Question from Cambridge (1 Viewer)

[unmentionable]

bahz this is still the yr11 book >.<
its from 14e question 13a
for people who have the book
anyways the question says:
Show that y=ln(x-1) and y=(2-x)/x intersect at the point (2,0)

 

[perfectionist]

ln (x-1) = (2 - x)/x

ln (x - 1) = 2/x - x/x

ln (x - 1) = 2/x -1

ln (x - 1) - 2/x + 1 = 0

Sub in x=2 in LHS

LHS = ln (x-1) - 2/x + 1

= ln ( 2 - 1) - 2/2 + 1

=ln 1 - 1 + 1

ln 1 = 0

:. LHS = 0

:. LHS = RHS

Hope this helps.

 

[Rahul]

you would also have to show, i presume, that when x=2, y=ln[(2)-1], y=ln[1], y=0. .'. [2,0] is the point of intersection for the two.

 

[perfectionist]

yeah may be...

 

[unmentionable]

oooooh thanks
i never thought of doing it that way >.<!
thanks again ! =P

 

[perfectionist]

no worries

jeez...i did a Maths Ext 1 Question after three months

 

[unmentionable]

ahaha yes how can you remember everything

 

[perfectionist]

nah... not really...just gave it a go and got it out.

 

[Grey Council]

pssst, i've seen perfectionist practising 3u maths everyday... he is trying to keep his 1337 skillzor for USYD commerce. teeheehee

 

[perfectionist]

LOL i am not a HE I am A SHE... LOL

 

[KeypadSDM]

Originally posted by unmentionable
bahz this is still the yr11 book >.<
its from 14e question 13a
for people who have the book
anyways the question says:
Show that y=ln(x-1) and y=(2-x)/x intersect at the point (2,0)

Let x = 2
:. y = Ln[2-1]
= 0
But y = (2 - 2)/2
= 0
:. y = ln[x-1] And y = (2 - x)/x
intersect at (2,0)

(Remember, you have to show, not prove)