question (1 Viewer)

[auzbabe4christ]

hey guys, im pretty sure u will b able to help with these few quesions... hope ya's can!!

ok.. here we go..

simplify: x^(p-1)/{[x^(p/q)]^(q-1)}

and..

solve: x=[e^y-e^(-y)]/2 for y in terms of x

hope ya can help. thanx

 

[Trev]

x=[e^y-e^-y]/2
2xe^y=e^2y-1
e^2y-2xe^y-1=0
Let e^y=u.
u²-2xu-1=0
(u-x)²-x²-1=0
u=x+/-√(x²+1)
e^y=x+/-√(x²+1)
∴y=ln|x+/-√(x²+1)|

 

[auzbabe4christ]

x=[e<SUP>y</SUP>-e<SUP>-y</SUP>]/2
2xe<SUP>y</SUP>=e<SUP>2y</SUP>-1
e<SUP>2y</SUP>-2xe<SUP>y</SUP>-1=0
Let e<SUP>y</SUP>=u.
u<SUP>2</SUP>-2xu-1=0
(u-x)<SUP>2</SUP>-x<SUP>2</SUP>-1=0
u=x+/-√(x<SUP>2</SUP>+1)
e<SUP>y</SUP>=x+/-√(x<SUP>2</SUP>+1)
∴y=ln|x+/-√(x<SUP>2</SUP>+1)|

ok i dont get wot u did to get to ur 2nd line... can u explain or expand it a bit please?

 

[Sober]

x=[e<SUP>y</SUP>-e<SUP>-y</SUP>]/2

Multiply by 2:

2x=e<SUP>y</SUP>- e<SUP>-y</SUP>

Mulitply by e<SUP>y</SUP>

2xe<SUP>y</SUP> = e<SUP>y</SUP>.e<SUP>y</SUP> - e<SUP>-y</SUP>.e<SUP>y</SUP> = e<SUP>2y</SUP> - e<SUP>0y</SUP> = e<SUP>2y</SUP> - 1

 

[Trev]

x^(p-1)/{[x^(p/q)]^(q-1)}
This one is a bit of a bitch, it's easier if you break it up into sections
Taking [x^(p/q)]^(q-1)
Same as [x^p/x^q]^(q-1)
{[x^p/x^q]^q}/[x^p/x^q]
[x^pq/x^q²]/[x^p/x^q]
[x^pq/x^q²]*[x^q/x^p]
Introducing this into the original question:
[x^p/x]/[x^pq/x^q²]*[x^q/x^p]
[x^p/x]*[x^q²/x^pq]*[x^p/x^q]
Blah, you can multiply and simplify the rest just remember your power rules etc.

 

[auzbabe4christ]

awesome thanx heaps guys!! u rock!

 

[acullen]

Just out of curiosity, why are you asking a question on 3 unit maths when you've done your HSC? Brushing up for uni or something?

 

[auzbabe4christ]

Just out of curiosity, why are you asking a question on 3 unit maths when you've done your HSC? Brushing up for uni or something?

yea we got a revision sheet for uni an stuff but my mind had a totally blank for a few questions.. so yea... its kinda sad. lol

 

[auzbabe4christ]

u=x+/-√(x<SUP>2</SUP>+1)

hey ive got a new question... what do u mean by the bit in bold?? cuz im reading it as "x + dvided by" an i think im wrong there.. so help?

 

[Trev]

y=ln|x+/-√(x²+1)| is quicker/easier than writing it as two separate answers:
y=ln|x+√(x²+1)| and y=ln|x-√(x²+1)|

 

[auzbabe4christ]

hey ive got 2 more new questions... i know what im supposed to be doing but the general rules for them have just slipped my mind... hope ya can help...

#1: simplify: 1/2[logaN-loga(N-1)]

#2: solve √2 sinx + sin 2x =0 for all values of x

as i said i know wot i am supposed to do, but the rules have just gone out of my head... so yea... hope u guys can help!!

 

[auzbabe4christ]

y=ln|x+/-√(x²+1)| is quicker/easier than writing it as two separate answers:
y=ln|x+√(x²+1)| and y=ln|x-√(x²+1)|

So with one answer its x + the square root, and with the other its timesed together??

 

[Trev]

So with one answer its x + the square root, and with the other its timesed together??

One is x plus the square root, the other is x minus the square root.
(u-x)²-x²-1=0
(u-x)²=x²+1
u-x=+/-√(x²+1)
u=x+/-√(x²+1)
See?

 

[Trev]

If you forget the rules just look them up on the net.
1/{2[log_a(N)-log_a(N-1)]}
1/{2[log_a(N/(N-1))]}
[log_a((N-1)/N)]/2

 

[auzbabe4christ]

One is x plus the square root, the other is x minus the square root.
(u-x)²-x²-1=0
(u-x)²=x²+1
u-x=+/-√(x²+1)
u=x+/-√(x²+1)
See?

sorry.. i couldnt see the minus sign...thats where i got confsed... i was in the process of writing it out..

 

[Trev]

sin2x-√2sinx=0 for all values of x
2sinxcosx-√2sinx=0
sinx(2cosx-√2)=0
So you must solve sinx=0 and cosx=(√2)/2 for all values of x.

For sinx=0:
x=...-3pi,-2pi,-pi,0,pi,2pi,3pi...
x=pi.n where n is an integer.

For cosx=(√2)/2
x=+/-pi/4,+/-9pi/4... ie. pi/4, -pi/4, pi/4+2pi, -pi/4-2pi...
x=+/-(pi/4+2pi.n) where n is a positive integer greater or equal than zero.

 

[auzbabe4christ]

hey i have a question which just isnt falling out for me..

its:

(x/x-l) + (2/2x+1)=2... now for this im supposed to make the bottoms the same arent i?? or am i just going about this the wrong way?? cuz i tried it an got to a stage where i had

[2(2x-1)(x+2)]/[(x-1)(2x+1)]=x^2-x-1.

i think i did something wrong but im not sure...

 

[auzbabe4christ]

hey u r guna hate me... but that sine question... i didnt mean to put the "-ve" sign in the question at the start... im sorry.. is it much different from what uve done anywayz?

 

[Trev]

You must of made a typo.
(x/(x-l)) + (2/(2x+1))=2
What is the x-l?

 

[auzbabe4christ]

ok i figured out the sine question, its alright...

oops.. its (x-1) i accidently put an "L" my bad