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**ganesh****Moderator**- Registered: 2005-06-28
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IS # 1

If a^x = b^y = c^z and b²=ac, show that 1/x + 1/z = 2/y.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Yo ho ho

I LOVE this

thanks

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**ganesh****Moderator**- Registered: 2005-06-28
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Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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OK. I'm ready with IS1.

Now making Tex input.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Soe here is it:

a^x=b^y=c^z=(ac)^(y/2);

a^x=(ac)^(y/2)

x=log(a,(ac)^(y/2))=y(1+loga,c)/2

z=y(1+logc,a)/2

M=1/z+1/x=2/y(1+log a,c)+2/y(1+log c,a)=2(log a,c +log c,a +2)/y(1+log a,c)(1+log c,a)=

=(2/y)(log a,c +log c,a +2)/(1+log a,c+log c,a+(log a,c)(log c,a))=

=(2/y)(log a,c +log c,a +2)/(1+log a,c+log c,a+(log a,c)(log c,a));

a^x=c =>c^(1/x)=a =>

(log a,c)(log c,a)=1

=>

M=2/y.

*Last edited by krassi_holmz (2006-02-26 18:50:04)*

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**ganesh****Moderator**- Registered: 2005-06-28
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That solution is too complicated. I am unable to check whether its correct right now. There's a much simpler solution, krassi_holmz. I shall post that a little later.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Here's the simpler solution:

1/x+1/z=y/2

(x+z)/xz=1/2y

xz=2y(x+z)

b^(2y(x+z))=(b^2yx)(b^2yz)=(c^2zx)(a^2zx)=ca^2zx=b^zx

2y(x+z)=zx

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Is it clear now?

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**ganesh****Moderator**- Registered: 2005-06-28
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In your post, I was not able to go far.

This is the solution I had in my mind:-

Let a^x = b^y = c^z = k,

Therefore,

a = k^(1/x), b=k^(1/y) and c=k^(1/z).

Since b²=ac,

{k^(1/y)}2 = [k^(1/x)*k^(1/z)]

k^(2/y) = k^[(x+z)/(xz)]

Since the bases of the LHS and RHS are the same, the exponents too are equal.

Hence,

2/y = (x+z)/xz

Therefore,

2/y = (x+z)/xz = 1/x + 1/z

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Yes, my second proof is just like yours, but instead plugging (1/x) and (1/z) in the power equation, I first solve it to get integer powers:

Now i'm poving that b^xz=b^2y(x+z):

*Last edited by krassi_holmz (2006-02-26 22:09:00)*

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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now, won't give second q?

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**ganesh****Moderator**- Registered: 2005-06-28
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IS # 2

Find the square root of 7 + 3√5.

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**rimi****Member**- Registered: 2006-02-21
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7+3√5 = (14+6√5) / 2 = ( 3 + √5 )² / 2

there fore , √(7+3√5) = ±(3 +√5)/√2 = (3√2 / 2) + (√10 / 2)

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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The first row is fine but I can't get the second:

*Last edited by krassi_holmz (2006-02-28 17:48:31)*

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**ganesh****Moderator**- Registered: 2005-06-28
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Well done, rimi and krassi_holmz!

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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the first is whitout +- and rimi gave correct rationalization of the denominator.

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**ganesh****Moderator**- Registered: 2005-06-28
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IS # 3

What are the values of A and B respectively, if

(√5-1)/(√5+1) + (√5+1)/(√5-1) = A + B√5 ?

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Is#3=3=3+0√5

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**ganesh****Moderator**- Registered: 2005-06-28
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Character is who you are when no one is looking.

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**ganesh****Moderator**- Registered: 2005-06-28
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IS # 4

If

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**ganesh****Moderator**- Registered: 2005-06-28
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IS # 5

If

then express z in terms of x and yCharacter is who you are when no one is looking.

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**navigator****Member**- Registered: 2007-03-21
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solution : -

3 ^ x = k

k ^(1/x ) = 3

k ^(1/y) =4

k ^ (1/z) =12

12 =3 *4 .

k ^ (1/z) = k ^(1/x ) *

k ^(1/y)

1/x + 1/y =1/z

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**JaneFairfax****Member**- Registered: 2007-02-23
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IS # 4

Cube both sides and *voilà*.

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