Make "x" the subject (1 Viewer)

[lacklustre]

I'm doing motion questions and I need to express "x" in terms of "t".

t = 0.5log_e(2x-1)

Some help?

 

[foram]

I'm doing motion questions and I need to express "x" in terms of "t".

t = 0.5log_e(2x-1)

Some help?

t=0.5 ln(2x -1)
2t = ln(2x -1)
e^2t = 2x -1
2x = e^2t +1
x = (e^2t + 1)/2

 

[lyounamu]

I'm doing motion questions and I need to express "x" in terms of "t".

t = 0.5log_e(2x-1)

Some help?

t = 0.5 ln (2x-1)

2t = ln (2x-1)

ln (2t) = 2x -1

2x = ln (2t) +1

x = ln(2t)+1/2

Mine is different from Foram.

 

[foram]

t = 0.5 ln (2x-1)

2t = ln (2x-1)

ln (2t) = 2x -1

2x = ln (2t) +1

x = ln(2t)+1/2

Mine is different from Foram.

how did you get the log from one side to the other?

 

[lacklustre]

t=0.5 ln(2x -1)
2t = ln(2x -1)
e^2t = 2x -1
2x = e^2t +1
x = (e^2t + 1)/2

Thank you man. My log laws are a little rusty.

Another problem I came across is the integration of 1/(cos²x) dx

Just having a little trouble (this topic requires good knowledge of the other topics *sigh*).

Cherios.

 

[tommykins]

Ooo Tension.

foram is correct

x = (e^2t+1)/2

 

[Mark576]

t = 0.5log_e(2x-1)
e^t = e^{0.5log_e(2x-1)}
e^t = e^{log_e(2x-1)0.5}
2x-1 = e^2t
x = (e^2t+1)/2

 

[tommykins]

Thank you man. My log laws are a little rusty.

Another problem I came across is the integration of 1/(cos²x) dx

Just having a little trouble (this topic requires good knowledge of the other topics *sigh*).

Cherios.

1/cos²x = sec²x

int sec²x dx = tan x + c

 

[lacklustre]

1/cos²x = sec²x

int sec²x dx = tan x.

*slaps head*

thanks smart 4U people.

 

[lyounamu]

Ooo Tension.

foram is correct

x = (e^2t+1)/2

LOL...

Stupid mistake. I didn't see that. Thanks.

 

[tacogym27101990]

回复: Make "x" the subject

wtf lyounamu wrong?!?!?!
something terrible has happened

 

[foram]

Re: 回复: Make "x" the subject

wtf lyounamu wrong?!?!?!
something terrible has happened

It means I've been studying better than him.

Try this, differentiate x^x

 

[tacogym27101990]

回复: Make "x" the subject

im goin to go with
though im not very confident
x^(2x-2)

my working:

y=x^x
y'= x.x^(x-1)
=x^2(x-1)
=x^(2x-2)
correct me if im wrong foram

 

[Mark576]

Re: 回复: Make "x" the subject

Spoiler

y = x^x
logy = xlogx
Differentiating implicitly:
(1/y)*dy/dx = logx + (1/x)*x
dy/dx*(1/y) = logx + 1
dy/dx = y(logx + 1) = x^x(logx + 1)

EDIT: Also -

y = x^x = e^xlogx
dy/dx = (logx + 1)e^xlogx = x^x(logx + 1)

 

[tacogym27101990]

回复: Make "x" the subject

aahh of course
that was stupid of me
i didnt think i could do it my way

 

[lacklustre]

Maybe I just haven't had enough sleep or something but I can't seem to make "y" the subject in the following (this is an inverse trig question):

x = 2y/(1+y²)

Help is appreciated, thanks!

 

[Mark576]

x = 2y/(1+y²)
x(1+y²) = 2y
x + xy² = 2y
xy² - 2y + x = 0
y = (2±√[4 - 4x²])/2x = (1±√[1 - x²])/x

 

[lacklustre]

x = 2y/(1+y²)
x(1+y²) = 2y
x + xy² = 2y
xy² - 2y + x = 0
y = (2±√[4 - 4x²])/2x

Oh so the quadratic formula was needed?

Well i'll be damned. Cheers mark

 

[lyounamu]

Maybe I just haven't had enough sleep or something but I can't seem to make "y" the subject in the following (this is an inverse trig question):

x = 2y/(1+y²)

Help is appreciated, thanks!

x + xy^2 = 2y

xy^2 - 2y +x =0

a=x, b= -2 and c=x

You use the quadratic formula

Therefore, y= (2 +_ root of (4-4x^2))/2x

EDIT: too late....

 

[lacklustre]

The answer in the book is:

g^-1(x) = 1-sqrt(1-x²)/x (i assume they just divided everything by 2)

I wonder why they only took the negative root and left out the other consideration?