Past Exam from my school 1 qu i don't get. (1 Viewer)

[currysauce]

If y = 1/2 (e^x -e^(-x) ) Show that x = log(e) (y+root(y²+1))

Thanks

 

[acmilan]

y = 1/2(e^x - e^-x)
2y = e^x - e^-x
2y = e^x - 1/e^x
2ye^x = e^2x - 1
e^2x - 2ye^x - 1 = 0

Let V = e^x

V² - 2yV - 1 = 0

V = (2y +- root(4y² + 4)) / 2
V = y +- root(y² + 1)

now
e^x = y + root(y² + 1) or e^x = y - root(y² + 1)
x ln e = ln (y + root(y² + 1) (ignore the second equation as it is always negative)
x = ln (y + root(y² + 1)

 

[currysauce]

V = y +- root(y2 + 1)

now
ex = y + root(y2 + 1) or ex = y - root(y2 + 1)
x ln e = ln (y + root(y2 + 1) (ignore the second equation as it is always negative)
x = ln (y + root(y2 + 1)


Can u explain each of these parts? how does V play a part in the ending?

 

[Trev]

V = y +- root(y2 + 1)

now
ex = y + root(y2 + 1) or ex = y - root(y2 + 1)
x ln e = ln (y + root(y2 + 1) (ignore the second equation as it is always negative)
x = ln (y + root(y2 + 1)


Can u explain each of these parts? how does V play a part in the ending?

First part; i'll fill in the steps inbetween:
V = (2y +- root(4y2 + 4)) / 2
V = (2y +- root[4(y² + 1)])/2
V = (2y +- 2root(y² + 1))/2
V = y +- root(y² + 1)

Sorry, I don't know any further - we have yet to do logs!

 

[currysauce]

thanks trev, anyone have an idea on the last part... i'm lost

 

[acmilan]

I just let V = e^x to make it more simpler to visualise. But other then that V means nothing and it is changed back to e^x once the quadratic is solved.

So i found what V equals and hence e^x equals the same thing as V was said to be equal to e^x. Since the solutions of V came from a quadratic equation there were 2 of them (which are shown by the +-). The - one was ignored as it gave a negative value for V (and hence e^x) and ln (natural logarithm) cannot be taken of a negative number.

So e^x = y + root(y² + 1)

then just ln both sides (ln = log(e) as you wrote it)

ln(e^x) = ln(y + root(y² + 1))
Using the basic log laws, the x moves down to the bottom so
xln(e) = ln(y + root(y² + 1))

now ln(e) = 1 so the final form is

x = ln(y + root(y² + 1))

 

[currysauce]

Thanks alot! I got it.