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C3 help needed

F(x) = ke^2
F^-1(x) = ln(x/k)

F(x) crosses the y-axis at k
F^-1(x) crosses the x-axis at ke^y ( can some1 check this please)

I need to find the distance between these 2 points.

So I did (ke^y)^2 + k^2

= k^2e^2y +k^2

How would I simplify this further?

And can some1 check my overall method is right?
(edited 10 years ago)
f^(-1) = ln(x/k) crosses the x-axis when y=0, and hence when x=k.

Also, when using Pythagoras, don't forget to root your answer.
Reply 2
Original post by PythianLegume
f^(-1) = ln(x/k) crosses the x-axis when y=0, and hence when x=k.

Also, when using Pythagoras, don't forget to root your answer.


How do I simplify k^2e^2y + k^2?
Original post by Vorsah
How do I simplify k^2e^2y + k^2?


You don't need to, because that isn't the distance. But if you did for some reason want to, you can factorise out k^2.
Reply 4
Original post by PythianLegume
You don't need to, because that isn't the distance. But if you did for some reason want to, you can factorise out k^2.


Oh, but I'm not sure what I should do next?
Original post by Vorsah
Oh, but I'm not sure what I should do next?


OK, I told you that you were wrong about where the second line (the inverse) crossed the x-axis.

0 = ln(x/k)
e^0 = x/k
1 = x/k
k=x

Hence you're looking for the distance between (0,k) and (k,0).

You can use Pythagoras' theorem to calculate this distance.
Reply 6
Original post by PythianLegume
OK, I told you that you were wrong about where the second line (the inverse) crossed the x-axis.

0 = ln(x/k)
e^0 = x/k
1 = x/k
k=x

Hence you're looking for the distance between (0,k) and (k,0).

You can use Pythagoras' theorem to calculate this distance.


Thanks man
Reply 7
Can someone help with this:

Find the intersection points of y=e^x+1 and y=6e^-x

I thought i had done it right but didn't get the same as the book
Original post by simonb451
Can someone help with this:

Find the intersection points of y=e^x+1 and y=6e^-x

I thought i had done it right but didn't get the same as the book


You're best starting new threads when you want maths help.

At the intersection, they are equal, so:
ex+1=6e(x) e^x + 1 = 6e^(-x)
e(2x)+ex=6 e^(2x) + e^x = 6
e(2x)+ex6=0 e^(2x) + e^x - 6 = 0
factorising the quadratic
(ex+3)(ex2) (e^x + 3)(e^x -2)
ex=2 e^x = 2
x=ln(2),y=3 x = ln(2) , y = 3
Reply 9
Original post by PythianLegume
You're best starting new threads when you want maths help.

At the intersection, they are equal, so:
ex+1=6e(x) e^x + 1 = 6e^(-x)
e(2x)+ex=6 e^(2x) + e^x = 6
e(2x)+ex6=0 e^(2x) + e^x - 6 = 0
factorising the quadratic
(ex+3)(ex2) (e^x + 3)(e^x -2)
ex=2 e^x = 2
x=ln(2),y=3 x = ln(2) , y = 3


Thanks i substituted e^x=y so thats where i went wrong as i got a y value of 2 instead of e^x

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