onur369's Methods Question Thread :)

[Water]

Hi kefoo.


x = 1 - e^(-y)


x - 1 = -e^(-y)


1-x = e^(-y)

loge(1-x) = -y


-loge(1-x) = y

[onur369]

y=1-e^-x
y=1-1/e^x  <- Make all indices positive before any other steps.

Substiute x and y. Therefore, x=1-1/e^y
Move values around 1/e^y=1-x
Move again. e^y= 1/1-x
Loge both sides it will equal y= log_e(1/1-x)
f^-1(x)= log_e(1/1-x)

[dooodyo]

Oh Onur you missed the negative value haha lol same thing happened to me 

And yeah Water's solution is right.

[m@tty]

Onur's result is the same as Water's:

[onur369]

Hi, I have another dilemma.

Calculus question: Show that the derivative of y = k, where k is a constant, is zero.

What do we do and why do we do it?

[Water]

Calculus question: Show that the derivative of y = k, where k is a constant, is zero.



Given that k is a constant . such as 6 , 7, 8 ,9 ,10


The gradient = rise/ run

                  When its (0,6), the rise is still (1,6) which is 6


Therefore, there is 0 rise. Run is 1

Gradient: 0/1 = 0

Dy/dx = 0


Reasoning: There is no rise, however there is still run. when y = a constant

[onur369]

Champion

[kefoo]

hey guys got a question. i have an idea on what to do but im stuffing up somewhere..

1.Find the values of a and b such that the graph of y = ae^(bx) goes through (3,10) and (6,50)
2.Find the values of a and b such that the graph of y = alog2(x+b) goes through the points (8,10) and (32,14)

[Greatness]

Well you would sub those 2 coordinates into the equations which will give you two equations which you can solve simultaneously. You may be able to use the solve function on the CAS to make it quicker (This would only be if its tech active)
1) 10 = a*e^(3b)    and    50 = a*e^(6b)

2)  10 = alog2(8+b)    and 14 = alog2(32+b)

[kefoo]

Well you would sub those 2 coordinates into the equations which will give you two equations which you can solve simultaneously. You may be able to use the solve function on the CAS to make it quicker (This would only be if its tech active)
1) 10 = a*e^(3b)    and    50 = a*e^(6b)

2)  10 = alog2(8+b)    and 14 = alog2(32+b)

yeah i can do it by CAS easily but im wondering where i stuffed up when doing it by hand xD

[Greatness]

Wanna type out your working??
Ill have a go at doing them

[Mao]

Well you would sub those 2 coordinates into the equations which will give you two equations which you can solve simultaneously. You may be able to use the solve function on the CAS to make it quicker (This would only be if its tech active)
1) 10 = a*e^(3b)    and    50 = a*e^(6b)

2)  10 = alog2(8+b)    and 14 = alog2(32+b)

The first one can be solved algebraically,

10 = a e^(3b) ----- [1]
50 = a e^(6b) ----- [2]

[2]/[1]: 5 = e^(3b) --> b = Ln(5)/3, a = 2


And I have a sneaking suspicion q2 should be y = a log2(x) + b, where a=2 and b=4. Otherwise it must be solved on a calculator.

[kefoo]

Find all values of x between 0 and 2pi for which:
sin(x) = -0.45

dunno what to do because there arent any examples ._.

[pi]

Find all values of x between 0 and 2pi for which:
sin(x) = -0.45

dunno what to do because there arent any examples ._.

I think you'll have t use a calc to find x (and sin-1(9/20) has no surd value from my knowledge) and then use symmetry of the unit circle to find all the solutions.

[onur369]

Hey guys I have a question attached below, the answer is E but how do we get the result?