Author Topic: VCE Methods Question Thread! (Read 5943262 times) Tweet Share

SE_JM · « **Reply #7095 on:** December 06, 2014, 02:48:36 pm »

I'm using my scientific calculator. it's on degree mode.
Maybe that's the problem? is it meant to be on a different mode?

I don't know how I am going to use my CAS well if I can't even operate a simple scientific calculator *sigh*

Thanks!

lArcdeTriomphe · « **Reply #7096 on:** December 06, 2014, 02:57:16 pm »

degree mode shouldn't affect your answer

perhaps you haven't put enough brackets in/put brackets in the wrong places? maybe try breaking up your calculations rather than just typing the whole expression in (i.e. evaluate the numerator by itself, then divide by the denominator)

SE_JM · « **Reply #7097 on:** December 06, 2014, 09:33:24 pm »

Thanks! I think I figured it out now

SE_JM · « **Reply #7098 on:** December 07, 2014, 11:36:59 am »

Hello!

I have two questions I want to ask

1. Prove that log_ba + log_cb+ log_ac = 1/(log_ab) + 1/(log_bc)+1/(log_ca)
but log_ba + log_cb+ log_ac cannot equal log_ab + log_bc+ log_ca

2. if u=log₉ x, find in terms of u:
log_x 81

thanks!!

achre · « **Reply #7099 on:** December 07, 2014, 11:56:34 am »

Quote from: SE_JM on December 07, 2014, 11:36:59 am

1. Prove that log_ba + log_cb+ log_ac = 1/(log_ab) + 1/(log_bc)+1/(log_ca)
but log_ba + log_cb+ log_ac cannot equal log_ab + log_bc+ log_ca

2. if u=log₉ x, find in terms of u:
log_x 81

Not sure on the first one, but by change of base, $\log_{9}(x)=\frac{\log_{x}(x)}{\log_{x}(9)}=\frac{1}{\log_{x}(9)}$ .

So $u=\frac{1}{\log_{x}(9)} \implies \frac{1}{u}=\log_{x}(9) \implies \frac{2}{u}=2\log_x(9)=\log_x(81)$ , as required.

Zues · « **Reply #7100 on:** December 07, 2014, 12:32:06 pm »

how did they get -100/2, i know you have to multiply it by 3 and that gets -300/36 but hows this -100/2 ?

thanks

Conic · « **Reply #7101 on:** December 07, 2014, 12:46:29 pm »

Quote from: SE_JM on December 07, 2014, 11:36:59 am

1. Prove that log_ba + log_cb+ log_ac = 1/(log_ab) + 1/(log_bc)+1/(log_ca)
but log_ba + log_cb+ log_ac cannot equal log_ab + log_bc+ log_ca

For the first part, using the change of base formula, we obtain

$\log_b (a)=\frac{\log_a (a)}{\log_a (b)}=\frac{1}{\log_a (b)},\: \log_c (b)=\frac{\log_b (b)}{\log_b (c)}=\frac{1}{\log_b (c)},\:\mathrm{and}\:\log_a (c)=\frac{\log_c (c)}{\log_c (a)}=\frac{1}{\log_c (a)}.$

So using this,

$\log_b (a) +\log_c (b) + \log_a (c) = \frac{1}{\log_a (b)}+\frac{1}{\log_b (c)}+\frac{1}{\log_c (a)},$

as required. Now for the second part, if for example a=b=c=2, then they are equal, so I don't think the statement is actually true.

SE_JM · « **Reply #7102 on:** December 07, 2014, 01:00:52 pm »

Quote from: Conic on December 07, 2014, 12:46:29 pm

For the first part, using the change of base formula, we obtain

$\log_b (a)=\frac{\log_a (a)}{\log_a (b)}=\frac{1}{\log_a (b)},\: \log_c (b)=\frac{\log_b (b)}{\log_b (c)}=\frac{1}{\log_b (c)},\:\mathrm{and}\:\log_a (c)=\frac{\log_c (c)}{\log_c (a)}=\frac{1}{\log_c (a)}.$

So using this,

$\log_b (a) +\log_c (b) + \log_a (c) = \frac{1}{\log_a (b)}+\frac{1}{\log_b (c)}+\frac{1}{\log_c (a)},$

as required. Now for the second part, if for example a=b=c=2, then they are equal, so I don't think the statement is actually true.

Hey
Thanks! I get the first bit now

Yeah, that's what I thought too. I think the statement is wrong. But the CAS says the statement is correct. Oh well $:-\$

thanks for your help!

