Logarithm question (1 Viewer)

[snoopy05]

g'day,

im stuck with this question, could someone help?

Show that the lines y=3log(2x - 1 ) and 3x + 2y = 3 , meet at (1,0)

, the answers on the back says that we shuld use subtitution to solve it...
:S but im really stuck with the logarithm.

 

[FinalFantasy]

Show that the lines y=3log(2x - 1 ) and 3x + 2y = 3 , meet at (1,0)

better solution,
y=3log(2x - 1)
when x=1, y=0
y=(3-3x)\2
when x=1, y=0
therefore the lines meet at (1,0)

 

[snoopy05]

yea, that works with lhs=rhs way,
but the point (1,0) is there, its given,

i dont think thats the actual way to find it, my friend also told me that. but its just.. dunno..
i'll ask my teacher tomorow and post back here tomorow. but yea...

 

[FinalFantasy]

but isn't it, if it says "show" u can just assume things.

 

[Slidey]

g'day,

im stuck with this question, could someone help?

Show that the lines y=3log(2x - 1 ) and 3x + 2y = 3 , meet at (1,0)

, the answers on the back says that we shuld use subtitution to solve it...
:S but im really stuck with the logarithm.

This is a show question. Simply 'show' that in both cases, when you solve for y and substitute in the x and y values of the point, that both equations satisfy the point.

You cannot solve the equation because it is transcendental. Use the Newton-Raphson method or solve it graphically, if you really insist on doing something other than showing by substitution.

 

[snoopy05]

well, i guess you guys are right.. ,
its a show question after all... ^^
thnx dude, cus this kinda freaked me out last night when i couldn't solve it simultaneously ^^;;

: slide rule
your sig says "where there's a will, there's a way

heheh

 

[Slidey]

Transcendental problems 'transcend' known mathematical techniques for solving exactly. However, the Newton-Raphson method will give you the approximated roots to any decimal place you desire, so yes, there's a way.