2 Questions (1 Viewer)

[wrxsti]

1) solve the inequality

1/(|x-3|) > 1/2

the | is absolute value sign............... i know how to do it without the absolute value sign... but i keep getting it wrong with the absolute value sign..


2) given a^x = b^y = (ab)^z prove 1/x + 1/y = 1/z




any help would be appreciated... thanks,

 

[ssglain]

Q1.
I think since |x - 3| is positive, it's safe to say |x - 3| < 2
Then x - 3 < 2 or - (x - 3) < 2
.: 1 < x < 5, x =/= 3.

Can't think of a way to solve Q2 at the moment.

 

[soisorce26]

I think since |x - 3| is positive, it's safe to say |x - 3| < 2

hey i think the question wants 0.5, not 2 lol

|x-3|>1/2
therefore x-3<-1/2 or x-3>1/2
hence x< 5/2 or x>7/2
not 100% sure im allowed to do that though :S lols

 

[ssglain]

hey i think the question wants 0.5, not 2 lol

Q1. 1/(|x-3|) > 1/2

 

[soisorce26]

lmao oops my baddddd hehe:rofl: your right ssglain, youre right

 

[loathi]

Hmm, i cant figure out the second one, first one is preliminary course stuff though.

 

[kony]

solution to 2:

<Img>http://www.geocities.com/sunnysnew/solutions.bmp</Img>

 

[ssglain]

solution to 2:

<IMG>http://www.geocities.com/sunnysnew/solutions.bmp</IMG>

Great stuff!

 

[soisorce26]

Q2. a^x=b^y=(ab)^z

find z in terms of y:
log(b)^y=log(ab)^z
ylog(b)=zlog(ab)
z=ylog(b) / log(ab)
therefore, z=ylog(b) / (loga+logb)

now, find x in terms of y:
log(a)^x=log(b)^y
xlog(a)=ylog(b)
x=ylog(b) / log(a)

take reciprocal,
therefore 1/x = log(a)/ ylog(b)
now add 1/y to both sides,

1/x + 1/y = log(a)/ylog(b) + 1/y
=[ log(a) +log(b) ] / ylog(b)
= 1/z as req'd

correct me if im wrong lol

 

[ngogiathuan]

yah for question 1, put |x-3|<2 as ssglain said
but then why don't you try sketching the graph of y=|x-3| and y=1/2, and see where the inequality holds. It won't take you long, and moreover, it's safe
you can take a shortcut to solve this as people said, but I think sketching is the safest option here.