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#1 2006-02-26 17:13:36

ganesh
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Indices and Surds

IS # 1

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


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#2 2006-02-26 18:13:32

krassi_holmz
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Re: Indices and Surds

Yo ho ho
I LOVE this
thanks


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#3 2006-02-26 18:26:20

ganesh
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Re: Indices and Surds

thank_you.jpg


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#4 2006-02-26 18:31:38

krassi_holmz
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Re: Indices and Surds

OK. I'm ready with IS1.
Now making Tex input.


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#5 2006-02-26 18:35:36

krassi_holmz
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Re: Indices and Surds

Soe here is it:

and
then
   

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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#6 2006-02-26 19:11:15

ganesh
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Re: Indices and Surds

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. up


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#7 2006-02-26 20:35:43

krassi_holmz
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Re: Indices and Surds

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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#8 2006-02-26 20:38:10

krassi_holmz
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Re: Indices and Surds

Is it clear now?


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#9 2006-02-26 21:29:48

ganesh
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Re: Indices and Surds

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
up


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#10 2006-02-26 21:42:02

krassi_holmz
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Re: Indices and Surds

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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#11 2006-02-26 22:10:20

krassi_holmz
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Re: Indices and Surds

now, won't give second q?


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#12 2006-02-28 16:19:16

ganesh
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Re: Indices and Surds

IS # 2

Find the square root of 7 + 3√5.


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#13 2006-02-28 17:28:50

rimi
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Re: Indices and Surds

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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#14 2006-02-28 17:43:27

krassi_holmz
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Re: Indices and Surds

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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#15 2006-02-28 19:52:26

ganesh
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Re: Indices and Surds

Well done, rimi and krassi_holmz!
bouquet-green1.jpg


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#16 2006-02-28 20:26:10

krassi_holmz
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Re: Indices and Surds


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#17 2006-02-28 20:31:34

krassi_holmz
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Re: Indices and Surds

the first is whitout +- and rimi gave correct rationalization of the denominator.


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#18 2006-03-01 20:59:17

ganesh
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Re: Indices and Surds

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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#19 2006-03-02 02:32:10

krassi_holmz
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Re: Indices and Surds


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#20 2006-03-02 02:34:45

krassi_holmz
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Re: Indices and Surds

Is#3=3=3+0√5


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#21 2006-03-02 02:55:21

ganesh
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Re: Indices and Surds

correct.gif


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#22 2006-03-09 03:34:26

ganesh
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Re: Indices and Surds

IS # 4

If

show that


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#23 2006-03-10 16:29:59

ganesh
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Re: Indices and Surds

IS # 5

If

then express z in terms of x and y


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#24 2007-03-21 03:53:47

navigator
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Re: Indices and Surds

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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#25 2007-03-21 04:44:00

JaneFairfax
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Re: Indices and Surds

IS # 4

Cube both sides and voilà.


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