Finding log([x+y]/√5) given x/y + y/x = 3

[lionely]

Homework Statement

I am having problem with number 16ii)

Homework Equations

log([x+y]/sqrt5)

The Attempt at a Solution

Sorry still not used to latex
So I tried this
if x/y + y/x =3 then(

(x^2 + y^2)/xy = 3xy
x^2 + y^2 = 3xy
(x-y)^2 = xy
(x-y) = sqrt(xy)
(x-y)(x+y) = sqrt(xy)(x+y)
(x+y) = (x^2-y^2)/sqrt(xy)

But subbing in the new expression for (x+y) did not help me at all I end up getting the same thing over and over.

 

[mfb]

x^2 + y^2 = 3xy

Good so far.

(x-y)^2 = xy

This leads to (i), for (ii) you can use the opposite direction. Getting (x-y)^2 again does not help I think.

 

[lionely]

By using the opposite direction you mean using the right side?

 

[mfb]

Instead of (x-y)^2, go for (x+y)^2.

 

[SammyS]

Homework Statement

I am having problem with number 16ii)

Homework Equations

log([x+y]/sqrt5)

The Attempt at a Solution

Sorry still not used to latex
So I tried this
if x/y + y/x =3 then(

What allows you to go from the above line to that below?

(x^2 + y^2)/xy = 3xy
x^2 + y^2 = 3xy
(x-y)^2 = xy
(x-y) = sqrt(xy)
(x-y)(x+y) = sqrt(xy)(x+y)
(x+y) = (x^2-y^2)/sqrt(xy)

But subbing in the new expression for (x+y) did not help me at all I end up getting the same thing over and over.

 

[mfb]

The uppermost line you quoted in the lower quote has a typo, the lines above and below are correct.

 

[SteamKing]

Homework Statement

I am having problem with number 16ii)

Homework Equations

log([x+y]/sqrt5)

The Attempt at a Solution

Sorry still not used to latex
So I tried this
if x/y + y/x =3 then

You start out OK.

(x^2 + y^2)/xy = 3xy

Whoops! You've jumped ahead of yourself here. Check your algebra. Remember, just because you add the terms on one side of an equation together, this does not imply that the other side is changed.

x^2 + y^2 = 3xy

Now, you've corrected yourself.

(x-y)^2 = xy

Whoops! Back in the weeds, again.

What happened to the factor of 3 on the RHS?
Is (x - y)² = x² + y² ?

(x-y) = sqrt(xy)
(x-y)(x+y) = sqrt(xy)(x+y)
(x+y) = (x^2-y^2)/sqrt(xy)

But subbing in the new expression for (x+y) did not help me at all I end up getting the same thing over and over.

You've got to carefully check your algebra.

 

[lionely]

Okay I see that mistake and I get (x-y) = sqrt(xy) the correct way now. So how I can I get (x+y) from the x/y + y/x = 3 ?

 

[SammyS]

So how I can I get (x+y) from the x/y + y/x = 3 ?

Yes. How did you get that? In other words, what is it that you intended to do in taking that step ?

 

[lionely]

I was hoping to get a new expression to sub in the log([x+y]/sqrt5) but the expression I get I think I showed it up the top and it doesn't help me

 

[SammyS]

I was hoping to get a new expression to sub in the log([x+y]/sqrt5) but the expression I get I think I showed it up the top and it doesn't help me

simple question:

What is the algebraic step you took there ?

 

[lionely]

x/y + y/x =3 then

(x^2 + y^2)/xy = 3
x^2 + y^2 = 3xy
x^2 - 2xy + y^2 = xy
(x-y)^2 = xy
(x-y) = sqrt(xy)
(x-y)(x+y) = sqrt(xy)(x+y) [ multiplying both sides by (x+y) ]
(x+y) = (x^2-y^2)/sqrt(xy)

 

[SammyS]

x/y + y/x =3 then

(x^2 + y^2)/xy = 3 OK: You used the common denominator to add fractions.
x^2 + y^2 = 3xy
x^2 - 2xy + y^2 = xy

At this point, I see that you subtracted 2xy . Why not add 2xy instead?

(x-y)^2 = xy
(x-y) = sqrt(xy)
(x-y)(x+y) = sqrt(xy)(x+y) [ multiplying both sides by (x+y) ]
(x+y) = (x^2-y^2)/sqrt(xy)

 

[lionely]

Oh because I'm an idiot sigh haha hmm wow . Okay thank you! Not sure why I couldn't see that...

 

[mfb]

x^2 + y^2 = 3xy

Now, you've corrected yourself.

(x-y)^2 = xy

Whoops! Back in the weeds, again.

What happened to the factor of 3 on the RHS?
Is (x - y)² = x² + y² ?

Two steps in one, but this is correct. Your two questions answer each other.