VCE Methods Question Thread!

[Orb]

how many hours u guys spend on meth a week/night?

I tend to find that 45mins-1hour per night is sufficient.

Less time over consistent period > 10 hour cram sesh night before the SAC

[wunderkind52]

I have some questions relating to logs -
1. If 10^a=2/3, 10^b=8/9, find log10(80/81) in terms of a and b. i've put a and b into log form but am having trouble finding a way to get 80/81!
2. Simplify into power form - log5(log5(log5(x)))

Many thanks in advance!

[Orb]

I have some questions relating to logs -
1. If 10^a=2/3, 10^b=8/9, find log10(80/81) in terms of a and b. i've put a and b into log form but am having trouble finding a way to get 80/81!
2. Simplify into power form - log5(log5(log5(x)))

Many thanks in advance!

For 1:
In terms of a
log10 (80/81) = log 10 (2/3)^4 + log 10 (5)
log10 (80/81) = 4log10(2/3) + log10 (5)
= 4a +log10 (5)

Now you want to find log10 (5) in terms of b

log10 (80/81) = log 10 (8/9)^2 + log10(10/9)
log10 (80/81) = 2log10(8/9) + log10(10/9)
= 2b + log10 (10/9)

[maurlock]

Can anyone help with any parts of this question? I am so confused!
Thanks

[Orb]

Can anyone help with any parts of this question? I am so confused!
Thanks

EDIT: solved by bottom post

[lzxnl]

Actually...I beg to differ. You're given that the origin is a stationary point (well, for the purpose of Methods it is one). Then, both A and B have zero gradient, so the derivative of your quartic is some multiple of x(x+4)(x-4)=x^3-16x. Now, as your quartic passes through the origin, it must be some constant times (1/4 x^4 - 16/3 x^2). You can work out what this constant is.
As for the second part, just find a cubic that is stationary at the origin and at x=4.

[Yacoubb]

The points (3,10) and (5,12) lie on the graph of the function with the rule f(x) = aloge(x-b) + c. The graph has a vertical asymptote at x = 1. Find the values of a, b and c.

Okay so b = 1.

Could someone show me how to deal with the simultaneous equations to find a and then c.

Thanks.

[Orb]

The points (3,10) and (5,12) lie on the graph of the function with the rule f(x) = aloge(x-b) + c. The graph has a vertical asymptote at x = 1. Find the values of a, b and c.

Okay so b = 1.

Could someone show me how to deal with the simultaneous equations to find a and then c.

Thanks.

vertical asymptote means b = 1 (as you said)

so you get the two simultaneous equations:

10=aloge(2) + c ....1
12=aloge(4) + c = 2aloge(2) + c ....2

2-1
2=aloge(2) (eliminated c) .... 3
just solve then sub it back to find c!

1-3
c=8

[maurlock]

Actually...I beg to differ. You're given that the origin is a stationary point (well, for the purpose of Methods it is one). Then, both A and B have zero gradient, so the derivative of your quartic is some multiple of x(x+4)(x-4)=x^3-16x. Now, as your quartic passes through the origin, it must be some constant times (1/4 x^4 - 16/3 x^2). You can work out what this constant is.
As for the second part, just find a cubic that is stationary at the origin and at x=4.

Thank you!

[Yacoubb]

How do I do this?

Find x:

e^ (0.5x -2) = loge(x+3)² + 7

[ETTH96]

Hey guys, I need help with this"

find all solutions in exact form, over the domain [0,2pi]

1. sinx=-cosx

2. sin(pi/3)=cos(pi/3)

3. sqrt3 sin(2x) = cos(2x)

thanks

[RKTR]

Hey guys, I need help with this"

find all solutions in exact form, over the domain [0,2pi]

1. sinx=-cosx

2. sin(pi/3)=cos(pi/3)

3. sqrt3 sin(2x) = cos(2x)

thanks

Hint: sin/cos =tan

[ETTH96]

Hint: sin/cos =tan

It says that in my text book  but I still don't get it..
for the first one can i divide both sides by cox or not? or do i move -cosx to the other side?!? how do i know what to do

[RKTR]

Divide both sides by cos

[IndefatigableLover]

Can someone give me a run down on 'Change of Base Law' for logs? Like the formula we use for it kind of confuses me a bit :/

So for questions like:
How would I go about using 'Change of Base' to find an answer?

(Finally Latex is working again :DD)