VCE Methods Question Thread!

[wunderkind52]

Have a few log questions -

1. Solve for x: x^-2/3 = 1/9
and
2. Solve (0.1)^x > 3 for x, in terms of a base 10 logarithm

Thanks!

[ETTH96]

hey guys how do i find the x intercept of circular function graphs, without using a calculator? I'm asking this because my teacher gave me a practice test that is TECH FREE and asks to draw a graph labeling the x intercepts as well.

Sketch the graph of y=2sin(2(x=(pi/6))+1, x E [0,2pi], showing all x intercepts.

i'm asking especially when there is a translation about the x or y intercept

thanks guys

[Lizzy7]

Have a few log questions -

1. Solve for x: x^-2/3 = 1/9
and
2. Solve (0.1)^x > 3 for x, in terms of a base 10 logarithm

Thanks!

Hello!

1) When I write log ___  for this question it is supposed to be log base 3

I don't know if this is the most efficient way but here's how I would attempt it:

log (x^-2/3) = log (3) ^-2
log (x^-2/3) = -2 log (3)    [Since I used log base 3 here, the log base 3 (3^-2) would simplify down to -2*1]
log (x^-2/3) = -2 
3^-2 = x^-2/3
1/9 = 1/x^2/3     [I then equated the denominators]

9= x^2/3
9= 3√x^2    [Cubed root x^2]
729 = x^2  [ Undo the cubed root by cubing 9 e.g. 9*9*9=729]
x=-27 or x=27 [ Undo x^2 , by square rooting]







2)  I don't know how to use 'latex' , if that's what you call it so please assume that when I type log, it's supposed to be log base 10

log (0.1)^x > log (3)

x log (0.1) > log (3)

 x * -1  >   log (3)   (since log (0.1) = -1)

-x  >   log (3)

x< - log (3)



Please let me know if these are the answers you were looking for

[achre]

hey guys how do i find the x intercept of circular function graphs, without using a calculator? I'm asking this because my teacher gave me a practice test that is TECH FREE and asks to draw a graph labeling the x intercepts as well.

Sketch the graph of y=2sin(2(x=(pi/6))+1, x E [0,2pi], showing all x intercepts.

i'm asking especially when there is a translation about the x or y intercept

thanks guys

Spoiler

And then next is the fun bit where you trial and error values of n until you find all the solutions that are inside the domain!

[Anchy]

If y=Ae^kt, and y=19.6 when t=2, and y=19.02, when t=5, find the value of the constants A and k. Give your answers correct to 2 decimal places

Thanks

[RKTR]

19.6=Ae^2k   
19.02=Ae^5k     

use equation 2 to divide by equation 1

e^3k =19.02/19.6
3k=loge(19.02/19.6)
k=1/3 loge(19.02/19.6)
  =-0.01

19.6=Ae^2k
19.6=Ae^2(1/3 loge(19.02/19.6)
19.6=Ae^[loge(19.02/19.6)^2/3]
19.6=A(19.02/19.6)^2/3
A=19.6(19.6/19.02)^2/3
  =20.00

[Jason12]

Find all values  between 0 and 360 for which: sin(theta) = 0.3, cos(theta) = 0.4

i've tried putting it in my cas as

solve(sin

• =0.3,x) but all i get is x=360 and n2 and n1

[Einstein]

all the other solutions have a 0x^2 in between the cubic value and linear value, this is the only solution without it. Is the solution wrong or am i wrong? i have attached the screesshots.

My answer was ,,,,,,,   5x^2-15x+58 r -171

[Lizzy7]

all the other solutions have a 0x^2 in between the cubic value and linear value, this is the only solution without it. Is the solution wrong or am i wrong? i have attached the screesshots.

My answer was ,,,,,,,   5x^2-15x+58 r -171

From the screenshots I saw, there was no ^3 value but just a ^2. I'm not too sure about the bit you're having trouble with... When I solved your question I got: 5x-2 + 7/(x+3)

You shouldn't get  5x^2-15x+58 r -171 , since the highest degree of polynomial in that question is ^2. You only add a 0x^(whatever)  if there are missing lower degrees like if you had :   x^5 + x^2 +x +1   [You would add in 0x^4, 0x^3 ]

Anyways I've attached how I attempted it..     I hope my explanation didn't confuse you

[Anchy]

Hi all,

Got a few questions which i'm having trouble with:

1. If log(base3)D = cy+log (base3)2, solve for 2 (sorry don't know how to use latex)

2. Solve 2^x-1 = 3^x+a for x, where a is a element of R

[RKTR]

Hi all,

Got a few questions which i'm having trouble with:

1. If log(base3)D = cy+log (base3)2, solve for 2 (sorry don't know how to use latex)

2. Solve 2^x-1 = 3^x+a for x, where a is a element of R

1. log(base3)D-log(base3)2=cy
log(base3)[D/2]=cy
D/2= 3^cy
2=D/(3^cy)

[Jason12]

help with these questions please

[souka]

From exact values Jason
x= - pi/6, - 5pi/6
And
x= 5pi/6

[alchemy]

f(x)=(ax+b)/(cx+d) where a,b,c,d are real numbers.
Find the values of a,b,c,d such that f^-1(x)=f(x)

[Phenomenol]

f(x)=ax+b/(cx+d) where a,b,c,d are real numbers.
Find the values of a,b,c,d such that f^-1(x)=f(x)

Interchange x and f(x) in the equation, and let's replace f(x) with f^-1(x). Then solve for f^-1(x). Hint: use the quadratic formula.

From there let f^-1(x)=f(x).

Disclaimer: This may not work.