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September 27, 2025, 06:40:09 pm

Author Topic: VCE Methods Question Thread!  (Read 5701205 times)  Share 

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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #3930 on: February 07, 2014, 07:34:08 pm »
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I don't know how you can have x=1 in the ln but anyway... I'll assume you meant minus

Ln(3x-2/x-1) = 0
3x-2 / x-1 = e^0
3x-2 / x-1 = 1
3x-2 = x-1
2x = 1
x = 1/2

Woops, typo! Thankyou!

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Re: VCE Methods Question Thread!
« Reply #3931 on: February 07, 2014, 07:42:25 pm »
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I don't know how you can have x=1 in the ln but anyway... I'll assume you meant minus

Ln(3x-2/x-1) = 0
3x-2 / x-1 = e^0
3x-2 / x-1 = 1
3x-2 = x-1
2x = 1
x = 1/2
Just hold up a min, if you substitute the back into the second log you'll get it to be undefined (you get the log of a negative number). The 'solution' is not within the intersection of the domains of the two logs. As a result there is no solution in the real number set to this equation, as evident from plotting the two curves.
https://www.desmos.com/calculator/uuoxq7iiro


That is of course assuming that the equals sign should be a negative.
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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #3932 on: February 07, 2014, 07:51:38 pm »
+1
Just hold up a min, if you substitute the back into the second log you'll get it to be undefined (you get the log of a negative number). The 'solution' is not within the intersection of the domains of the two logs. As a result there is no solution in the real number set to this equation, as evident from plotting the two curves.
https://www.desmos.com/calculator/uuoxq7iiro
(Image removed from quote.)

That is of course assuming that the equals sign should be a negative.

Is actually a plus haha. Failure on my part :)

Stick

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Re: VCE Methods Question Thread!
« Reply #3933 on: February 07, 2014, 08:14:31 pm »
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Thanks!

Now for
ln(3x-2) = ln(x=1)

Was I right in making it ln(3x-2/x=1)
What do I do now?

Either that way, or just equate what's in the brackets. :)
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rhinwarr

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Re: VCE Methods Question Thread!
« Reply #3934 on: February 07, 2014, 08:22:48 pm »
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I feel so dumb. I didn't even think of that D: I blame the equals sign haha

Rod

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Re: VCE Methods Question Thread!
« Reply #3935 on: February 07, 2014, 10:23:49 pm »
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How do I find the inverse of a 3x3 matrix?
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Re: VCE Methods Question Thread!
« Reply #3936 on: February 07, 2014, 10:29:22 pm »
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you do it on a cas.
you will never be asked to find the inverse of a 3x3 matrix by hand in vce. save that for linear algebra in uni :)

Rod

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Re: VCE Methods Question Thread!
« Reply #3937 on: February 07, 2014, 10:32:05 pm »
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you do it on a cas.
you will never be asked to find the inverse of a 3x3 matrix by hand in vce. save that for linear algebra in uni :)
Thank you ;)
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Re: VCE Methods Question Thread!
« Reply #3938 on: February 08, 2014, 12:24:52 pm »
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I've got a question from the Maths Quest 11 Mathematical Methods CAS textbook. I have attached my own working out as well as the relevant pages from the textbook.

Thanks in advance for your help

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Re: VCE Methods Question Thread!
« Reply #3939 on: February 08, 2014, 01:35:05 pm »
+1
Dilation away from the x-axis is the same as saying 'dilation in the y direction' or 'parallel to y-axis' since it's going away from the x-axis and going closer to y

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Re: VCE Methods Question Thread!
« Reply #3940 on: February 08, 2014, 01:35:59 pm »
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There's a way to find the inverse (of a 3x3 matrix) by hand but it's incredibly annoying and time-consuming.  :(
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Einstein

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Re: VCE Methods Question Thread!
« Reply #3941 on: February 08, 2014, 01:42:37 pm »
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already upto chapter 2? :O

We have our chapter 1 SAC this Thursday lol

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Re: VCE Methods Question Thread!
« Reply #3942 on: February 08, 2014, 02:06:43 pm »
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If g(x) = x^3+2 and f(x)=sqrt(x+3) then what is the largest possible range of g for f[g(x)] to be defined?

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Re: VCE Methods Question Thread!
« Reply #3943 on: February 08, 2014, 02:35:03 pm »
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How would I go about finding the implied domain for ?

EDIT: LOL awks the actual question was but I've worked it out using the examples from below!
« Last Edit: February 08, 2014, 04:20:05 pm by IndefatigableLover »

Zealous

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Re: VCE Methods Question Thread!
« Reply #3944 on: February 08, 2014, 02:48:43 pm »
+1
How would I go about finding the implied domain for ?
Let the denominator equal 0 then solve for x. The denominator of a fraction cannot be 0 as anything divided by 0 is undefined.





(You can solve this by using hybrid functions, or simply by using symmetry)

Therefore the domain is all real numbers excluding plus or minus 3/2.

\ {}
« Last Edit: February 08, 2014, 02:51:33 pm by Zealous »
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