Indices and Surds

Jai Ganesh · 2006-02-26 17:13:36

IS # 1

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

krassi_holmz · 2006-02-26 18:13:32

Yo ho ho
I LOVE this
thanks

Jai Ganesh · 2006-02-26 18:26:20

krassi_holmz · 2006-02-26 18:31:38

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

krassi_holmz · 2006-02-26 18:35:36

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)

Jai Ganesh · 2006-02-26 19:11:15

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.

krassi_holmz · 2006-02-26 20:35:43

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

krassi_holmz · 2006-02-26 20:38:10

Is it clear now?

Jai Ganesh · 2006-02-26 21:29:48

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

krassi_holmz · 2006-02-26 21:42:02

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)

krassi_holmz · 2006-02-26 22:10:20

now, won't give second q?

Jai Ganesh · 2006-02-28 16:19:16

IS # 2

Find the square root of 7 + 3√5.

rimi · 2006-02-28 17:28:50

7+3√5 = (14+6√5) / 2 = ( 3 + √5 )² / 2
there fore , √(7+3√5) = ±(3 +√5)/√2 = (3√2 / 2) + (√10 / 2)

krassi_holmz · 2006-02-28 17:43:27

The first row is fine but I can't get the second:

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

Jai Ganesh · 2006-02-28 19:52:26

Well done, rimi and krassi_holmz!

krassi_holmz · 2006-02-28 20:26:10

krassi_holmz · 2006-02-28 20:31:34

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

Jai Ganesh · 2006-03-01 20:59:17

IS # 3

What are the values of A and B respectively, if
(√5-1)/(√5+1) + (√5+1)/(√5-1) = A + B√5 ?

krassi_holmz · 2006-03-02 02:32:10

krassi_holmz · 2006-03-02 02:34:45

Is#3=3=3+0√5

Jai Ganesh · 2006-03-02 02:55:21

Jai Ganesh · 2006-03-09 03:34:26

IS # 4

If

show that

Jai Ganesh · 2006-03-10 16:29:59

IS # 5

If

then express z in terms of x and y

navigator · 2007-03-21 03:53:47

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

JaneFairfax · 2007-03-21 04:44:00

IS # 4

Cube both sides and voilà.

Math Is Fun Forum

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

Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

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

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

Re: Indices and Surds

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

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