SE_JM · « **Reply #7103 on:** December 07, 2014, 01:03:56 pm »

Quote from: Conic on December 07, 2014, 12:46:29 pm

For the first part, using the change of base formula, we obtain

$\log_b (a)=\frac{\log_a (a)}{\log_a (b)}=\frac{1}{\log_a (b)},\: \log_c (b)=\frac{\log_b (b)}{\log_b (c)}=\frac{1}{\log_b (c)},\:\mathrm{and}\:\log_a (c)=\frac{\log_c (c)}{\log_c (a)}=\frac{1}{\log_c (a)}.$

So using this,

$\log_b (a) +\log_c (b) + \log_a (c) = \frac{1}{\log_a (b)}+\frac{1}{\log_b (c)}+\frac{1}{\log_c (a)},$

as required. Now for the second part, if for example a=b=c=2, then they are equal, so I don't think the statement is actually true.

Do you mind helping me on this too?

solve the equation:

log₅x=16* log_x5

Thank you so much in advance!!

SE_JM · « **Reply #7104 on:** December 07, 2014, 01:06:01 pm »

Quote from: Zues on December 07, 2014, 12:32:06 pm

how did they get -100/2, i know you have to multiply it by 3 and that gets -300/36 but hows this -100/2 ?

thanks

That's really strange! I would think that you are right. But maybe tell us the whole question? That might help clear it up a bit more!

Don't give up!

Zues · « **Reply #7105 on:** December 07, 2014, 01:13:55 pm »

they have asked to graph this, which involves completing the square. And their y value of the turning point is wrong, my answer is -25/3

y = 3x^ 2 + 8x − 3

SE_JM · « **Reply #7106 on:** December 07, 2014, 01:21:19 pm »

Quote from: Zues on December 07, 2014, 01:13:55 pm

they have asked to graph this, which involves completing the square. And their y value of the turning point is wrong, my answer is -25/3

y = 3x^ 2 + 8x − 3

Hello~!

I solved it myself, and I also got the y-value as -25/3

I think it's safe to assume that we're right. Perhaps it's an error made by the textbook/resource?

Conic · « **Reply #7107 on:** December 07, 2014, 01:24:22 pm »

Quote from: SE_JM on December 07, 2014, 01:03:56 pm

solve the equation:

log₅x=16* log_x5

You can also use the change of base formula here. We have

$\log_5 (x) = \frac{\log_x (x)}{ \log_x (5)} = \frac{1}{\log_x (5)},$

so using this in the equation gives

$\frac{1}{\log_x (5)} = 16\:\log_x(5) \implies 1 = 16\:(\log_x (5))^2$

$\implies (\log_x (5))^2 = \frac{1}{16}$

$\implies \log_x (5) = \pm \frac{1}{4}.$

Now if $\log_x (5) = \frac{1}{4}$ , then that means $x^{\frac{1}{4}}=5\implies x=5^4=625$ . If $\log_x (5) = -\frac{1}{4}$ then $x^{-\frac{1}{4}}=5 \implies x= 5^{-4} =\frac{1}{625}$ . So $x = 625$ or $x = \frac{1}{625}$ .

SE_JM · « **Reply #7108 on:** December 07, 2014, 01:30:01 pm »

Quote from: Conic on December 07, 2014, 01:24:22 pm

You can also use the change of base formula here. We have

$\log_5 (x) = \frac{\log_x (x)}{ \log_x (5)} = \frac{1}{\log_x (5)},$

so using this in the equation gives

$\frac{1}{\log_x (5)} = 16\:\log_x(5) \implies 1 = 16\:(\log_x (5))^2$

$\implies (\log_x (5))^2 = \frac{1}{16}$

$\implies \log_x (5) = \pm \frac{1}{4}.$

Now if $\log_x (5) = \frac{1}{4}$ , then that means $x^{\frac{1}{4}}=5\implies x=5^4=625$ . If $\log_x (5) = -\frac{1}{4}$ then $x^{-\frac{1}{4}}=5 \implies x= 5^{-4} =\frac{1}{625}$ . So $x = 626$ or $x = \frac{1}{625}$ .

THANK YOU!!
It helped so much

knightrider · « **Reply #7109 on:** December 07, 2014, 04:39:19 pm »

i dont understand this?

when i enter $(-1-x)^2$ on my cas it automatically gives me $(x+1)^2$ why is that and why does the cas do this?

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