?
And I thought it was
$\sqrt{x^2+y^2-4x+6y-14}=0$ but when I checked the answer, it was a lot simpler. Can anyone help? Thanks

DBA-144 · « **Reply #17877 on:** April 21, 2019, 10:27:48 pm »

Quote from: aspiringantelope on April 21, 2019, 10:00:05 pm

This sounded a little easy to me when I read it but I'm not just don't seem to know why?
For the circle $x^2+y^2-4x+6y=14$, find the equation of the diameter which passes through the origin.

I graphed it up but it the diameter doesn't go pass the origin?
And I thought it was
$\sqrt{x^2+y^2-4x+6y-14}=0$ but when I checked the answer, it was a lot simpler. Can anyone help? Thanks

CAS is out of charge rn so just doing it by hand roughly, is the equation y=-1.5x?

S_R_K · « **Reply #17878 on:** April 22, 2019, 09:00:07 am »

Quote from: aspiringantelope on April 21, 2019, 10:00:05 pm

This sounded a little easy to me when I read it but I'm not just don't seem to know why?
For the circle $x^2+y^2-4x+6y=14$, find the equation of the diameter which passes through the origin.

I graphed it up but it the diameter doesn't go pass the origin?
And I thought it was
$\sqrt{x^2+y^2-4x+6y-14}=0$ but when I checked the answer, it was a lot simpler. Can anyone help? Thanks

For any circle in the plane, there is a diameter that passes through the origin – this is simply the line that passes through the centre of the circle and the origin.

Write the circle in the form (x – h)^2 + (y – k)^2 = r^2, and the equation of the diameter through the origin follows quickly.

Quote from: DBA-144 on April 21, 2019, 10:27:48 pm

CAS is out of charge rn so just doing it by hand roughly, is the equation y=-1.5x?

Yes, this is correct.

aspiringantelope · « **Reply #17879 on:** April 22, 2019, 10:02:07 am »

Quote from: S_R_K on April 22, 2019, 09:00:07 am

For any circle in the plane, there is a diameter that passes through the origin – this is simply the line that passes through the centre of the circle and the origin.

Write the circle in the form (x – h)^2 + (y – k)^2 = r^2, and the equation of the diameter through the origin follows quickly.

Yes, this is correct.

How do you get from $\left(x-2\right)^2+\left(y+3\right)^2=27$ to the diameter?
root 27 times 2?

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