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#1 2006-02-27 16:13:36
- 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 2006-02-27 17:13:32
- krassi_holmz
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Re: Indices and Surds
Yo ho ho
I LOVE this
thanks
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#4 2006-02-27 17:31:38
- 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 2006-02-27 17:35:36
- krassi_holmz
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Re: Indices and Surds
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 (2006-02-27 17:50:04)
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#6 2006-02-27 18:11:15
- 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. 
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#7 2006-02-27 19:35:43
- 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 2006-02-27 19:38:10
- krassi_holmz
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Re: Indices and Surds
Is it clear now?
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#9 2006-02-27 20:29:48
- 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
Character is who you are when no one is looking.
#10 2006-02-27 20:42:02
- 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 (2006-02-27 21:09:00)
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#11 2006-02-27 21:10:20
- krassi_holmz
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Re: Indices and Surds
now, won't give second q?
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#12 2006-03-01 15:19:16
- ganesh
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Re: Indices and Surds
IS # 2
Find the square root of 7 + 3√5.
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#13 2006-03-01 16:28:50
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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)
#14 2006-03-01 16:43:27
- 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 (2006-03-01 16:48:31)
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#15 2006-03-01 18:52:26
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Re: Indices and Surds
Well done, rimi and krassi_holmz!
Character is who you are when no one is looking.
#16 2006-03-01 19:26:10
- krassi_holmz
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Re: Indices and Surds
IPBLE: Increasing Performance By Lowering Expectations.
#17 2006-03-01 19:31:34
- 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 2006-03-02 19:59:17
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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 ?
Character is who you are when no one is looking.
#19 2006-03-03 01:32:10
- krassi_holmz
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Re: Indices and Surds
IPBLE: Increasing Performance By Lowering Expectations.
#20 2006-03-03 01:34:45
- krassi_holmz
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Re: Indices and Surds
Is#3=3=3+0√5
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#22 2006-03-10 02:34:26
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Re: Indices and Surds
IS # 4
If
Character is who you are when no one is looking.
#23 2006-03-11 15:29:59
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Re: Indices and Surds
IS # 5
If
Character is who you are when no one is looking.
#24 2007-03-22 02:53:47
- 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
